1. Engineering Mechanics Full Material

Chapter 4 • Rectilinear Motion | 3.55 Motion can also be defined as the change in the position of a body with respect to a given object. The position of a point P at any time t is expressed in terms of the distance x from a fixed origin O on the reference x-axis, y-axis or z-axis and can be taken as positive or negative as per the usual sign convention. P + x −x P1 X1 X2 P2 X • • • • O 2. Average velocity: The average velocity vav of a point P, in the time interval between t + Dt and t, i.e., in the time interval Dt, during which its position changes from x to x + Dx is defined by v x t av = D D . O P P1 x • • • t t + Dt Dt 3. Instantaneous velocity and speed: The instantaneous velocity v of a point P at time t is the limiting value of the average velocity as the increment of time approaches zero as a limit. Mathematically it can be expressed as: v x t dx t dt = = → Limit D D 0 D The velocity v is positive if the displacement x is increasing and the particle is moving in a positive direction. The unit of velocity is metre per second (m/s). If s is the distance covered by a moving particle at time t, then speed = ds dt . The unit of speed is the same as that of the velocity. 4. Average acceleration: The average acceleration aav of a point P, in the time interval between t + Dt and t, i.e., in the time interval Dt, during which its velocity changes from v to v + Dv is defined by a v t av = D D . 5. Instantaneous acceleration: The instantaneous acceleration of a point P is the limiting value of the average acceleration as the increment of time approaches zero. Mathematically it can be expressed as: a v t v t dt = = → Limit D D 0 D d Then, a dv dt d x dt = = 2 2 Now, a dv dt dv dx dx dt dv dx = = × = × v Acceleration is positive when velocity is increasing. A positive acceleration means that the particle is either moving further in a positive direction or is slowing down in the negative direction. Retardation or deceleration of a body in motion is the negative acceleration, i.e., retarding acceleration. Acceleration is the rate of increase in the velocity and deceleration is the rate of decrease in the velocity. Uniform Motion When a particle moves with a constant velocity so that its acceleration is zero, then the motion is termed as uniform motion. Uniformly Accelerated Motion When a particle moves with a constant acceleration, then the motion is termed as a uniformly accelerated motion. Motion at a Uniform Acceleration Let the uniform acceleration be ‘a’. Then v = u + at And, v2 = u2 + 2as s = ut + 1 2 at2 sn = u + a n - ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 Where, v – Velocity at any time instant t (secs) u – Initial velocity s – Distance travelled during the time t (secs) sn – Distance travelled at the nth second For motion under constant retardation or deceleration, assign negative sign for acceleration (a). NOTE Vertical Motion Under Gravity A body in motion above the ground will be under influence of the gravitational force of attraction (g). If the body moves upwards then it is subjected to gravitational retardation, i.e., a = -g. Then, the equations for the upward motion of a body under gravity will be v = u - gt v2 = u2 – 2gs s = ut - 1 2 gt2 sn = u – g n - ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2

3.56 | Part III • Unit 1 • Engineering Mechanics If the body moves downwards then it is subjected to gravitational attraction and hence an acceleration, i.e., a = g. Then, the equations for the downward motion of a body under gravity will be v = u + gt v2 = u2 + 2gs s = ut + 1 2 gt2 sn = u + g(n – 1 2 ) 1. For a body that is just dropped, a = g and u = 0. 2. The final vertical velocity of a body thrown upwards as it reaches the maximum height, will be zero, i.e., v = 0. NOTES Motion Curves These are the graphical representation of displacement, velocity and acceleration against time. t dt dv v dt ds s a a v s Considering the general case of acceleration not being a constant, the above graphical representation is made. The slope of the displacement-time curve → Velocity The slope of the velocity-time curve → Acceleration The area under the velocity-time curve → Displacement The area under the acceleration-time curve → Velocity Solved Examples Example 1: A particle has two velocities v1 and v2. Its resultant is v1 in magnitude. When the velocity v1 is doubled, the new resultant is (A) Perpendicular to v2 (B) Parallel to v2 (C) Equal to v2 (D) Equal to 2 v2 Solution: Applying the principle of vector, the magnitude of the resultant between   v v 1 2 + Given that    v v 1 2 + = v1 Squring both side,    v v 1 2 v 2 1 2 + = ∴ + ( )⋅ + ( ) = .       v v 1 2 v v 1 2 v v 1 1         v v 1 1 2v v 1 2 v v 2 2 v v 1 1 ⋅ + ⋅ + ⋅ = ⋅ 2 0 1 2 2 2     v v⋅ + v v⋅ = ( ) 2 0. 1 2 2    v v + ⋅ v = Dot product zero means the new resultant between 2v1 and v2 is at right angles to v2. Example 2: If the two ends of a train, moving with a constant acceleration, pass a certain point with velocities u and v respectively, the velocity with which the middle point of the train passes through the same point is (A) u v + 2 (B) u v u v 2 2 + + (C) u - v (D) u v 2 2 2 + Solution: We have the relation v2 = u2 + 2as (1) If v is the velocity with which the mid point of the train crosses the point, we have v2 = u2 + 2 a s 2 (2) Eliminating s from (1) and (2), We have, v2 - u2 = as And, v2 - u2 = 2as Now, v u v u 2 2 2 2 1 2 - - = ∴ 2v2 - 2u2 = v2 - u2 Or, 2v2 = v2 + u2 Now, v v u 2 2 2 2 = + ∴ = + v v u 2 2 2 . Direction for examples 3 and 4: The motion of a particle is defined as s = 2t 3 – 6t 2 + 15, where s is in metres and t is in seconds. Example 3: The acceleration when the velocity is zero is (A) 2 m/ 2s (B) 8 m/s2 (C) 6 m/s2 (D) 4 m/s2 Solution: s = 2t 3 – 6t 2 + 15 ds dt = 6 1 t t 2 2 – a ds dt = = t - 2 2 12 12

Chapter 4 • Rectilinear Motion | 3.57 When velocity is zero, 6t 2 – 12t = 0, ∴ t = 2 sec Then acceleration is, a = 12 × 2 – 12 = 12 m/s2 Example 4: The minimum velocity is (A) -2 m/s (B) 6 m/s (C) -6 m/s (D) 2 m/s Solution: Velocity is minimum when dv dt = 0, i.e., when 12t – 12 = 0, ∴ t = 1 sec (Velocity)min = = 6 12 6 12 = -6 2t t – – m/s Example 5: The velocity of a particle along the x-axis is given by v = 5x3/2 where x is in metres and v is in m/s. The acceleration when x = 2m is (A) 300 m/ 2s (B) 200 m/ 2s (C) 180 m/ 2s (D) 150 m/ 2s Solution: Given v = 5x3/2, differentiating with respect to t, we have dv dt x dx dt = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ - 5 3 2 3 2/ 1 = = 15 2 1 2 x dx dt but dx dt v / , ∴ = a x × = x x 15 2 5 75 2 1 2/ / 3 2 2 When m x a = = 2 × = /s 75 2 4 150 2 , . Example 6: A particle is moving in a straight line starting from rest. Its acceleration is given by the expression a = 50 – 36t 2, where t is in seconds. The velocity of the particle when it has travelled 52 m can be (A) 2 3. m/s (B) 4 m/s (C) 6 7. m/s (D) 8 m/s Solution: Given, a = 50 – 36t 2 So, dv dt = 50 – 36t 2 Or, dv = 50dt – 36t 2 dt Integrating the above equation, we have v t t = - 50 36 + = C t - + t C 3 50 12 3 3 When t = 0, v = 0 \ C = 0 \ v = 50t – 12t 3 ds dt = - 50t t 12 3 Integrating, s t t = - 50 +C 2 12 4 2 4 1 = 25t 2 – 3t 4 + C1 When t = 0, s = 0 \ C1 = 0 s = 25t 2 – 3t 4 Here we can find the time when s = 52 m. \ 25t 2 – 3t 4 = 52 Let t 2 = u, then 25u – 3u2 = 52 3u2 - 25u + 52 = 0 u = 25 ± - 625 624 6 u = ± = 25 1 6 26 6 24 6 or Case 1: when t 2 24 6 = = 4 \ t = 2 sec v = 50t – 12t 3 = 50 × 2 - 12 × 8 = - 100 96 = 4 m/s Case 2: when t 2 26 6 = = 4.333 \ t = 2.08 sec The value of the velocity calculated with this t value is not available in the options provided. Example 7: A body dropped from a certain height covers 5 9 th of the total height in the last second, the height from which the body is dropped is (A) 36.8 m (B) 40.3 m (C) 44.1 m (D) 50.6 m Solution: Let ‘h’ be the height and let ‘n’ be the time taken for the fall. Then, s u = + a n - ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 5 9 0 1 2 h g = + n - ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 5 9 1 2 h g = -n ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (1) Also, h = un + 1 2 an2 h = 0 + 1 2 gn2 (2)

3.58 | Part III • Unit 1 • Engineering Mechanics Putting (2) in (1), 5 9 1 2 1 2 2 × = - ⎛ ⎝ ⎜ ⎞ ⎠ gn g n ⎟ \ 5n2 – 18n + 9 = 0 5n2 – 15n – 3n + 9 = 0 5n(n - 3) – 3(n - 3) = 0 \ (5n – 3)(n – 3) = 0 ∴ = n n = > n 3 5 or 3 1 , but ∴ n = 3 ∴ h = 1 2 gn2 = 1 2 × 9.81 × 9 = 44.1 m. Example 8: A stone falls past a window 2 m high in a time of 0.2 seconds. The height above the window from where the stone has been dropped is (A) 4.15 m (B) 5.23 m (C) 5.87 m (D) 6.32 m Solution: window A h The stone is dropped from A. Let the body reach the top of the window with a velocity of u m/s. Then, u2 = 02 + 2gh u2 = 2gh (1) Falling with an initial velocity u, it covers the window 2 m high in 0.5 seconds. s = ut + 1 2 at2 2 = u × 0.2 + 1 2 × 9.81 × 0.22 2 = 0.2u + 1 2 × 9.81 × 0.04 2 = 0.2u + 9.81 × 0.02 u = 9.019 m/s From (1), u2 = 2gh, ∴ × h = 9 019 2 9 81 4 145 2 . . . . m Example 9: A ball is projected vertically upwards with a velocity of 49 m/s. If another ball is projected in the same manner after 2 seconds, and if both meet t seconds after the second ball is projected, then t is equal to (A) 3 s (B) 10 s (C) 5 s (D) 6 s Solution: Let both the balls meet T seconds after the first ball is projected. Therefore when the balls meet, for the first ball, h = 49 × T – 1 2 gT2 For the second ball, h = 49 × (T - 2) – 1 2 g(T - 2)2 Equating, 49T – 1 2 gT2 = 49 (T - 2) – 1 2 g(T - 2)2 ∴ T = 11.99 sec ∴ t = T – 2 = 9.99 sec ≈ 10 sec Example 10: Two bodies are moving uniformly towards each other. The distance between them decreases at a rate of 6 m/s. If both the bodies move in the same direction with the same speeds, then the distance between them increases at a rate of 4 m/s. The respective speeds of the bodies are (A) 3 m/s and 1 m/s (B) 5 m/s and 1 m/s (C) 4 m/s and 2 m/s (D) 3 m/s and 5 m/s Solution: Let u and v be the velocities of the bodies. From the statement of the problem, u + v = 6 u – v = 4 ∴ u = 5 m/s and v = 1 m/s. Example 11: Two cars are moving in the same direction each with a speed of 45 km/h. The distance separating them is 10 km. Another vehicle coming from the opposite direction meets these two cars in an interval of 6 minutes. The speed of the vehicle is (A) 45 km/h (B) 50 km/h (C) 55 km/h (D) 60 km/h Solution: The distance between the cars moves with a velocity of 45 km/h. If the speed of the vehicle is u, then its velocity relative to the moving distance is 45 + u m/s. It takes 6 minutes to cover the distance of 10 km. ∴ + ( ) 45 × = 6 60 u 10 ∴ 45 + u = 100 u = 55 km/h. Motion under Variable Acceleration In practical conditions a body may very often move with variable acceleration. The rate of change of velocity will not remain constant. We know that acceleration, a dv dt dv ds ds dt = = ⋅ Or a v dv ds = ⋅ Also when displacement can be expressed as a third degree or higher degree equation in time, the acceleration becomes a variable with respect to time.

Chapter 4 • Rectilinear Motion | 3.59 For example, if s = 4t 3 + 3t 2 + 5t + 1 ds dt = + 12t t 6 5 + 2 d s dt t 2 2 = + 24 6 The velocity and displacement are evaluated by integration. Example 12: A body is starting from rest and moving along a straight line whose acceleration is given by f = 10 – 0.006x2 where x is the displacement in m and f is the acceleration in m/s2. The distance travelled by it when it comes to rest is (A) 70.7 m (B) 68.3 m (C) 62.6 m (D) 58.5 m Solution: Given that f = 10 – 0.006x2 dv dt = - 10 0 006x2 . dv dx dx dt ⋅ = 10 - 0 006x2 . v dv dx ⋅ = 10 - 0 066x2 . vdv = (10 – 0.006x2)dx integrating v v x x C 2 3 10 0 006 3 = - . + when x = 0, v = 0 ∴ C = 0 v x x 2 3 2 10 0 006 3 = - . v2 = 20x – 0.004x3 when v = 0; 20x – 0.004x3 = 0 ∴ 0.004x2 = 20 (note that the solution of x = 0 is also possible for the above equation, but the value of x > 0 is sought for) ∴ x = 70.7 m. Direction for examples 13 and 14: An electric train starting from rest has an acceleration f in m/s2. which vary with time as shown in the table below: t(secs) 0 6 12 18 f 2 m s       12 10 9.5 8 Example 13: The velocity at the end of the first 6 seconds is (A) 18 m/s (B) 27 m/s (C) 43 m/s (D) 66 m/s Solution: During the first 6 seconds, the average acceleration = + = 12 10 2 11 m/ 2s ∴ Increase in velocity during this interval of 6 seconds = average acceleration × 6 = 66 m/s ∴ Velocity at the end of 6 second = 66 m/s. Example 14: The distance travelled during these six seconds is (A) 242 m (B) 218 m (C) 198 m (D) 124 m Solution: Average velocity during this interval = + = 0 66 2 33 m/s ∴ Distance travelled during this interval = 33 × 6 = 198 m. Example 15: At any instant, the acceleration of a train starting from rest is given by f u = + 10 1 where u is the velocity of the train in m/s. The distance at which the train will attain a velocity of 54 km/h is (A) 123.7 m (B) 185.4 m (C) 214.4 m (D) 228.2 m Solution: It is given f u = + 10 1 u du dx u ⋅ = + 10 1 u(u + 1)du = 10dx Integrating we have, u u x c 3 2 3 2 + = 10 + when x = 0, u = 0. ∴ c = 0 u u x 3 2 3 2 + = 10 when u = 54 km/h = 54 × 5/18 = 15 m/s 15 3 15 2 10 3 2 + = x 1125 + 112.5 = 10x ∴ x = 123.7 m. Example 16: The motion of a particle is given by the equation a = t 3 – 3t 2 + 5, where ‘a’ is acceleration in m/ s2 and t is time in seconds. It is seen that the velocity and displacement of the particle at ‘t’ = 1 sec are 6.25 m/s and 8.3 m respectively. Then the displacement at time t = 2 sec is (A) 17.3 m (B) 15.6 m (C) 14.8 m (D) 12.6 m

3.60 | Part III • Unit 1 • Engineering Mechanics Solution: Given a = t 3 – 3t 2 + 5 dv dt = + t t 3 2 – 3 5 Integrating, v t t = - + +t c 4 3 4 3 3 5 at t = 1 sec, v = 6.25 m/s i.e., 6 25 1 4 . = -1 5 + + c = 4.25 + c ∴ c = 2 ∴ = v - + + t t t 4 3 4 5 2 ds dt t = - t t + + 4 3 4 5 2 Integrating, s t t t = - + ⋅ + +t c 5 4 2 20 4 5 2 2 , at t = 1, s = 8.3 m 8 3 1 20 1 4 5 2 . , = -++ 2 + c 8 3 1 20 . . = + 4 25 + c, c = 8.3 – 4.25 – 0.05 = 4.05 – 0.05 = 4 s t t t = - + ⋅ + +t 5 4 2 20 4 5 2 2 4 s at t = 2 sec is s = - + + + 32 20 16 4 1044 = + = 32 20 14 15. . 6 m Example 17: In the figure shown, AB is the diameter ‘d’ of the circle and AC is the chord of the same circle? A B C a Making an angle α with AB. Two particles are dropped from rest one along AB and the other along AC. If t 1 is the time taken by the particle to slide along AB and t 2 is the time taken to slide along AC, then t 1: t 2 is (A) 1:cosa (B) 1:seca (C) 1:1 (D) 1:15 Solution: Let AB = l, AC = l cosa Consider sliding along AC, acceleration is gcosa We have, s u = +t at 1 2 2 Now, l g cos c α α = +0 os t 1 2 2 2 ∴t t = = l g l g 2 2 2 2 2 or Consider sliding along AB, I g = +0 t 1 2 1 2 t l g 1 2 = ∴ t 1:t 2 = 1:1. Relative Velocity The motion of one body with respect to another moving body is known as relative motion. Take the case of two bodies P and Q moving along the same straight line. The position of the bodies is specified with reference to an origin O. xP and xQ are measured from the origin O. The difference xQ - xP defines the relative position of Q with respect to P. It is denoted as xQ/P = xQ - xP ∴ xQ = xP + xQ/P Consider the rate of change of displacement, then x x x Q P Q xQ xP vQ/P = vQ - vP ∴ vQ = vP + vQ/P Similar relations hold good for acceleration also, i.e., ∴ aQ = aP + aQ/P Working rule: Let two particles A and B move with velocities v1 m/s and v2 m/s respectively in directions as shown in the following figure. vA = v1 vB = v2 m/sec If we want to find out the velocity of A relative to B, the velocity of B is to be made zero. For that we provide velocity v2 in the reverse direction of OB and find the vector sum with v1 = OA.

Chapter 4 • Rectilinear Motion | 3.61 D C V O B A v v1 1 v1 The vector OD    gives both the magnitude and direction of the velocity of A relative to B. Another method is to resolve their velocities into their components with sign. Then evaluate the relative velocity in the x-direction and in the y-direction. Find their resultant vector. This vector will be the relative velocity, both in magnitude and in direction. Example 18: Two boats start from a point at the same time. The boat A has a velocity of 30 km/h and move in the direction N 30° W. The boat moves in the south west direction with a velocity of 40 km/h. The distance between the boats after half an hour is (A) 27.9 km (B) 32.3 km (C) 36.7 km (D) 42.3 km Solution: Method 1: O N E S W 40 km/h 30 km/h 45° 30° Resolving along the x-axis, ( ) v i ( sin ) A x = - 30 30°  for A and ( ) v i B x = -( c 40 os 45°) , B  for where  i is a unit vector along the x-axis. (vA/B)x = (vA)x - (vB)x = ° –(30sin ) 30 – (– c 40 os 45°)   i i = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ - 40 2 15  i = 13.28 i km/h Similarly, (vA/B) y = (vA) y - (vB) y = ° ( c 30 os30 ) ( - - 40sin ) 45° ,   j j Where  j is a unit vector along the y-axis. = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + = 40 2 30 3 2 54 26   j j . km/h vA B/ = = 54. . 26 +13 28 . 2 2 55 86 km/h The relative distance after half an hour = × 55 86 = 1 2 . . 27 9 km. Method II: O N E B B′ A C W 40 km/h 30 km/h 45° 30° The vector OC    is the resultant velocity vector. Velocity of B is reversed and considered. Therefore the resultant is the velocity of A relative to B. OC    = + 40 30 + ×2 40 3 × ×0 75° 2 2 cos = 1600 + + 900 40 × × 60 0. . 258 = 55 86 m/s Relative distance after half and hour = × 55 86 = 1 2 . . 27 9 km. Example 19: A vessel which can steam in still water with a velocity of 48 km/h is steaming with its bow pointing due east. It is carried by a current which flows northward with a speed of 14 km/h. The distance it would travel in 12 minutes is (A) 14 km (B) 12 km (C) 10 km (D) 8 km Solution: N E 14 km/h 14 km/h 48 km/h To find the velocity of the steamer relative to the flow, the flow velocity is reversed and vector sum is found. Relative velocity = k 48 14 50 m/h 2 2 + = Distance after 12 minutes = × 50 12 60 = 10 km. Example 20: A man keeps his boat at right angles to the current and rows across a stream 0.25 km broad. He reaches the opposite bank 0.125 km below the point opposite to the starting point. If the speed of the boat in rowing alone is 6 km/ph, the speed of the current is (A) 5 km/h (B) 4 km/h (C) 3 km/h (D) 2 km/h Solution: The speed responsible for reaching the opposite side is the rowing velocity 6 km/h. Due to the velocity of the current by the time the boat can cross the stream with its absolute

3.62 | Part III • Unit 1 • Engineering Mechanics velocity, the boat flows down 0.125 km due to the speed of the current. Time for crossing = = 0 25 6 0 04166 . . hr Let the stream velocity be v m/s. ∴ Resultant speed = v + 2 2 6 0.125 km 0.25 km O A The distance covered by the boat within this time is OA = + 0 25 0 125 2 2 . . ∴ × 0 04166 + = 6 0 25 + 0 125 2 2 2 2 . . v . ∴ v = 3 km/h. Example 21: A boat of weight 45 kg is initially at rest. A boy of 32 kg is standing on it. If he jumps horizontally with a speed of 2 m/s relative to the boat, the speed of the boat is (A) 2 m/s (B) 3.42 m/s (C) 4.92 m/s (D) 5.36 m/s Solution: Given vA/B = 2 m/s It is relative velocity of the boy with respect to the boat. vA/B = vA - vB 2 = vA - vB ∴ vA = 2 + vB By conservation of momentum, 0 = 32 (2 + vB) – 45 vB = 64 – 13 vB ∴ vB = 4.92 m/s. Example 22: A stream of water flows with velocity of 1.5 km/h. A swimmer swims in still water with a velocity of 2.5 km/h. If the breadth of the stream is 0.5 km, the direction in which the swimmer should swim so that he can cross the stream perpendicularly is (A) 26° with the vertical (B) 29.4° with the vertical (C) 32.5° with the vertical (D) 36.8° with the vertical Solution: 0.5 km 1.5 km/h 1.5 km/h 2.5 km/h O A q The swimmer must swim in the direction OA with velocity 2.5 m/s so that he can cross the stream at right angles. From geometry 2.5 sinθ = 1.5 ∴ = sinθ = 1 5 2 5 0 6 . . . θ = 36.8°. Example 23: An aeroplane is flying in a horizontal direction with a velocity of 1800 km/h. At a height of 1960 metres, when it is above a point A on the ground, a body is dropped from it. If the body strikes the ground at point B, then the distance AB is (A) 18 km (B) 15 km (C) 10 km (D) 8 km Solution: The time taken by the body to fall down the distance 1960 m is h g = t 1 2 2 1960 1 2 9 8 2 = . t 2 1960 9 8 2 × = . t 400 = t 2; t = 20 sec AB = ×v t = × × = 1800 60 60 20 10 km Example 24: Two ships leave the port at the same time. The first ship A steams north-west at 32 km/h and the second ship B 40° south of west at 24 km/h. The time after which they will be 160 km apart is (A) 2.15 hrs (B) 2.86 hrs (C) 3.46 hrs (D) 4.19 hrs Solution: Let us find the velocity of the second ship relative to the first. For that consider the velocity of the first ship in the reverse direction and evaluate the vector sum of the velocities. Resultant or velocity of B relative to A is = + 24 32 + ×2 32 2 × ° 4 95 2 2 cos = = 1466 38. k 3 m/h O N E B S A W 24 km/h 32 km/h 45° 40° 95° Time for two ships to be 160 km apart = = 160 38 3 4 19 . . hrs.

Chapter 4 • Rectilinear Motion | 3.63 Example 25: A particle is accelerated from (1, 2, 3), where it is at rest, according to the equation a t = + 6 2 i t 4 j 2   – 10 2  km/s , where    i j , and k are unit vectors along the x, y and z axes. The magnitude of the displacement after the lapse of 1 second is (A) 5 m (B) 30 m (C) 6 m (D) 47 m Solution: It is given that a t = - 6 2 i t 4 1 j k + 0 2    ∴ = v t 3 8 i t – j t + + 10 k c 2 3    when t = 0, v = 0 ∴ c = 0 ∴ = v t 3 8 i t – j t +10 k 2 3    dx dt = - 3 8 t i t j +10tk 2 3    x t i t j t = - 3 + + k C 3 8 4 10 2 3 4   2  x t = -i t j t + + k C 3 4 2 2 5    when t = 0, position of the particle is at (1, 2, 3) i.e., at t = 0, x i = + 1 2 j k + 3    ∴ = C i 1 2 + +j k3    ∴ = x t i t - +j t k i + + j k + 3 4 2 2 5 1 2 3       = + ( ) t i - - ( ) t j + + ( ) t k 3 4 2 1 2 2 3 5    When t = 1, x i = + 2 8k   \ Displacement vector = + 2 8 - + 1 2 + = 3 1 - + 2 5         i k ( ) i j k i j k Magnitude of the displacement vector = +1 4 + = 25 30 m. Example 26: If a particle, moving with uniform acceleration, travels the distances of 8 and 9 cms in the 5th sec and 9th sec respectively, then its acceleration will be (A) 1 cm/s2 (B) 5 cm/s2 (C) 25 cm/s2 (D) 0.5 cm/s2 Solution: s in the nth sec = + u - a n 2 ( ) 2 1 8 2 = + u 2 5 × -1 4 = + 5 a ( ) u a . (1) 9 2 = + u 2 9 × -1 8 = + 5 a ( ) u a . (2) Subtracting Eq. (1) from Eq. (2), 1 = 4a or a = 0.25 cm/s2. Example 27: The acceleration due to gravity on a planet is 200 cm/s2. If it is safe to jump from a height of 2 m on earth, then the corresponding safe height on the planet is (A) 2 m (B) 9.8 m (C) 10 m (D) 8 m Solution: Let hse and h sp denote the safe heights on the earth and the planet. On the earth, v2 = 2ghse = 2 × 9.8 × 2 = 39.2 m2/s2 On the planet, v2 = 2 × 2 × h sp For a safe jump the final velocity (v) should be same on earth and the planet, hence, 2 × 2 × h sp = 39.2 ∴ h sp = 9.8 m. Example 28: A ball weighing 500 gm is thrown vertically upwards with a velocity of 980 cm/s. The time that the ball will take to return back to earth would be (A) 1 s (B) 2 s (C) 3 s (D) 4 s Solution: For the upward journey, u = u0 – gt 0 = 980 × 10-2 - 9.8 t ⇒ t = 1 s v2 - u2 = 2gs ⇒ 0 – 9.82 = -2 × 9.8 s s = 4.9 m For the downward journey, s u = +t gt 1 2 2 4 9 0 1 2 9 8 2 . . = + × t t = 1 s Total time taken to return back to earth = 1 + 1 = 2 s. Kinetics of a Particle Kinetics can be used to predict a particle’s motion, given a set of forces (acting upon the particle) or to determine the forces necessary to produce a particular motion of the particle. Kinetics of the rectilinear motion of a particle are governed mainly by Newton’s three laws of motion. Newton’s first law: Every body continues in its state of rest or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it. This law is sometimes called as the law of inertia. From Newton’s first law, it follows that any change in the velocity of a particle is the result of a force. The question, of the relationship between this change in the velocity of the particle and the force that produces it, is answered by the second law of motion which is as follows. Newton’s second law: The acceleration of a given particle is proportional to the force applied to it and takes place in the direction of the straight line in which the force acts.

3.64 | Part III • Unit 1 • Engineering Mechanics Newton’s third law: Every action there is always an equal and opposite reaction, or the forces of two bodies on each other are always equal and directed in opposite directions. General Equation of Motion for a Particle From Newton’s second law, the relationship between the acceleration ‘a’ produced in a body of mass ‘m’ (mass is always assumed to be invariant with time) by a resultant, ‘F’, of all the forces acting on the body can be derived as follows: F = ma, which is the general equation of motion for a particle. For a stationary body lying on a surface (body with no motion), there is a force (F) exerted by the body on the surface which is equal to the weight of the body (W), i.e., F = W = mg, where ‘m’ is the mass of the body and ‘g’ is the acceleration due to gravity. There is an equal and opposite force exerted by the surface on the body (consequence of Newton’s third law). Note that the weight of a body is obtained by multiplying the mass of the body by the acceleration due to gravity. Differential Equation of Rectilinear Motion The general equation of motion for a particle can be applied directly to the case of the rectilinear translation of a rigid body, since all the particles of the rigid body have the same velocity and acceleration (same motion) where the particles move in parallel straight lines. Here, the rigid body is considered as a particle concentrated at the center of gravity of the rigid body. Whenever such a body or particle moves under the action of a force applied at its center of gravity and having a fixed line of action, acceleration of the body is produced in the same direction, and if any initial velocity of the body is also directed along this line, then the motion corresponding to this case is known as rectilinear translation. If the line of motion of a particle is taken to be along the x-axis (i.e., displacement at a time t is denoted by x), x d x dt = 2 2 represents the acceleration and F represents the resultant force acting, then the differential equation of the rectilinear motion of the particle is given by F m= x. Two types of problems that can be solved by the above equation are: (a) Determination of the force necessary to produce a given motion of the particle where the displacement x is given as a function of time t, (b) Determination of the motion of a particle given a force F acting on the particle, i.e., to determine a function relating x and t such that the above equation is satisfied. Motion of a Particle Acted upon by a Constant Force A particle, acted upon by a force of constant magnitude and direction, will move rectilinearly in the direction of the force subjected to a constant acceleration. Let us consider a particle moving along the x-axis (see figure) where the initial (at t = 0) displacement and velocity of the particle is x0 and x0 respectively. O C X X0 X F D If F is the magnitude of the constant force acting on the particle, then from the differential equation of rectilinear motion, x F m = = a, where a is the constant acceleration produced in the particle due to the constant force. The equation x a = can be written as d x dt a ( ) .  = Integration of the above equation with the initial value condition, at t x = = 0 x0   , gives: x x   = + at 0 Which is the general velocity-time equation for the rectilinear motion of a particle under the action of a constant force F producing the constant acceleration a in the particle. With x dx dt = , equation (1) can be rewritten as follows: dx dt = + x a  t 0 . Integration of the above equation with the initial value condition, at t = 0, x = x0, gives: x x = + x t + at 0 0 2 1 2  ,which is the general displacement-time equation for the rectilinear motion of a particle under the action of a constant force F producing the constant acceleration a in the particle. Freely falling body The force acting on a freely falling body is the weight of the body (assuming no friction in the motion) and therefore the acceleration produced in the body is the acceleration due to gravity, i.e., F = W = mg and \ a = g. Hence, the velocitytime and displacement-time equations for a freely falling body are respectively as follows: x x   = + gt 0 x x = + x t + gt 0 0 2 1 2  If the freely falling body starts to fall from a resting position, i.e., it has an initial velocity of zero ( ( x 0 0 ) ) = and if the origin of displacement of the body is taken to coincide with the initial position of the body, i.e., it has an initial displacement of zero (x0 = 0), then the above equations reduce to: x g  = t x g = t 1 2 2

Chapter 4 • Rectilinear Motion | 3.65 Force as a Function of Time If the force acting on the particle is a function of time t, i.e., the acting force = F(t), then the acceleration a(t), velocity x t ( ) and displacement x(t) of the particle at time t (with initial time, t = 0) is given by the following respective equations. a t F t m ( ) ( ) = x t  a t dt t ( ) = ( ) ∫ 0 x t x t dt t ( ) = ( ) ∫  0 Dynamics of a Particle D’Alembert’s Principle Let ΣFi , where Fi denotes the ith force, be the resultant of a set of forces acting on a particle in the x-axis direction. From the 0 differential equation of the rectilinear motion of a particle, we have ∑ - F m i x = 0 or ∑ + F m i ( ) - = x 0 From the above equation, it can be seen that if a fictitious force ( ) -mx is added to the system of forces acting on the particle, then an equation resembling equilibrium is obtained. The force ( ) -mx which has the same magnitude as mx but opposite in direction is called as the inertia force. Hence, it can be seen that if an inertia force is added to the system of forces acting on a particle, then the particle is brought into an equilibrium state called as dynamic equilibrium. This is called as D’Alembert’s principle. The above equation thus represents the equation of dynamic equilibrium for the rectilinear translation of a rigid body. Let us consider, now any system of particles connected between them and so constrained that each particle can have only a rectilinear motion. To exemplify such a system, the case of two weights W1 and W2 attached to the ends of a flexible but inextensible string overhanging a pulley, as shown in the figure below, is considered. S W W1 2 m2 m2 m1 m1 S X X The inertia of the pulley and the friction on its axle are assumed to be negligible. If the motion of the system is assumed to be in the direction as shown by the arrow on the pulley, an upward acceleration x of the weight W2 and a downward acceleration x of the weight W1 is obtained. The inertia forces acting on the corresponding weights are shown in the above figure. By adding the inertia forces to the real forces such as (W1 and W2 and the string reactions S), a system of forces in equilibrium is obtained for each particle. Hence, the entire system of forces can be concluded to be in equilibrium. An equation of equilibrium can be written for the entire system (instead of separate equilibrium equations for the individual weights) by equating to zero the algebraic sum of moments of all the forces (including the inertia forces) with respect to the axis of the pulley or by using the principle of virtual work. In either case, the internal forces ‘S’ of the system need not be considered and the following equation of equilibrium can be obtained for the entire system. W m x W m x or x W W W W 2 2 1 1 g 1 2 1 2 + = - = - + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟    Momentum and Impulse The differential equation of the rectilinear motion of a particle may be written as m dx dt F d mx Fdt  = = or ( ) (1) X X B C 0 D t t dt It is assumed that the force ‘F’ is known as a function of time and is given by the force-time diagram as shown in the above figure. The right hand side of equation (1) is then represented by the area of the shaded elemental strip of height F and width dt in the force-time diagram. This quantity is called as the impulse of the force F in the time interval dt. The expression mx on the left hand side of the equation is called as the momentum of the particle. The equation states that the differential change of the momentum of the particle during the time interval dt is equal to the impulse of the acting force during the same time interval. Impulse and momentum have the same dimensions of the product of mass and velocity. Integrating equation (1), we get mx mx Fdt t   - = 0 ∫ 0 , where x0 is the velocity of the particle at time t = 0. Thus the total change in the momentum of a particle during a finite time interval is equal to the impulse of the acting force during the same time interval. This impulse is

3.66 | Part III • Unit 1 • Engineering Mechanics represented by the area OBCD of the force-time diagram. The equation of momentum-impulse is particularly useful when dealing with a system of particles, since in such cases the calculation of the impulse can often be eliminated. As a specific example, consider the case of a gun and shell as shown in the figure below, which may be considered V1 V2 F F as a system of two particles. During the extremely short interval of explosion, the forces ‘F’ acting on the shell and gun and representing the gas pressure in the barrel are varying in an unknown manner and a calculation of the impulses of these forces would be extremely difficult. However, the relation between the velocity of the shell and velocity of recoil of the gun can be obtained without calculation of the impulse. Since the forces ‘F’ are in the nature of action and reaction between the shell and gun, they must at all times be equal and opposite, and hence their impulses for the interval of explosion are equal and opposite since the forces act exactly the same time ‘t’. Let m1 and m2 be the masses of the shell and gun respectively. If the initial velocities of the shell and gun are assumed to be zero and if the external forces are neglected, then m v m v v v m m 1 1 2 2 2 1 1 2 = = , . i.e , The velocities of the shell and gun after discharge are in opposite directions and inversely proportional to the corresponding masses. Internal forces in a system of particles always appear as pairs of equal and opposite forces and need not be considered when applying the equation of momentum and impulse. Thus it may be stated that, for a system of particles on which no external forces are applied, the momentum of the system remains unchanged, since the total impulse is zero. This is sometimes called as the principle of conservation of momentum. Moment and Couple Moment or moment of a force is the turning effect caused by the force. It is the force acting at a perpendicular distance d Moment of a force = Force × Perpendicular distance. • F x Moment = F × x. Couple Tow equal and opposite forces with separate lines of action present in a system of forces constitute a couple. Both forces create their own moment of force. The net moment of the couple is independent of the location of the point considered. Moment of couple = Force × Perpendicular distance between the forces. F F x Moment of couple = F. x • Moment is the measure of the turning effect produced by a force about a point. Couple consists of two forces, equal and opposite, acting in two different but parallel lines of action. • Moment of a couple is independent of the location of the pivot or point considered. Work and Energy The differential equation of the rectilinear motion of a particle can be written in the following form: m dx dt F  = Multiplying both sides of the above equation by x and with suitable modifications, the above equation can be written as follows: d mx Fdx 2 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = (2) It is assumed that the force F is known as a function of the displacement x of the particle and is represented by the below force-displacement diagram. X X0 B C A D h dx The right hand side of equation (2) is represented by the area of the elemental strip of the height h and width dx in the above figure. This quantity represents the work done by the force F on the infinitesimal displacement dx, and the expression in the parenthesis on the left hand side of equation (2) is called the as kinetic energy of the particle. Equation (2) thus states that the differential change in the kinetic energy of a moving particle is equal to the work done by the acting force on the corresponding infinitesimal displacement dx. Work and kinetic energy have the same dimensions of the product of force and length. They are usually expressed in the unit of Joules (J).

Chapter 4 • Rectilinear Motion | 3.67 Integrating equation (2), with the assumption that the velocity of the particle is xo when the displacement is x0, we have mx mx Fdx x x   2 0 2 2 2 0 - = ∫ (3) The definite integral on the right hand side of the above equation is represented by the area ABCD of the force-displacement diagram. This is the total work of the force F on the finite displacement of the particle from x0 to x. The work of a force is considered positive if the force acts in the direction of the displacement and negative if it acts in the opposite direction. The total change in the kinetic energy of a particle during a displacement from x0 to x is equal to the work of the acting force on the displacement. The equation of work and energy is especially useful in cases where the acting force is a function of displacement and where the velocity of the particle as a function of displacement is of interest. For example, the velocity with which a weight W falling from a height h strikes the ground is to be determined. In this case, the acting force F = W and the total work is Wh. Thus if the body starts from rest, the initial velocity x0 = 0 and hence equation (3) becomes mx Wh 2 2 = (4) Which yields x v  = = 2 . gh Let the same body slide, without friction, along an inclined plane AB starting from an elevation h above point B as shown in the figure below. W B a A W sin a h The equation of work and energy can be used to determine the velocity of the body when it reaches point B. Here only the component W sin a of the gravity force does work on the displacement and the component perpendicular to the inclined plane is at all times balanced by the reaction of the plane. In short, the resultant of all the forces acting on the body is F = Wsina in the direction of motion, and this force acts through the distance h sin . α ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ The work of the force acting on the body is = × W = h sin Wh sin α α and hence velocity at the point B (derived from equation (4 2 )), v g = h. Hence, the velocity is the same as that gained in a free fall through the height h. If (is the coefficient of friction between the block and the inclined plane, then the work of friction has to be considered in equation (3). In such a case, the resultant acting force, in the direction of motion F = Wsina - mWcosa. Then through the displacement h sin α ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ between the points A and B, the work done is = - Wh m α Whcot Equation (3) would then yield v g = - 2 1 h( c m α ot( )) When α π = 2 , the above equation agrees with the velocity equation derived for a freely falling body and when m = 0, the above equation agrees with the velocity equation derived for the inclined plane motion of the body with no friction. Also from the above equation, it can be noted that to obtain a real value for the velocity, m α < tan , otherwise the block would not slide down. Work done by Torque Consider a light rod of length l pin joined at one end and is turned by an angle q by the force F from the position A to B. Work done by the constant torque is the product of the torque and the angle turned by the rod. ∴ Work done = F.s. = F.r.q = T.q B F S A l O q Work Energy Formulations • Kinetic energy of a body/particle in translation = 1 2 mv2. • Kinetic energy of a body/particle in rotation and rotating about a point = 1 2 2 IW . • Work Energy principle for a body/particle in translation. Work done on body/particle between points 1 and 2 is W Fxdx x x 1 2 1 2 - = ∫ ∑ . (0,0) • • • X2 X1 F1 F2 t1 V1 V2 t2 Y R μR X 1 W 2 Change in kinetic energy from the positions 1 to 2 is ( ) DK E⋅ = m v( ) - v . 1 2- 2 2 1 2 1 2

3.68 | Part III • Unit 1 • Engineering Mechanics ∴ = W F ∫ ∑ xdx = m v( ) v x x 1 2 2 2 1 2 1 2 1 2 – – . Work energy principle for a body/particle in rotation. IO w2 w1 q = q2 q = q1 IO 1 2 • O • (q1 − q2) Work done from 1 to 2 is given by. W M 1 2 O d 1 2 - = ∫ ∑ θ θ θ . Change in kinetic energy from 1 to 2 is KE1 2 IO 1 2 2 2 1 2 – = - ( ) ω ω ∴ Work done W MO O d I 1 2 1 2 2 2 1 2 1 2 – = = ∑ ( ) - ∫ θ ω ω θ θ 1. Work done by a force is zero if either displacement is zero or the force acts normal to the displacement for example, gravity force does no work when a body moves horizontally. 2. Work done by a force is positive if the direction of force and the direction of displacement are same. For example, work done by force of gravity is positive when a body moves from a higher elevation to a lower elevation. A position work can be said as the work done by a force and negative work as the work done against a force. 3. Work is a scalar quantity and has magnitude but no direction. 4. Work done by a force depends on the path over which the force moves except in the case of conservative forces. Forces due to gravity, spring force are conservative forces, where as friction force is a non-conservative force. NOTES Example 29: If a bucket of water weighing 15 N is pulled up from a well of 25 m depth by a rope weighing 1.5 N/m, then the work done is (A) 843.75 Nm (B) 500 Nm (C) 575 Nm (D) 600 Nm Solution: The work done to pull the rope = × - ∫ 1 5 25 0 25 . ( ) h dh (h is the tip of the rope from the bottom of the well) = × 1 5 = 25 2 468 75 2 . . Nm Total work done = Work done to pull the bucket + work done to pull the rope = 15 × 25 + 468.75 = 843.75 Nm. Example 30: A uniform chain of length 10 m and mass 100 kg is lying on a smooth table such that one third of its length is hanging vertically down over the edge of the table. If g is the acceleration due to gravity, then the work required to pull the hanging part of the chain is (A) 50g (B) 55.55g (C) 100g (D) 150g Solution: Work done = potential energy change in the raising of the centre of mass over the distance L 6 . = = × × = = m g L g g 3 6 100 10 18 1000 18 55. . 55g Alternate method: dx W mg L x dx mg L x L L = = ⎛ ⎝ ⎜ ⎞ ⎠ ∫ ⎟ 2 0 3 0 3 2 = × = × mg mg L L L 2 18 18 . Ideal Systems: Conservation of Energy m2 m3 m1 x1 O a The method of work and energy for a single particle can be extended to apply to a system of connected particles as shown in the above figure. In doing so, it is to be noted that the attention is limited to ideal systems with one degree of freedom. That is, it is assumed that the system has frictionless constraints and inextensible connections and that its configuration can be completely specified by one coordinate such as x1 in the below figure. In the case shown in the above figure, for example, the assumptions involve a smooth inclined plane, frictionless bearings, inextensible

Chapter 4 • Rectilinear Motion | 3.69 strings and neglecting entirely the rotational inertia of the pulleys. Then the system may be regarded simply as three particles, m1, m2 and m3, each of which performs a rectilinear motion. From kinematics, the displacements and velocities of all the three masses can be expressed in terms of one variable, say the coordinate x1 of the particle m1. During motion of the system, an infinitesimal interval of time dt is considered during which the system changes its configuration slightly and each particle is displaced by a length of dxi , along its line of motion. If Fi is the resultant force acting on any particle mi , then the total increment of work of all the forces during such a displacement, dU = ∑F di i x (5) For the system of particles, it can be shown that dT = dU (6) Where T m = ∑ i i x T 1 2 2 ( )  , is the total kinetic energy of the system of particles with the mass and velocity of the ith particle being mi and xi respectively. Equation (6) states that the differential change in the total kinetic energy of the system when it changes its configuration slightly is equal to the corresponding increment of work of all forces. Consider any two configurations of the system denoted by the subscripts A and B, then from equation (6) we have dT dU or T T x x A B A B = ∫ ∫ T T B A dU x x A B - = ∫ (7) This is the equation of work and energy for a system of particles. It states that the total change in the kinetic energy of the system when it moves from configuration A to configuration B is equal to the corresponding work of all the forces acting upon it. In the case of an ideal system, the reactive forces will produce no work and work of all the internal forces which occur in equal and opposite pairs will cancel out each other. Thus for such systems, only the work of active external forces are to be considered on the right hand side of equation (7). The potential energy of a system in any configuration (A or B) is defined as the work which will be done by the acting forces if the system moves from that configuration (A or B) back to a certain base or reference configuration (O). If VA and VB are the potential energies of the system in configurations A and B respectively, then V d A U V dU A B B = = ∫ ∫ 0 0 and If a particle of weight w is at an elevation x above a chosen datum plane, then the potential energy of the particle, V = mx. Similarly for a system of particles at an elevation, the potential energy of the system,V w = ∑ i i x W= xc , NOTE Where wi and xi are the weight and elevation above a chosen datum plane for the ith particle, W is the total weight of the system and xc is the elevation of the center of gravity of the system above the chosen datum plane. For the system of particles moving from the configuration A to the configuration B, it can be shown that TB + VB = TA + VA. Law of Conservation of Energy That is, as the system moves from one configuration to another, the total energy (kinetic + potential) remains constant. Kinetic energy may be transformed into potential energy and vice versa but the system as a whole can neither gain nor lose energy. This is the law of conservation of energy as it applies to a system of particles with ideal constraints. Such systems are sometimes called conservative systems. Impact The impact between two moving bodies refers to the collision of the two bodies that occurs in a very small time interval and during which the bodies exert a very large force (active and reactive force) on each other. The magnitudes of the forces and the duration of impact depend on the shapes of the bodies, their velocities, and their elastic properties. Consider the impact of two spheres of masses m1 and m2 as shown in the below figure. Let the spheres have the respective velocities of u1 and u2, where u1 > u2, before impact and the respective velocities of v1 and v2 after impact. Before impact x m1 u1 u2 After impact x m1 u1 u m2 2 It is assumed that these velocities are directed along the line joining the centers of the two spheres and are considered to be positive if they are in the positive direction of the x-axis. This is called the case of direct central impact. Two equal and opposite forces, i.e., action and reaction, are produced at the point of contact during impact. In accordance with the law of conservation of momentum, such forces cannot change the momentum of the system of two balls and hence, m u1 1 + = m u2 2 m v1 1 + m v2 2 (8)

3.70 | Part III • Unit 1 • Engineering Mechanics Elastic Impact In an elastic impact, the momentum and kinetic energy is conserved. If the kinetic energy is conserved during impact, then 1 2 1 2 1 2 1 2 1 1 2 2 2 2 1 1 2 2 2 m u m u m v m v2 + = + (9) Since momentum is conserved, equation (8) is also applicable in this type of impact. From equations (8) and (9), it can be shown that v1 - v2 = - (u1 - u2) (10) This equation represents a combination of the law of conservation of momentum and conservation of energy. It states that for an elastic impact the relative velocity after impact has the same magnitude as that before impact but with reversed sign. For two bodies of equal masses undergoing an elastic impact, from equations (8) and (10) it can be shown that they will exchange their velocities, i.e., v1 = u2 and v2 = u1. If the second body was at rest before the impact, i.e., u2 = 0, then it would seem that the striking body stops, i.e., v1 = 0, after having imparted its velocity to the other ball. This phenomenon can be observed in the case of a moving billiard ball which squarely strikes one that was at rest. Again, if the two balls were moving toward each other with equal speeds before impact, an exchange of velocities will simply mean that they rebound from one another with the same speed with which they collided. As another special case, we assume that m2 = ∞ while m1 remains finite and further u2 = 0. This will represent the case of an elastic impact of a ball against a flat immovable obstruction, such as the dropping of a ball on a cement floor. In this case, it is obtained that v1 = -u1, i.e., the striking ball rebounds with the same speed with which it hits the obstruction. Plastic or Inelastic Impact In a plastic or inelastic impact, the momentum is conserved but the kinetic energy is not (part of the kinetic energy is converted to a different form of energy). In a perfectly plastic impact, the colliding bodies will stick to each other after collision and will move with a common velocity. If v is the common velocity of two colliding bodies after a perfectly plastic impact, then from equation (8), we have v m u m u m m = + + 1 1 2 2 1 2 Newton’s experimental law of colliding bodies: Newton proposed an experimental law that describes how the impact of moving bodies was related to their velocities and found that: Speed of separation Speed of approach = e e = coefficient of restitution e satisfies the condition 0 ≤ e ≤ 1. If e = 1 ⇒ the collision is perfectly elastic. If e = 0 ⇒ the collision is inelastic If 0 < e < 1 ⇒ the collision is said to be elastic. Energy loss due to impact: The energy lost in impact when e ≠ 1, i.e., when the collision is not perfectly elastic is given by Loss in kinetic energy = + - - 1 2 1 1 2 1 2 1 2 m m 2 2 m m ( ) u u ( ) e . ∴ When e = 1 the loss is zero. Coefficient of restitution: It is defined as the ratio of the relative velocity of the impacting bodies after impact to their relative velocity before impact. The coefficient of restitution ‘e’ is given by the following equation. e v v u u = - - ( ) ( ) 2 1 1 2 Example 31: A bullet travelling with a velocity of 800 m/s and weighing 0.25 N strikes a wooden block of weight 50 Nresting on a horizontal floor. The coefficient of friction between floor and the block is 0.5. Determine the distance through which the block is displaced from its initial position. Solution: Velocity of the bullet before impact, va = 800 m/s Velocity of the block before impact, vb = 0 m/s Mass of the bullet, . m g a = 0 25 kg Mass of the block, m g b = 50 kg The bullet after striking the block remains buried in the block and both move with a common velocity v. Applying the principle of conservation of momentum, mava + mbvb = (ma + mb)v 0 25 800 50 0 . . 0 25 50 g g g g × + × = + v ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ v = 3.98 m/s To find the distance travelled by the block, apply the priniciple of work and energy. Kinetic energy lost by the block with the bullet buried = Work done to overcome the frictional force If s is the distance travelled by the block, then 1 2 (ma + mb)v2 = mR s = mg(ma + mb) s (∴ R = g(ma + mb)) ∴ = × × s = 3 98 2 9 81 0 5 1 61 2 . . . . m

Chapter 4 • Rectilinear Motion | 3.71 Example 32: Three spherical balls of masses 3 kg, 9 kg and 12 kg are moving in the same direction with velocities 12 m/s, 4 m/s and 2 m/s respectively. If the ball of mass 3 kg impinges with the ball of mass 9 kg which in turn impinges with the ball of mass 12 kg. Prove that the balls of masses 3 kg and 9 kg will be brought to rest by the impacts. Assume the balls to be perfectly elastic. 12 m/s 4 m/s 2 m/s 3 kg 9 kg 12 kg Solution: For perfectly elastic balls, e = 1 ma = 3 kg, mb = 9 kg, mc = 12 kg Impact of balls A and B: Conservation of momentum gives, mava + mbvb = mav′ a + mbv′ b 3 × 12 + 9 × 4 = 3v′ a + 9v′ b (1) e v v v v b a b a = - ′ - ′ - v′ b - v′ a = e(va - vb) = 1 × (12 - 4) = 8 (2) Solving Eqs. (1) and (2), we get v′ b = 8 m/s and v′ a = 0 m/s, i.e., the ball of mass 3 kg is brought to rest. Impact of balls B and C: Consider now the impact of the ball B, of mass 9 kg and moving with the initial velocity of 8 m/s, with the ball C, of mass 12 kg and moving with the velocity of 2 m/s. Conservation of momentum gives mbvb + mc vc = mbv′ b + mc v′ c 9 × 8 + 12 × 2 = 9v′ b + 12v′ c (3) e v v v v c b c b = - ′ - ′ - v′ c - v′ b = e(vb – vc ) = 1 × (8 - 2) = 6 (4) Solving Eqs. (3) and (4), we get v′ c = 6 m/s and v′ b = 0 m/s, i.e., the ball of mass 9 kg is brought to rest. Direction for questions 33 and 34: The blocks 1 and 2, having a weight of 1 kg each and the respective velocities of 10 m/s and 4 m/s, undergo a perfect inelastic collision. Example 33: The final velocity of the blocks is (A) 7 m/sec (B) 6 m/sec (C) 3 m/sec (D) 4 m/sec Solution: V M V M V m m = = + + = × + × + 1 1 2 2 1 2 1 10 4 1 1 1 7 m/s. Example 34: The energy converted into heat as a result of the collision is (A) 40 J (B) 9 J (C) 50 J (D) 54 J Solution: The original kinetic energy was, K1 1 2 1 100 1 2 = × × + × ×1 16 5 = 8 J The final kinetic energy is, K2 1 2 = × 2 4 × = 9 49 J Loss of Kinetic energy = 58 - 49 = 9 J (converted to heat energy). Exercises Practice Problems 1 Direction for questions 1 to 10: Select the correct alternative from the given choices. 1. A car starts with an acceleration of 2 m/s2. Another car starts from the same point after 5 seconds and chases the first car with a uniform velocity of 20 m/s. The time at which the second car, after it starts, will overtake the first car is (A) 5 sec (B) 7 sec (C) 9 sec (D) 11 sec 2. A body is moving with uniform acceleration. In the 4th second of its travel it covers 20 m and 30 m in the 8th second. The distance travelled at the 10th second is (A) 24 m (B) 35 m (C) 43 m (D) 52 m 3. A block is made to slide down an inclined plane which is smooth. It starts sliding from rest and takes a time of t to reach the bottom of the plane. An identical body is freely dropped from the same point. The time the body takes to reach the bottom of the plane is (A) t (B) t 2 (C) t 3 (D) t 4 4. A stone is dropped into a well. The sound of the splash is heard 3.63 seconds later. Assume the velocity of the sound to be 331 m/s. The depth of the surface of water from the ground is (A) 46.38 m (B) 51.36 m (C) 58.39 m (D) 64.62 m

3.72 | Part III • Unit 1 • Engineering Mechanics 5. A motor cycle starts from rest from point A, 2 seconds after a car, speeding at a constant velocity of 120 km/h, passes that point. The motor cycle accelerates at a rate of 6 m/s2 until the motor cycle attains a maximum speed of 150 km/h. The distance from the starting point to the point at which the motor cycle overtakes the car is (A) 912 m (B) 1024 m (C) 1286 m (D) 1356 m 6. A rail road coal car of tare weight mo is moving at a constant speed v while being loaded with coal at a constant rate of w per second. The force necessary to sustain the constant speed, neglecting friction, is (A) w2v (B) wv (C) w v2 2 (D) w2v2 7. A 10 kg shell is fired with a velocity of 800 m/s at an angle of 30° from an old 2000 kg gun. Assuming that barrel and frame can recoil freely, the reaction of the gun, if the shell leaves the barrel 10 milliseconds after firing, is (A) 400 kN (B) 450 kN (C) 600 kN (D) 550 kN 8. A baggage truck pulls two carts A and B. If the mass of the truck is 400 kg and the carts A and B carry 800 and 400 kg respectively, and the truck develops a tractive force of 2 kN. The horizontal forces between the truck and the cart A and between the two carts, respectively, are (A) 1200 N and 400 N (B) 1000 N and 450 N (C) 1500 N and 500 N (D) 500 N and 500 N 9. A body of weight 200 N is placed on a rough horizontal plane. The coefficient of friction, if a horizontal force of 80 N just causes the body to slide over the horizontal plane, is (A) 0.6 (B) 0.1 (C) 0.2 (D) 0.4 10. A body of weight 400 N is pulled up along an inclined plane having an inclination of 30° to the horizontal at a steady speed. The pulling force applied on the body is parallel to the inclined plane. The coefficient of friction between the body and the plane is 0.25. If the distance travelled by the body is 10 m along the plane, then the work done on the body is (A) 3412 J (B) 2866 J (C) 1002 J (D) 4956 J Practice Problems 2 Direction for questions 1 to 10: Select the correct alternative from the given choices. 1. A boat goes 30 km down the stream in 75 minutes and the same distance up the stream in 90 minutes. The speed of the stream is (A) 0.8 km/h (B) 1.2 km/h (C) 1.6 km/h (D) 2 km/h 2. The motion of a body is explained by the equation: s = t 3 – 3t 2 – 9t + 12, where s is the displacement in metres at any time t in seconds. The acceleration of the particle when its velocity is zero is (A) 4.5 m/s2 (B) 6.2 m/s2 (C) 8 m/s2 (D) 12 m/s2 Direction for questions 3 and 4: There are three marks along a straight road at a distance of 100 m. A vehicle starting from rest and accelerating uniformly passes the first mark (P) and takes 10 seconds to reach the second mark (Q). Further it takes 8 seconds to reach the third mark (R). 3. The velocity of the car at Q is (A) 11.38 m/s (B) 13.5 m/s (C) 14.8 m/s (D) 15.5 m/s 4. The distance of mark P from the starting point is (A) 218 m (B) 183 m (C) 156 m (D) 134 m 5. An aircraft is flying at an elevation of 1500 m above the ground horizontally. The velocity is 100 km/h, horizontal and uniform. The aircraft releases a bomb at this elevation. If the target on the ground was just below the plane at the time of releasing the bomb, the distance away from the target, the bomb will hit the ground is (A) 2.35 km (B) 3.42 km (C) 4.86 km (D) 5.32 km Direction for questions 6 and 7: A pile of mass 400 kg is driven by a distance of d into the ground by the blow of a hammer of mass 800 kg through a height of h onto the top of the pile. Assume the impact between the hammer and pile to be plastic. h δ δ M 3 2 1 Given M = 800 kg, m = 400 kg, h = 1.2 m, d = 10 cm.

Chapter 4 • Rectilinear Motion | 3.73 6. The work done is (A) 5.28 kJ (B) 6.278 kJ (C) 7.126 kJ (D) 6.8 kJ 7. The kinetic energy of the whole system in the position 3 is (A) 0 J (B) 10 J (C) 100 J (D) 20 J Direction for questions 8 and 9: A gun of mass 2000 kg fires horizontally a shell of mass 50 kg with a velocity of 300 m/s. 8. The velocity with which the gun will recoil is (A) –7.5 m/s (B) –8.4 m/s (C) 9.2 m/s (D) 10 m/s 9. The uniform force required to stop the gun in 0.6 m is (A) 55310 N (B) 46875 N (C) 55475 N (D) 82750 N 10. A tennis ball is having a velocity of 40 m/s at an angle of 30° with the horizontal just after being struck by the player. The radius of curvature of its trajectory is (A) 188.2 m (B) 198.6 m (C) 200 m (D) 168.2 m 1. A 1 kg mass of clay, moving with a velocity of 10 m/s, strikes a stationary wheel and sticks to it. The solid wheel has a mass of 20 kg and a radius of 1m. Assuming that the wheel and the ground are both rigid and that the wheel is set into pure rolling motion, the angular velocity of the wheel immediately after the impact is approximately [2005] 20 kg 10 m/s 1 m 1 kg + (A) Zero (B) 1 3 rad/s (C) 10 3 rad/s (D) 10 3 rad/s 2. During inelastic collision of two particles, which one of the following is conserved? [2007] (A) Total linear momentum only (B) Total kinetic energy only (C) Both linear momentum and kinetic energy (D) Neither linear momentum nor kinetic energy 3. A block of mass M is released from point P on rough inclined plane with inclination angle θ, shown in the figure below. The coefficient of friction is μ. If μ < tan θ, then the time taken by the block to reach another point Q on the inclined plane, where PQ = s, is [2007] g P q Q (A) 2s g cos ( θ θ tan ) - m (B) 2s g cos ( θ θ tan ) + m (C) 2s g sin ( θ θ tan ) - m (D) 2s g sin ( θ θ tan ) + m 4. Match the approaches given below to perform stated kinematics/dynamics analysis of machine. [2009] Analysis Approach (P) Continuous relative rotation (1) D’ Alembert’s principle (Q) Velocity and acceleration (2) Grubler’s criterion (R) Mobility (3) Grashoff’s law (S) Dynamicstatic analysis (4) Kennedy’s theorem (A) P–1, Q–2, R–3, S–4 (B) P–3, Q–4, R–2, S–1 (C) P–2, Q–3, R–4, S–1 (D) P–4, Q–2, R–1, S–3 5. The coefficient of restitution of a perfectly plastic impact is [2011] (A) 0 (B) 1 (C) 2 (D) ∞ 6. A truck accelerates up a 10° incline with a crate of 100 kg. Value of static coefficient of friction between the crate and the truck surface is 0.3. The maximum value of acceleration (in m/s2) of the truck such that the crate does not slide down is [2014] 7. A mass m1 of 100 kg travelling with a uniform velocity of 5 m/s along a line collides with a stationary mass m2 of 1000 kg. After the collision, both the masses travel together with the same velocity. The coefficient of restitution is [2014] (A) 0.6 (B) 0.1 (C) 0.01 (D) 0 Previous Years’ Questions

3.74 | Part III • Unit 1 • Engineering Mechanics 8. A swimmer can swim 10 km in 2 hours when swimming along the flow of a river. While swimming against the flow, she takes 5 hours for the same distance. Her speed in still water (in km/h) is _____. [2015] 9. A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 m. Ball hits the ground and bounces off the ground. Upon impact with the ground, the velocity reduces by 20%. The height (in m) to which the ball will rise is _____. [2015] 10. A small ball of mass 1 kg moving with a velocity of 12 m/s undergoes a direct central impact with a stationary ball of mass 2 kg. The impact is perfectly elastic. The speed (in m/s) of 2 kg mass ball after the impact will be ______. [2015] 11. The initial velocity of an object is 40 m/s. The acceleration a of the object is given by the following expression: a = –0.1v, where v is the instantaneous velocity of the object. The velocity of the object after 3 seconds will be _______. [2015] 12. A bullet spins as the shot is fired from a gun. For this purpose, two helical slots as shown in the figure are cut in the barrel. Projections A and B on the bullet engage in each of the slots. Gun Barrel 0.5 m A Bullet B Helical slots are such that one turn of helix is completed over a distance of 0.5 m. If velocity of bullet when it exists the barrel is 20 m/s, it spinning speed in rad/s is _____. [2015] 13. A point mass having mass M is moving with a velocity V at an angle θ to the wall as shown in the figure. The mass undergoes a perfectly elastic collision with the smooth wall and rebounds. The total change (final minus initial) in the momentum of the mass is: [2016] θ V x, i ^ y, j ^ (A) –2MV cos θ ˆj (B) 2MV sin θ ˆj (C) 2MV cos θ ˆj (D) –2MV sin θ ˆj 14. An inextensible mass less string goes over a frictionless pulley. Two weights of 100 N and 200 N are attached to the two ends of the string. The weights are released from rest, and start moving due to the gravity. The tension in the string (in N) is _______. [2016] 200 N 100 N Answer Keys Exercises Practice Problems 1 1. A 2. B 3. B 4. C 5. A 6. B 7. A 8. C 9. D 10. B Practice Problems 2 1. D 2. D 3. A 4. D 5. C 6. B 7. A 8. A 9. B 10. A Previous Years’ Questions 1. B 2. A 3. A 4. B 5. A 6. 1 to 1.3 7. D 8. 3.5 9. 0.64 10. 7.8 to 8.2 11. 29.5 to 29.7 12. 251 to 252 13. D 14. 133–134

Curvilinear Motion Kinematics of Curvilinear Translation Motion of a particle describing a curved path is called as curvilinear motion, Velocity and Acceleration The curvilinear motion of a body P may be imagined as the resultant of two rectilinear motions of its projections Px and P y on Ox and O y axis respectively. Velocity Let us consider a body moving through a distance ds from position P to P1 along a curved path in time dt. ds dy dl x s V V + dv V + dv dv V (dv)y (dv)x Px x O O y y Consider PP1 as a chord instead of an arc, we have V s t av = d d Its projections on the x and y co-ordinates are ( ) V s t x s x t av x= = d d d d d d ( ) v s t y s y t av y = = d d d d d d Now d d d d x t y t and are the average velocities of the projections Px and P y respectively in the direction of their respective co-ordinates. If dt approaches zero, vav becomes the instantaneous velocity. Instantaneous velocity at P v d t d s t s t , = = lim d → d 0 d and its direction will be tangential to the path at P. Similarly v dx dt x = , v dy dt y = Total velocity v v = + x y v 2 2 Acceleration The average acceleration during the interval t is a v t av = d d The direction will be same as that of the change of velocity dv. The projections of aav on x and y co-ordinates will be d d v dv dt x t y and respectively. ☞ Kinematics of Curvilinear Translation ☞ Projectile Motion ☞ Equations of the Path of Projectile ☞ Motion of a Projectile on an Inclined Plane ☞ Kinematics of Rotation ☞ Angular Displacement and Angular Velocity ☞ Angular Acceleration ☞ Equations of Motion Along a Circular Path ☞ Curvilinear and Rotary Motion Kinetics of Curvilinear and Rotary Motion ☞ Laws for Rotary Motion ☞ Angular Momentum or Moment of Momentum ☞ Conservation of Angular Momentum ☞ Simple Harmonic Motion and Free Vibrations ☞ Oscillation, Amplitude, Frequency and Period ☞ Velocity and Acceleration ☞ Frequency of Vibration of a Spring Mass System ☞ Oscillations of a Simple Pendulum CHAPTER HIGHLIGHTS Chapter 5

3.76 |  Part III  •  Unit 1  •  Engineering Mechanics When dt approaches zero, the instantaneous acceleration, a v t dv t dt = = → lim d d 0 d a d dt ds dt d s dt = = 2 2 Similarly the components of the instantaneous acceleration a are, a d x dt x = 2 2 , a d y dt y = 2 2 We get, a a = +x y a 2 2 Tangential and Normal Acceleration A particle moves on a curved path and from position P, moves a distance ds to position P1, in the time interval dt, such that at P the instantaneous velocity is v and that at P1 it is (v + dv) ds dq dq dy v + dv v p q q′ P r O P1 Resolving the acceleration into two components: 1. Tangential to the path at the position P. 2. Normal to the path at position P. Let r be the radius of the curved path PP1 and dq, the angle subtended at the centre O. Let q be the angle included between the normals at P1 and P. From the figure we see that P p = instantaneous velocity v at P. Resolving dV into two components (pq) in the direction tangential at P and qq′ in the direction normal at P as shown. Tangential acceleration a t pq t t t = = → → lim lim d d 0 0 d d tangential change in velocity From the triangle Pqq′; pq = Pq - Pp = (v + dv) cos dq - v = v + dv - v = dv (dq being very small, cos dq = 1) Then a v t dv dt t t = = → lim d d 0 d Now normal acceleration a qq t n t = ′ → lim d 0 d qq′ = pq sin dq = (v + dv) dq (dq being small dq = dq in radians) = vdv + dvdq = vdq (dq and dv being very small, their product will be negligible) From above figure OPP1 dq d = = PP r s r 1 qq v s r ′ = d Substituting qq′ in equation we have a v s r n t t = → lim d d 0 d a v r ds dt ds dt v n = × = ,� But ∴ = a v r n 2 Normal acceleration is also known as ‘centripetal acceleration’. During the motion of a particle along a curved path there is a change in the direction of its velocity from instant to instant with or without any change in magnitude. When both magnitude and direction of velocity change, the particle has the tangential and normal acceleration. When there is only change in the direction of velocity, the particle has only normal acceleration. NOTE Solved Examples Example 1: The equation of motion of a particle moving on a circular path, radius 400 m, is given by S = 18t + 3t 2 + 2t 3. Where S is the total distance covered from the starting point, in metres, till the position reached at the end of t seconds. (i) The acceleration at the start is (A) 6m/s2 (B) 5m/ 2s (C) 10m/ 2s (D) 7m/s2 (ii) The time when the particle reaches its maximum velocity is (A) 0.5 s (B) 0.6 s (C) 0.8 s (D) 0.95 s (iii) The maximum velocity of the particle is (A) 19.58 m/s (B) 20.53 m/s (C) 18.65 m/s (D) 13.5 m/s Solution: (i) Given, s = 18t + 3t 2 - 2t 3 v ds dt = = 18 + - 6 6 t t 2

Chapter 5  •  Curvilinear Motion  | 3.77 From equation, a d s dt = = - t 2 2 6 12 At the starting point when t = 0, Acceleration a = 6 – 0 = 6 m/s2. (ii) For determining the condition for maximum velocity, we have d s dt t 2 2 = - 6 12 0 = = 0 5. s s ec (iii) When t = 0.5 s, vmax = + 18 3 1- = . . 5 19 5m/s Example 2: A particle moving along curved path has the law of motion vx = 2t - 4, v y = 3t 2 - 8t + 8 where vx and v y are the rectangular components of the total velocity in the x and y co-ordinates. The co-ordinates of a point on the path at an instant when t = 0, are (4, -8). The equation of the path is (A) x2 + 3x - 2 (B) x3 + 4x + 2 (C) x x 1 2 + + 3 2 (D) x x 3 2 1 + + 4 2 2 Solution: vx = 2t - 4 v y = 3t 2 - 8t + 8 Integrating both sides, we have v dt t dt x = - ∫ ∫( ) 2 4 x t = ×2 - +t C = - t t +C 2 4 4 2 1 2 1 v dt t t dt y = - + ∫ ∫( ) 3 8 8 2 y t t = ×3 - × + +t C = - t t + +t C 3 8 2 8 4 8 3 2 2 3 2 2 Where C1 and C2 are constants Given x = 4, y = -8 when t = 0 Substituting for x, y and t in equation 4 = 0 - 0 + C1 ∴ C1 = 4 -8 = 0 - 0 + 0 + C2 ∴ C2 = -8 Now the equations of displacement are x = t 2 - 4t + 4 and y = t 3 - 4t 2 + 8t - 8 x = (t - 2)2 x t 1 2 = - 2 t x = + 1 2 2 (1) y = t 3 - 4t 2 + 8t - 8 (2) Substituting the value of t from (1) in (2), we get y x = + x x + 3 2 1 4 2 2 Projectile Motion Definitions 1. Projectiles: A particle projected at a certain angle is called projectile. 2. Angle of Projection: Angle between the direction of projection and the horizontal plane through the point of projection is called as the angle of projection. It is denoted by a. 3. Trajectory: The path traced out by the projectile is called the trajectory of the projectile. 4. Velocity of projection (u): The initial velocity of projectile is the velocity of projection. 5. Time of flight (T): The total time taken by the projectile is termed as the time of flight. 6. Horizontal range (R): It is the distance between the point of projection and the point where the trajectory meets the horizontal plane. Equations of the Path of Projectile a C(vertex) P O B A X Y x u y P is the position occupied by the projectile after t sec and x and y are the two co-ordinates of P along the x-axis and y-axis respectively. Along the x-axis, ux = u cos a. Along the y-axis u y = u sin a The component ux remains constant all throughout u y retards due to the action of gravitational force. We know S = vt, for horizontal motion x = u cos a xt t x u = cos a s u = +t at 1 2 2 , for vertical motion Therefore y u = - sin at gt 1 2 2 Substituting value of t we can write y u x u g x u = - sin cos cos a a a 1 2 2 2 2 y x gx u = - tan cos a a 2 2 2 2 This is the equation of the path of a projectile which represents a parabola.

3.78 |  Part III  •  Unit 1  •  Engineering Mechanics Horizontal range, R u g u g = = 2 2 2 2 sin cos sin a a a Time of Flight, T u g = ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 sin a Maximum height when the vertical component of the velocity is zero. v y u g y y = = 0 2 2 . max y u g max u u y sin = = , (sin sin ) 2 2 2 a ce a Co-ordinates of vertex C u g u g 2 2 2 2 2 2 sin , a a sin Motion of a Projectile on an Inclined Plane Consider the motion of projectile with an initial velocity u and making an angle a with the horizontal on an inclined plane of inclination q, taking the coordinate axes x, y the expressions for the distance r and height h can be derived. h = r sin q r cos q a r u q x = u(cosa)t = r cosq y u = - (sin a q ) s t gt h = = r in 1 2 2 By eliminating t, we get r r gr u sin cos tan cos cos q q a q a = - 2 2 2 2 2 ⇒ =r - u g 2 2 2 cos cos (tan tan ). a q a q (1) ∴ The distance r is given by equation (1) and thus the height h and the distance on the horizontal plane can be found. i.e., h = r sinq and x = r cosq. The maximum range possible on the inclined place is found out by differentiation equation 1 with respect to a and equating it to zero. ∴ tan 2a = – cotq. ∴ for maximum range the angle made by the velocity vector a should be equal to (45° + q/2) with the horizontal. Example 3: Find the least initial velocity which a projectile may have so that it may clear a wall of 3.6 m high and 6 m distant and strike the horizontal plane through the foot of the wall at a distance of 3.6 m beyond the wall. The point of projection being at the same level as the foot of the wall. Take g = 9 8. m1 /sec2 (A) 10.2 m/s (B) 11 m/s (C) 12 m/s (D) 13.5 m/s Solution: Let u be the least initial velocity of the projectile and a be the angle of projection with the horizontal plane. Horizontal range of projectile, R = 6 + 3.6 = 9.6 m R u g = 2 2 sin c a a os ∴ = 9 6 2 2 . u sin cos g a a Now, u g 2 9 6 2 = . sin c a a os Putting value, u g 2 2 4 8 = . s × ec tan a a (1) Equation for the path of projectile y x gx u max tan cos = - a a 2 2 2 2 3 6 6 6 2 2 2 2 . tan cos = - a a g u Substituting for u2, we have, 3 6 6 6 9 6 2 . tan tan . = - a a 3 6 6 6 9 6 2 . tan . = - ⎡ ⎣ ⎢ ⎤ ⎦ a ⎥ 3 6. . = 2 25tan a tan . . a = = . 3 6 2 25 1 6 a = 57.9° From equation (1) u g g 2 2 4 8 57 9 57 9 4 8 3 54 1 594 = 104 57 × = × = . sec . tan . . . . . u = 10.2m/s Example 4: An aeroplane is moving horizontally at 108 km/h at an altitude of 1000 m towards a target on the ground which is intended to be bombed.

Chapter 5  •  Curvilinear Motion  | 3.79 B 108 mph 1000 m (a) The distance from the target where the bomb must be released in order to hit the target is (A) 428.35 m (B) 450.54 m (C) 580.2 m (D) 800 m (b) The velocity, with which the bomb hits the target is (A) 143 m/s (B) 148 m/s (C) 150 m/s (D) 161.2 m/s Solution: (a) Let B be the point of target and A be the position of the aeroplane and the bomb is released from A to hit at B. The horizontal component of the bomb velocity, which is uniform, is v = = × × 108 = 108 1000 60 60 km/h 30 m/sec. Considering the vertical component of bomb velocity, At m/sec2 A u, , = = 0 9 g .81 S g = t 1 2 2 Let t be the time required to hit B, then 1000 1 2 9 81 2 = × . × t t 2 2000 9 81 = = 203 87 . . t = 14.278 sec Horizontal distance covered by the bomb S = Vt = 30 × 14.278 = 428.35 m i.e., the bomb is released from plane when the horizontal distance is 428.35 from B (b) Vertical component velocity at B = u + gt = 0 + 9.81 × 14.278 = 140.06 m/sec Resultant velocity at B = + 30 140 06 2 2 . = = 20518.8 143 m/sec Example 5: A ball weighing 10 N starts from the position A as shown in figure and slides down a frictionless chute under its own weight. After leaving the chute 1 at point D, the ball hits the wall as depicted in the figure. C B D A Wall Ball V cos 60 2.5 m 1.5 m 1.5 m 1 m 60° V sin 60 (a) The time interval of the ball’s travel from the point D to the point of hit is (A) 0.88 s (B) 0.92 s (C) 0.733 s (D) 0.898 s (b) The distance on the wall above the point D to the point of hit is (A) 0.21 m (B) 0.158 m (C) 0.32 m (D) 0.168 m Solution: (a) The ball starts from point A. The vertical distance from A to C is equal to 3 m. Considering the motion of ball from A to C, V as 2 = 2 Since initial velocity is zero, a = g = 9 8. 1m/sec2 or vC 2 = ×2 9.81× 3 vC = 7 6. 7 m/s, This is the velocity of the ball at C. The motion of the ball from C to D v v D C as 2 2 2 = - 2 7. . 67 = ×2 9 81×1 5. = 58.82 - 29.43 = 29.39 vD = 5 4. 2 m/s On reaching the point D, the horizontal component of the velocity of the ball = ° v cos . 60 = × 5 42 = . 1 2 2 71 m/s Let t be the time taken by the ball to hit the wall from point D. Then, t = = 2 5 2 71 0 922 . . . sec.

3.80 |  Part III  •  Unit 1  •  Engineering Mechanics (b) Finally considering the vertical motion of the ball beyond the point D s u = -t gt 1 2 2 Here u v m s = = D 5 4. 2 = × 5 42 0 922 - × 1 2 9 81 0 922 2 . . . ( . ) = 4.327 - 4.169 = 0.158 m. Hence the ball will hit the wall 0.158 m above the point D after 0.922 sec Example 6: From the top of a tower 60 m high, a bullet is fired at an angle of 60° with the horizontal, with an initial velocity of 120m/s as shown in figure. Neglect air resistance. (a) The maximum height from the ground that would be attained by the bullet, is (A) 528 m (B) 611 m (C) 680 m (D) 720 m (b) The velocity of bullet, 12 seconds after it is fired, is (A) 55 m/s (B) 58 m/s (C) 61 m/s (D) 80 m/s 120 m/sec 60 m 60 B D v C A h Tower q Solution: (a) Height h u g = 2 2 2 sin a = × × × = × × × × 120 120 60 2 9 81 120 120 3 2 3 2 2 9 81 2 (sin ) . . = × = 10800 2 9 81 551 . m Maximum height above ground = 551 + 60 = 611 m. (b) Time of travel upto highest point B is given by, t u g = = × = sin sin . . sec. a 120 60 9 81 10 6 Let D be the point reached by the bullet, 12 seconds after it is fired. Time taken by the bullet to reach point B from A (point at which it is fired from) = 10.6 sec. ∴ Time taken by the bullet to travel from point B to point D = 12 – 10.6 = 1.4 sec. Horizontal velocity at B, vH = 120 cos 60° = × 120 0 5. m = 60 /s The vertical velocity after 1.4 sec of travel from point B, vv = +0 × × = 1 2 9 81 1 4 9 62 2 . . . m/s Velocity at point D v v = + H v v = + = 2 2 2 2 60 9 6. . 2 60 8 m/s Kinematics of Rotation When a moving body follows a circular path it is known as circular motion. In circular motion the centre of rotation is stationary. Angular Displacement and Angular Velocity Angular displacement is defined as the change in angular position (usually referred to as the angle θ), with respect to time. Angular velocity is defined as the rate of change of angular displacement with respect to time. Let a body, moving along a circular path, be initially at P and after time t seconds be at Q. Let ∠POQ = θ Then angular displacement = ∠POQ = θ O Q r P q Time taken = t Angular velocity = = Angular displacement Time q t Mathematically, it is expressed as d dt q . It is denoted by the symbol w w q = d dt It is measured in radian/sec or rad/sec

Chapter 5  •  Curvilinear Motion  | 3.81 Relation between Linear Velocity and Angular Velocity Let v = linear velocity = Linear displacement Time But linear displacement = Arc PQ = OP × q = rq v r t = r × = × q Angular velocity ∵ q t = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ angular velocity v r = × w Where w = angular velocity Angular Acceleration It is defined as the rate of change of angular velocity. It is measured in radians per sec2 and written as rad/sec2 and is denoted by the symbol a. a = Rate of change of angular velocity a w q w q q = = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = d dt d dt d dt d dt d dt ∵ 2 2 . Also d dt d d d dt d d d d w w q q w q w w w q = × = × = It has two components: Normal component = = V r r 2 2 w and tangential component = = = dV dt r d dt r w a If a is the linear acceleration, then a r = a Equations of Motion Along a Circular Path a w w = - 0 t q w a w w aq = + - = 0 2 2 0 2 1 2 2 t t If N is the r.p.m. w p w p p = = = × = 2 60 2 60 60 N v r N r DN radians/sec m/s Where, w0 = initial angular velocity in cycles/sec, w = final angular velocity in cycles/sec, t = time (in seconds) during which angular velocity changes from w0 to w, v = linear speed in m/s, The rotational speed is N revolutions per minute or N r.p.m. Example 7: A wheel rotates for 5 seconds with a constant acceleration and describes during the time 100 radians. It then rotates with a constant angular velocity and during the next 5 seconds, it describes 70 radians. The initial angular velocity and angular acceleration respectively are, (A) 15 rad/s, 2.5 rad/s2 (B) 13 rad/s, 2 rad/s2 (C) 15 rad/s, -2 rad/s2 (D) 26 rad/s, -2.4 rad/s2 Solution: Angular velocity w q = = = t 70 5 14 rad/s a is constant angular acceleration and w0 be initial angular velocity. q w= + 0 a 2 1 2 t t 100 5 1 2 0 52 = + ( ) w a× 5w0 + 12.5a = 100 (1) w = w0 + at 14 = w0 + 5a (2) Solving equations (1) and (2) w0 = 26 rad/sec a = -2.4 rad/sec2 (Retardation) Example 8: A wheel rotating about a fixed axis at 20 r.p.m. is uniformly accelerated for 80 seconds during which time it makes 60 revolutions. (a) The angular velocity at the end of the time interval is (A) 7.294 rad/s (B) 8.384 rad/s (C) 6.812 rad/s (D) 7.829 rad/s (b) The time required for the speed to reach 100 r.p.m. (A) 3.65 min (B) 2.14 min (C) 1.85 min (D) 2.58 min Solution: (a) q w= + 0 a 2 1 2 t t w0 = initial angular velocity w p 0 2 20 60 = 2 094 × = . 2 60 2 094 80 1 2 80 2 p a × = ( . × +) ( ) 2p × 60 = 167.52 + 3200a a p = - = 2 60 167 52 3200 0 065 . . rad/sec2

3.82 |  Part III  •  Unit 1  •  Engineering Mechanics Let w be the angular velocity at the end of 80 seconds in rad/sec. Then w = w0 + at w = 2.094 + (0.065 × 80) = 7.294 rad/sec. (b) 7 294 2 60 . = p × N N = 69.65 r.p.m w1 = w0 + at 1 Where w p 1 2 100 60 = × rad/sec = 10.466 rad/sec 10.466 = 2.094 + 0.065 × t 1 t1 8 372 0 065 = = 128 8 2 = 14 . . . s ec . min. Curvilinear and Rotary Motion Kinetics of Curvilinear and Rotary Motion For a particle or a body moving in a curved path with particular emphasis to the circular path comes under this section. In order to maintain the circular motion, an inward radial force called ‘centripetal force’ is acted upon the body, which is equal and opposite to the centrifugal force that is directed away from the centre of curvature. If r is the radius of the circular path, v is the linear velocity, w is the angular velocity and t is the time, then Angular acceleration = d dt w Tangential acceleration = r d dt w , Normal acceleration = = v r r 2 2 w , Centripetal or centrifugal force = × = W g v r W g r 2 2 w . Laws for Rotary Motion First Law It states that a body continues in its state of rest or of rotation about an axis with constant or uniform angular velocity unless it is compelled by an external torque to change that state. Second Law It states that the rate of change of angular momentum of a rotating body is proportional to the external torque applied on the body and takes place in the direction of the torque. I = Mk2 where M = mass of the body and k = radius of gyration = moment of inertia × initial angular velocity Initial angular momentum = Iw0 Final angular momentum = Iw Change of angular momentum = I (w – w0) Rate of change of angular momentum = change of angular momentum Time = - = = - = ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ I t I t ( ) w w a a 0 0 w w ∵ angular acceleration From second law of motion of rotation, Torque a rate of change of angular momentum T = Ia T = KIa, where K is a constant of proportionality. SI unit of torque is Nm. Angular momentum or moment of momentum: Moment of momentum of the body about O = Iw, Where the rigid body undergoes rotation about O. Angular momentum is the moment of linear momentum Rotational kinetic energy: Rotational kinetic energy = 1 2 2 Iw Angular impulse or impulsive torque: Angular impulse or impulsive torque = I dw Work done in rotation: Work done in rotation = T × q Kinetic energy in combined motion: Kinetic energy due to translatory motion = 1 2 mv2 Kinetic energy due to rotation = 1 2 2 Iw Kinetic energy due to combined motion = + 1 2 1 2 mv2 2 Iw . Conservation of Angular Momentum The law of conservation of angular momentum states that the angular momentum of a body or a system will remain unaltered if the external torque acting on it is zero. D’alemberts’ Principle for Rotary Motion D’Alemberts’ principle for rotary motion states that the sum of the external torques (also termed as active torques) acting on a system, due to external forces and the reversed active torques including the inertia torques (taken in the opposite direction of the angular momentum) is zero. Suppose a disc of moment of inertia I rotates at an angular acceleration a under the influence of a torque T, acting in the clockwise direction. Inertia torque = Ia (acting in the anti-clockwise direction) From D’Alemberts’ principle, T - Ia = 0, the dynamic equation of equilibrium for a rotating system.

Chapter 5  •  Curvilinear Motion  | 3.83 Rotation caused by a weight W attached to one end of a string passing over a pulley of weight W0 From D’Alemberts’ principle, it can be shown that, a gW W W = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0 2 , when the pulley is considered as a disc. Rotation caused due to two weights W1 and W2 attached to the two ends of a string which passes over a rough pulley of weight W0 a g W W W W W = - + + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ( ) 1 2 1 2 0 2 Example 9: In a pulley system shown in figure the pulley weighs 20 N and its radius of gyration is 40 cm. A 200 N weight is attached to the end of a string and a 50 N is attached to the end of the other string as shown in the figure. 50 N 14 cm 42 cm 200 N (a) The torque to be applied to the shaft to raise the 200 N weight at an acceleration of 1.5 m/s2 is (A) 6812 Ncm (B) 9136 Ncm (C) 700 Ncm (D) 7832 Ncm. (b) The tensions in the strings are respectively (A) 170.4 N, 35.6 N (B) 180 N, 40 N (C) 190.2 N 35 N (D) 180.6 N, 42.34 N Solution: (a) Moment of inertia of the pulley I W g = k 2 I = × = 20 981 40 32 62 2 ( ) Ncm 2 2 . Ncm T1 = Torque produced by 200 N = 200 × 42 = 8400 Ncm T2 = Torque developed by 50 N = 50 × 14 = 700 Ncm Inertia torque due to angular rotation of the pulley with angular acceleration a = Ia = 32.62a Ncm. Torque due to inertia force on 200 N = = ( ) ma r r r = × × ( ) 200 981 200 981 42 2 a a = 359.63a Ncm Torque due to inertia force on 50 N = × × = 50 981 14 9 99 2 a a . Ncm Let T be the torque applied to the shaft for dynamic equilibrium ΣT = 0 T + 700 = 8400 + 32.62a + 359.63a + 9.99a T = 8400 + 312.33 = 9136 Ncm, Since r a = = ad/s 150 42 3 57 2 . . (b) Let F1 and F2 be the tensions in the strings. Applying D’Almberts’ principle for linear motion, we get F1 200 200 9 8 - - × = 1 5 0 . . F F 2 2 50 50 9 8 + - = ×1 5 . . F1 200 200 9 8 = + × = 1 5 200 + 22 96 . . . = 180.6 N F2 50 9 8 50 1 5 9 8 = 42 34 × - × = . . . . N Simple Harmonic Motion and Free Vibrations Simple harmonic motion: It is defined as the type of motion in which the acceleration of the body in its path of motion, varies directly as its displacement from the equilibrium position and is directed towards the equilibrium point. Oscillation, Amplitude, Frequency and Period Y y Y′ P w X1 M X x O In the above figure, when a particle P is describing a circular path, M being the projection of P, it describes a simple harmonic motion. The motion of M from X to X′ and back to X is called an oscillation or simple harmonic motion. OX = OX′ is the amplitude. This amplitude is the distance between the centre of simple harmonic motion and the point where the velocity is zero. The period of one complete oscillation is the period of simple harmonic motion.

3.84 |  Part III  •  Unit 1  •  Engineering Mechanics Thus the period of simple harmonic motion is the time in which M describes 2p radians at w radians/sec. T = 2p w , where T is the time period in seconds. Velocity and Acceleration The simple harmonic displacement X = r sin wt v r = - w x 2 2 Acceleration = = - d x dt r t 2 2 2 w w sin a = -w2x Frequency = 1 2p a x Frequency of Vibration of a Spring Mass System Consider a helical spring subjected to a load W. The static equilibrium position is 0-0. Let S be the stiffness of the spring which is defined as force required to cause one unit extension. If the weight is displaced and stretched to position 1-1′ by an amount ‘y’, as shown in the below figure, then the acceleration with which the load springs back, w g a s = - y ∴ = × - a ⋅ s g W y This is of the form a y = -wn 2 Where w d n sg w g 2 = = , d w being s Frequency f n g = = w 2p p d 1 2 . w w w 2 0 1 2 0 1 y Oscillations of a Simple Pendulum Period of oscillation T f l g = = 1 2p (for 2 beats) l = length of pendulum. Half of an oscillation is called a beat or swing. A pendulum executing one half oscillation per second is called seconds pendulum. Time of one beat or swing = = p l g T 2 . For n number of beats, time = n l g p . For a compound pendulum T K h gh G = + 2 2 2 p Where h is the distance between the point of suspension and centre of gravity. Where kG = radius of gyration about O, the centre of suspension. A compound pendulum is a rigid body free to oscillate about a smooth horizontal axis passing through it. A simple pendulum whose period of oscillation is the same as that of a compound pendulum is called as a simple equivalent pendulum 1 2 = + k h h G . Example 10: A body performing simple harmonic motion has a velocity 12 m/s when the displacement is 50 mm and 3 m/s when the displacement is 200 mm, the displacement being measured from the mean position. (a) Calculate the frequency of the motion. (A) 35 cycles/sec (B) 40.5 cycles/sec (C) 31.8 cycles/sec (D) 35.5 cycles/sec (b) What is the acceleration when the displacement is 75 mm. (A) 15 m/s2 (B) 16.5 m/s2 (C) 13.8 m/s2 (D) 15.6 m/s2 Solution: (a) In simple harmonic motion V2 = w2(r2 - x2) V = velocity, r = amplitude x = distance from mid positions x1 = 50 mm, x2 200 mm V1 = 12 m/s V2 = 3 m/s 12 50 1000 2 2 2 2 = - ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ w r 3 200 1000 2 2 2 2 = - ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ w r (1) Dividing we get 144 9 1 400 4 100 2 2 = - - r r 16 1 400 4 100 2 2 = - - r r

Chapter 5  •  Curvilinear Motion  | 3.85 16 16 4 100 1 400 2 2 r r - × = - 15 16 2 50 1 400 2 r = × - 15 2 64 4 400 1 400 511 400 2 r = × × - = r2 511 400 15 = 0 085 × = . r = 0.29, m = 290 mm. Putting the value of r2 in equation (1), we get 9 0 085 0 04 2 = - w [ . . ] Or, w2 9 0 045 = . We get, w = 200 rad/s So, f = = = w 2p p 200 2 31. 83 cycles/sec. (b) If a be the acceleration when displacement x = 75 mm a x = = × ⎛ ⎝ ⎜ ⎞ ⎠ w ⎟ = 2 9 0 045 75 1000 15 . m/s . 2 Example 11: The amount of seconds a clock would loose per day, if the length were increased in the ratio 800 : 801 is (A) 48 s (B) 54 s (C) 50 s (D) 60 s Solution: Given I = 800 units I + dI = 801 units dI = 1 unit We get, dl I I = 800 dn n dI I I = - = 2 1600 dn n = - = - = - 1600 86400 1600 54 Where n = 86400, as a seconds pendulum will beat 86400 times/day. The clock will loose 54 seconds a day.) Exercises Practice Problems 1 Direction for questions 1 to 10: Select the correct alternative from the given choices. Direction for questions 1 to 3: A force of 2t Newton, where t in seconds, acts on a mass of 100 kg initially at rest, for a period of 20 seconds. 1. The impulse on the mass is (A) 400 Ns (B) 300 Ns (C) 350 Ns (D) 500 Ns 2. The velocity of the mass is (A) 1 m/s (B) 2 m/s (C) 3 m/s (D) 2.5 m/s 3. The average force, which would have resulted in the same velocity, is (A) 15 N (B) 30 N (C) 20 N (D) 10 N 4. A car of mass 1500 kg descends a hill of 1 in 5 incline. The average braking force required to bring the car to rest from a speed of 80 km per hour in a distance of 50 m is (take the frictional resistance as 300 N) (A) 10 N (B) 15 N (C) 8 N (D) 12 N Direction for questions 5 and 6: A thin circular ring of mass 200 kg and radius 2 m resting flat on a smooth surface is subjected to a sudden application of a force of 300 N at a point of its periphery. 5. The angular acceleration is (A) 0.75 rad/s2 (B) 1.5 rad/s2 (C) 2 rad/s2 (D) 2.5 rad/s2 6. The acceleration of mass centre is (A) 1 m/s2 (B) 1.5 m/s2 (C) 2 m/s2 (D) 3 m/s2 7. A particle traveling in a curved path of radius of curvature 500 m with a speed of 108 km/h and a tangential acceleration of 4 m/s2. The total acceleration of the particle is (A) 4.38 m/s2 (B) 5 m/s2 (C) 3.5 m/s2 (D) 8 m/s2 Direction for questions 8 and 9: A solid cylinder 80 cm in diameter is released from the top of an inclined plane 2.0 m high surface and rolls down the inclined surface without any loss of energy due to friction. 8. The energy equation for the system is (A) mgh m = v 1 2 2 (B) mgh m = v 1 3 2 (C) mgh m = v 3 4 2 (D) mgh m = v 2 3 2

3.86 |  Part III  •  Unit 1  •  Engineering Mechanics Practice Problems 2 Direction for questions 1 to 10: Select the correct alternative from the given choices. 1. A bullet is projected so as to graze the top of two walls each of height 20 m located at distances of 30 m and 180 m in the same line from the point of projection as shown in figure. The angle and speed of projection of the bullet, respectively, are O a P1 P2 V0 108 m 30 m 20 m 20 m (A) 34.1° and 44 m/s (B) 38.2° and 48 m/s (C) 35.29° and 49.5 m/s (D) 37.87° and 46.1 m/s 2. For a given value of initial velocity for a projectile, the maximum range, on an inclined plane inclined to the horizontal at an angle of b (in degrees), can be obtained if the angle of projection is (A) 45° (B) 90° – 0.5b (C) 45° + 0.5b (D) 45° – 0.5b 3. A shell bursts on contact with the ground and the pieces of it fly off in all directions with speeds up to 40 m/s. A person, standing 40 m away from the point of burst, can be hit by a piece in a time duration of (A) 1.5 sec (B) 1 sec (C) 2 sec (D) 3 sec 4. The coefficient of restitution is defined as the (A) Negative of the ratio of the velocity of separation to the velocity of approach (B) Ratio of the velocity components in the line of impact (C) Ratio of the velocity vectors before and after collision (D) Negative of the ratio of the energies of the bodies before and after the impact 5. A cylinder of radius of r and mass m rest on a rough horizontal rug. If the rug is pulled from under it with an acceleration, a perpendicular to the axis of the cylinder, the angular acceleration of the centre of mass of the cylinder, assuming that it does not slip, is (A) 2 3 A r (B) 1 3 A r (C) 3 4 A r (D) 2 3 A Direction for questions 6 to 8: A soldier positioned on a hill fires a bullet at an angle of 30° upwards from the horizontal as shown in the figure. The target lies 60 m below him and the bullet is fired with a velocity of 200 m/s. 60 m 200 m/s 30° 6. The maximum height, to which the bullet will rise above the position of the soldier, is (A) 615 m (B) 490 m (C) 509.7 m (D) 710.6 m 7. The velocity with which the bullet will hit the target is (A) 202.9 m/s (B) 245.3 m/s (C) 312.7 m/s (D) 343.6 m/s 8. The time required to hit the target is (A) 21.7 sec (B) 20.97 sec (C) 15.6 sec (D) 23 sec 9. A carpet of mass m made of an inextensible material is rolled along its length in the form of a cylinder of radius R and is kept on a rough horizontal floor. When a small push, of negligible force, is given to the carpet, it starts 9. The linear and angular speeds, at the bottom respectively are (A) 6.1 m/s and 12.75 rad/s (B) 5.5 m/s and 34 rad/s (C) 5.1 m/s and 12.75 rad/s (D) 6.1 m/s and 34 rad/se 10. A disc shaped frictionless pulley I M = R 1 2 2 has a mass of 80 kg and radius of 2 m. A rope is wound round the pulley and supports a 4 kg mass. The angular acceleration of the pulley ( ) g = 10 m/s2 is (A) 1 4 2 rad/s (B) 1 2 2 rad/s (C) 1 2 rad/s (D) 3 4 2 rad/s

Chapter 5  •  Curvilinear Motion  | 3.87 Previous Years’ Questions 1. A circular disk of radius R rolls without slipping at a velocity v. The magnitude of the velocity at point P (see figure) is [2008] R P 30° V (A) 3v (B) 3 2 v (C) v 2 (D) 2 3 v 2. An annular disc has a mass m, inner radius R and outer radius 2R. The disc rolls on a flat surface without slipping. If the velocity of the centre of mass is v, the kinetic energy of the disc is [2014] (A) 9 16 mv2 (B) 11 16 mv2 (C) 13 16 mv2 (D) 15 16 mv2 3. Consider a steel (Young’s modulus E = 200 GPa) column hinged on both sides. Its height is 1.0 m and cross-section is 10 mm × 20 mm. The lowest Euler critical buckling load (in N) is ______. [2015] 4. A point mass M is released from rest and slides down a spherical bowl (of radius R) from a height H as shown in the figure below. The surface of the bowl is smooth (no friction). The velocity of the mass at the bottom of the bowl is: [2016] M R H (A) gH (B) 2gR (C) 2gH (D) 0 5. A mass of 2000 kg is currently being lowered at a velocity of 2 m/s from the drum as shown in the figure. The mass moment of inertia of the drum is 150 kg-m2. On applying the brake, the mass is brought to rest in a distance of 0.5 m. The energy absorbed by the brake (in kJ) is _______. [2016] 2 m/s 2 m 2000 kg 6. A circular disc of radius 100 mm and mass 1 kg, initially at rest at position A, rolls without slipping down a curved path as shown in figure. The speed v of the disc when it reaches position B is _______ m/s. [2016] Acceleration due to gravity g = 10 m/s2. unrolling without sliding on the floor. The horizontal velocity of the axis of the cylindrical part of the carpet is 63 3 gR when the radius of the carpet reduces to (A) 3 4 R (B) R 4 (C) R 2 (D) R 5 10. A small sphere rolls down without slipping from the top most point of a track, with an elevated section and a horizontal part, as shown in the following figure, in a vertical plane. The horizontal part is 2 m above the ground level and the top of the track is 8.3 m above the ground. The distance on the ground, with respect to the point B (which is vertically below the end of the track), where the sphere would land is 8.3 m 2 m A B C D (A) 6 m (B) 10 m (C) 3 m (D) 2 m

3.88 |  Part III  •  Unit 1  •  Engineering Mechanics A 30 meters v B 7. A rigid rod (AB) of length L = 2 m is undergoing translational as well as rotational motion in the x-y plane (see the figure). The point A has the velocity V1 = +ˆ ˆ i j 2 m/s. The end B is constrained to move only along the x direction. V2 B V1 A θ = 45º ∧ y, j x, i ∧ The magnitude of the velocity V2 (in m/s) at the end B is ______. [2016] Answer Keys Exercises Practice Problems 1 1. A 2. B 3. C 4. A 5. A 6. B 7. A 8. C 9. C 10. A Practice Problems 2 1. D 2. C 3. B 4. A 5. A 6. C 7. A 8. B 9. B 10. A Previous Years’ Questions 1. A 2. C 3. 3285 to 3295 4. C 5. 14.1 to 14.3 6. 20 7. 3

Test Engineering Mechanics Time: 60 Minutes Direction for questions 1 to 30: Select the correct alternative from the given choices. 1. Two equal and opposite co-planar couples (A) Balance each other. (B) Produce a couple and unbalanced force. (C) Cannot balance each other. (D) Give rise to a couple of double the magnitude. 2. In a perfect frame, the number of members are (A) 2j – 3 (B) 2j + 3 (C) 2j – 2 (D) 2j – 1 Where j = number of joints. 3. The state of equilibrium of a body implies that the body must (with respect to some inertial frame) be: (A) At rest or with uniform acceleration. (B) Uniform velocity or uniform acceleration. (C) At rest or with uniform velocity. (D) At rest or with uniform velocity or uniform acceleration. 4. The distance of the centroid of a semicircle of radius ‘r’ from its base is (A) 4r 3p (B) 3p 4r (C) 4p 3r (D) 2p 3r 5. A machine requires an effort of 10 Kg to lift a load of 200 Kg an effort of 12 Kg for a load of 300 Kg. The effort required to lift a load of 500 Kg will be (A) 16 Kg (B) 15 Kg (C) 14 Kg (D) 17 Kg 6. The moment of a force (A) Ocures about a point (B) Measures the capacity to do useful work. (C) Occurs only when bodies are in motion (D) Measures the abilities to turning or twisting about axes. 7. The required condition of equilibrium of a body is that (A) The algebraic sum of horizontal components of all the forces must be zero. (B) The algebraic sum of the vertical components of all the forces must be zero. (C) The algebraic sum of moments about a point must be zero. (D) All the above. 8. The unit of the moment of Inertia of an area is (A) Kg-m (B) Kg-m2 (C) Kg-m4 (D) m4 9. Moment of Inertia of a square of side ‘a’ about an axis passing through its C. G is equal to (A) a3 12 (B) a4 12 (C) a3 36 (D) a4 36 10. According to the law of the machine, the relation between effort ‘P’ and load W is given by (A) W = mP + C (B) W = mP - C (C) P = mW + C (D) P = mW - C 11. Weight of 150 kN is being supported by a tripod whose leg is of the length of 13 m. If the vertical height of the point of attachment of the load is 12, the force on the tripod leg would be (A) 48.24 N (B) 54.16 N (C) 50.8 N (D) 45.3 N 12. The resultant of two forces 4P and 3P is R. If the first force is doubled the resultant is also doubled. The angle between the two forces is (A) 48.25° (B) 95.73° (C) 32.5° (D) 45.53° 13. In the truss shown the force in the member BC is B A 60° 60° D P P C (A) 0 (B) 0.577 P(T) (C) 0.577 P (comp) (D) 0.866 P (comp) Direction for questions 14 and 15: A body is weighing 500 N is just moved along a horizontal plane by a pull of 100 2 N making 45° with horizontal. 14. Find the value of normal reaction R (A) 300 N (B) 400 N (C) 200 N (D) 500 N 15. Find the coefficient of friction (A) 0.32 (B) 0.33 (C) 0.25 (D) 0.28 Direction for question 16, 17, 18: For the mass-pulley system shown, the mass m2 = 5 Kg is placed on a smooth inclined plane of inclination θ where as mass m1 = 5 Kg is a hanging force. If acceleration of the system is 1.5 m/s2.

3.90 |  Part III  •  Unit 1  •  Engineering Mechanics M2-5 Kg T 5 Kg a = 1.5 m/s2 M1 q 16. The inclination of the plane will be (A) 41.52° (B) 35.50° (C) 52.15° (D) 43.96° 17. The tension in the string will be (A) 41.55 N (B) 35.15 N (C) 21.5 N (D) 25.28 N 18. How the acceleration of the system would be affected of each mass is doubted (A) 3 m/s2 (B) 2 m/s2 (C) 1.5 m2 (D) 2.5 m/s2 19. A block is sliding down an incline of 30° with an acceleration g 4 .Then the kinetic coefficient of friction is (A) 3 2 (B) 1 3 (C) 1 2 (D) 1 2 3 Direction for question for 20 and 21: A 600 N weight is suspended by flexible cables as shown in figure 30° 60° 90° = 600 N C W A B 20. The tension in the wire BC will be (A) 519.6 (B) 613.4 (C) 318 (D) 435.5 21. The tension in the wire AC will be (A) 256 (B) 300 (C) 311 (D) 288 22. The smallest angle θ for equilibrium of the homogenous ladder of length l is, when coefficient of friction for all surfaces is assumed as m: (A) tan- ⎛ - ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 1 2 m m (B) tan-1 2 2 m (C) tan- - ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 1 m m (D) tan- ⎛ - ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 1 2 m 23. The reaction at the hinge when a rigid rod of mass ‘m’ and length ‘L’ is subjected to a force ‘P’ as shown O L P 2L 3 (A) –P (B) 0 (C) P 3 (D) 2 3 P 24. In the figure shown tension in the member QR is P 105° 45° 30° Q R F (A) 0.732 F (B) 0.63 F (C) 0.433 F (D) 0.75 F 25. Force in member QR (A) 0.633 F (B) 0.75 F (C) 0.732 F (D) 0.433 F 26. A force of 600 N is applied to the brake drum of 0.6 m diameter in a band brake. System as shown in below figure, where the wrapping angle is 180°c. If the coefficient of friction between the drum and band is 0.25, the breaking lorque applied, in Nm is 600 N (A) 97.8 N (B) 16 N (C) 22.1 N (D) 15.7 N 27. A circular roller of weight 200 N and radius of 0.8 m hangs by a tie rod of length 2 m and rests on a smooth vertical wall as shown in figure. The tension ‘T’ in the tie rod will be (A) 219.78 (B) 239.2 (C) 310.30 (D) 250.5

Test  | 3.91 28. A mass of 50 kg is suspended from a weight less bar ‘AB’ which is supported by a cable BC and pinned at ‘A’ as shown in figure. The Pin reactions at ‘A’ on the bar AB are 125 mm A B T C 75 mm 50 kg (A) Rx = 343.4 N, R y = 755.4 N (B) Rx = 343.4 N, R y = 0 (C) Rx = 1080 N; R y = 0 (D) Rx = 755. N, R y = 0 Direction for questions 28 and 29: All the forces acting on a particle are situated at the origin of the two dimensional reference frame. One force has a magnitude of 10 N acting in the positive ‘X’ direction, whereas the other has a magnitude of 5 N acting at an angle of 120° directed away from the origin 29. The value of the resultant force will be. (A) 5.88 N (B) 7.2 N (C) 7.98 N (D) 8.66 N 30. The value of a made by resultant with the horizontal force will be (A) 43° (B) 30° (C) 78° (D) 80° Answer Keys 1. D 2. A 3. C 4. A 5. A 6. A 7. D 8. D 9. B 10. C 11. B 12. B 13. C 14. B 15. C 16. D 17. A 18. C 19. D 20. A 21. B 22. A 23. C 24. A 25. A 26. A 27. A 28. C 29. D 30. B

Directions for questions 1 to 35: Select the correct alternative from the given choices 1. A body is moving in a curved path with speed of 10 m/s and tangential acceleration of 3 m/s2 . If radius of curvature be 25 m, the total acceleration of body in m/s2 is (A) 3 (B) 4 (C) 5 (D) 6 2. A stone of mass m at the end of a string of length ‘↓’ is whirled in a vertical circle at a constant speed. The tension in the string will be maximum when the stone is (A) at the top of the circle (B) half way down from the top (C) quarter – way down from the top (D) at the bottom of the circle 3. For maximum range of a projectile, the angle of projection should be (A) 30° (B) 45° (C) 60° (D) 90° 4. The resultant of two forces P and Q inclined at angle q will be inclined at following angle with respect to P. (A) 1 sin tan Cos Q P Q q q −       + (B) 1 Cos tan Sin Q Q P q q −       + (C) 1 Sin tan Cos P Q P q q −       + (D) 1 Cos Tan Sin P Q P q q −       + 5. If n = number of members and j = number of joints, then for a perfect frame n = (A) j – 2 (B) 2j – 1 (C) 2j – 3 (D) 3j – 2 6. A body of mass 15 kg moving with velocity of 2 m/s is acted upon by a force of 75 N for two seconds. The final velocity will be (A) 10 m/s (B) 11 m/s (C) 12 m/s (D) 15 m/s 7. Two blocks with masses 10 kg and 5 kg are in contact with each other and are resting on a horizontal frictionless floor as shown in figure. When horizontal force 600 N is applied to the heavier, the blocks accelerate to the right. The force between the two blocks is 5 kg 10 kg 600 N (A) 300 N (B) 200 N (C) 100 N (D) 50 N 8. Two particles with masses in the ratio 1 : 9 are moving with equal kinetic energies. The magnitude of their linear momentums will conform to the ratio. (A) 1 : 3 (B) 1 : 9 (C) 3 : 1 (D) 3 : 1 9. Match List – I with List – II LIst – I List – II P. Collision of particles 1. Euler’s equation of motion Q. Stability 2. Minimum kinetic energy R. Satellite motion 3. Minimum potential energy S. Spinning top 4. Impulse momentum principle P Q R S (A) 1, 2, 3, 4 (B) 4, 2, 1, 3 (C) 3, 1, 4, 2 (D) 4, 3, 2, 1 10. A spring scale indicates a tension to 10 N in the right hand cable of the pulley system shown in the figure. Neglecting the mass of the pulleys, ignoring friction between the cable and pulley the mass m is (Take g = 10 m/s2 ) m T Spring scale (A) 10 kg (B) 40 kg (C) 1 kg (D) 4 kg 11. Two bodies of mass m1 and m2 are dropped from different heights h1 and h2 respectively. Neglecting the effect of friction, the ratio of time taken to drop though the given heights would be (A) 1 2 m m (B) 1/2 1 2 h h       (C) 2 1 2 h h       (D) 1/2 2 1 h h       Engineering Mechanics Test 1 Number of Questions 35 Time:60 min.

3.6 | Engineering Mechanics Test 1 12. A sphere ‘M’ impinges directly on to another identical sphere ‘N’ at rest. If the co–efficient of restitution is 0.5, the ratio of velocities N M V V after the impact would be (A) 1 : 3 (B) 3 : 1 (C) 1 : 2 (D) 2 : 1 13. A block of steel is loaded by a tangential force on its top surface while the bottom surface is held rigidly. The deformation of the block is due to (A) Shear only (B) Torsion only (C) Bending only (D) Shear and bending 14. The co-efficient of friction depends on (A) Nature of the surface (B) Area of contact (C) Strength of surface (D) All of the above 15. The force induced in member PQ due to load W in figure will be Q W θ R P (A) W secq (B) W cosq (C) W tanq (D) W cosecq 16. A 10 m long ladder is placed against a smooth vertical wall with its lower end 3 m from the wall. For ladder to remain in equilibrium as shown in figure, what should be the co-efficient of friction between ladder and floor? A 10 m 3 m B W µW θ (A) 0.16 (B) 0.25 (C) 0.36 (D) 0.45 17. Three forces acting on a particle in equilibrium are 2P and 3 P. Angle between 2P and P is 120°. What will be angle between P and 3 P. (A) 45° (B) 60° (C) 90° (D) 135° 18. A plane lamina of 200 mm radius is shown in figure given below. The centre of gravity of lamina from the point O 200 60° O (A) 150 mm (B) 140 mm (C) 128 mm (D) 108 mm 19. A uniform rod PQ remains in equilibrium position resting on a smooth inclined plances PO and QO which are at an angle of 90° as shown in figure. α Q P 90° θ O If the plane QO makes angles of a with the horizontal, then what is the inclination q of the rod PQ with the plane PO (A) equal to a (B) less than a (C) greater than a (D) equal to 90° 20. A uniform wheel of 500 mm diameter, weighing 5 kN rests against a rigid rectangular block of 100 mm height as shown in figure. The least pull, through the centre of the wheel, required just to turn the wheel over the corner A of the block is 500 mm A O 100 mm (A) 2 kN (B) 3 kN (C) 4 kN (D) 5 kN 21. A hollow semicircular section has it outer and inner diameter of 200 mm and 150 mm respectively shown in figure. The moment of inertia about the base AB in mm4 is 150 mm 200 mm (A) 5 × 106 (B) 10 × 106 (C) 17 × 106 (D) 27 × 106

Engineering Mechanics Test 1 | 3.7 22. What is the maximum load (W) which a force P equal to 6 kN will hold up, if the co–efficient of friction at C is 0.2 in the arrangement shown in figure, neglect other friction and weight of the member? 1m 40mm 60mm W 0.5m P (A) 2.7 kN (B) 3.5 kN (C) 4.0 kN (D) 5.0 kN 23. A load of 3 kN is to be raised by a screw jack with mean diameter of 60 mm and pitch of 10 mm. The co– efficient of friction between the screw and nut is 0.075. The efficiency of screw jack is (A) 38.54% (B) 41.24% (C) 42.25% (D) 44.15% 24. A system of masses connected by string, passing over pulley A and B is shown in figure. A B 5 kg 20 kg 7 kg The acceleration of mass 20 kg is (A) 2.45 m/s2 (B) 2.01 m/s2 (C) 1.89 m/s2 (D) 1.255 m/s2 25. A solid body A of mass 12 kg, when it is being pulled by another body B of mass 6 kg along a smooth horizontal plane as shown in figure. The tension in the string is (Take g = 9.8 m/s2 ) A 12 kg B 6 kg (A) 39.2 N (B) 25.0 N (C) 12.5 N (D) 6 N 26. A simple pendulum consists of a 500 mm long chord and a bob of mass 2 kg is suspended inside a train, accelerating smoothly on a level track at the rate of 3.2 m/s2 . Find the angle which the chord will make with the vertical. (A) 12° (B) 14° (C) 16° (D) 18° 27. A body of mass 0.6 kg oscillates about an axis at a distance 300 mm from the centre of gravity. If the mass moments of inertia about the centroidal axis, parallel to the axis of rotation, be 0.125 kg-m2 , the length of the equivalent simple pendulum is. (A) 0.6 m (B) 0.8 m (C) 0.9 m (D) 1.0 m 28. A conical pendulum 2 m long is revolving at 35 revolutions per minute. Find the angle which the string will make with the vertical, if the bob describes a circle of 500 mm radius. (A) 15.2° (B) 29.8° (C) 32.5° (D) 35° 29. A rod of length 2 m is sliding in a corner, as shown in figure. At an instant when the rod makes an angle of 55 degrees with the horizontal plane, the angular velocity of the rod is 5 rad/s. The velocity of the rod at point B is 2 m A 55° B

3.8 | Engineering Mechanics Test 1 (A) 1.50 m/s (B) 2.50 m/s (C) 5.01 m/s (D) 8.19 m/s 30. A mass 40 kg is suspended from a weightless bar AB which is supported by a cable CB and a point at A, as shown in figure. The tension in the cable is C B m 200 mm 100 mm A (A) 876.5 N (B) 755.5 N (C) 654.5 N (D) 500 N 31. An elevator weighting 1000 kg attains an upward velocity of 4 m/sec in two sec with uniform acceleration. The tension in the supporting cables will be (A) 2000 kg (B) 1200 kg (C) 1000 kg (D) 800 kg Common Data for Questions 32 and 33: A body of weight 600 N is lying on a rough plane inclined at an angle of 25° with the horizontal. It is supported by an effort (P) parallel to the plane as shown in figure. The angle of friction is 20° 600 N P F 25° 32. The minimum value of P for which the equilibrium can exist. (A) 45.25 kN (B) 55.65 kN (C) 85.55 kN (D) 105.25 kN 33. The maximum value of P for which the equilibrium can exist. (A) 115.5 kN (B) 250.5 kN (C) 350.5 kN (D) 451.5 kN Linked Answer for Questions 34 and 35: The figure below shown a pair of pin jointed gripper tongs holding an object weighting 1500 N. The co-efficient of friction (m) at the gripping surface is 0.1. X – X is the line of action of the input force and Y – Y is the line of application of gripping force. Assuming pin joint is friction less. 200 mm X F F X 100 mm Pin Y 150 N Y 34. The reaction force at the gripping surface is (A) 10,000 N (B) 7500 N (C) 5000 N (D) 3750 N 35. The magnitude of force F required to hold the weight is (A) 7500 N (B) 5000 N (C) 3750 N (D) 2000 N Answer Keys 1. C 2. D 3. B 4. A 5. C 6. C 7. B 8. A 9. D 10. D 11. B 12. B 13. D 14. A 15. A 16. A 17. C 18. C 19. A 20. C 21. D 22. A 23. B 24. A 25. A 26. D 27. C 28. B 29. D 30. A 31. B 32. B 33. D 34. B 35. C

Hints and Explanations 1. Radial acceleration ar = V2 /r = 2 10 25 = 4 m/s2 Tangential acceleration at = 3 m/s2 ∴ Total acceleration a = 2 2 a a r t + = 2 2 4 3 + = 5 m/s2 Choice (C) 4. Tan q = Sin Cos Q P Q q + q Choice (A) 6. Force = mass × acceleration 75 = (15) × a ; a = 5 m/s2 Velocity after 2 seconds v = u + at = 2 + 5 × 2 = 2 + 10 = 12 m/s Choice (C) 7. Let N force b/w the block. Form free body diagram. 600 – N = (10)a (i) N = (5)a (ii) From (i) & (ii) 600 – N = 10 5   N     3N = 600 N = 200 N Choice (B) 8. KE1 = 2 1 1 2 m v KE2 = 2 2 2 2 m v Given that KE1 = KE2 1 2 1 9 m m = m1 v1 2 = m2 v2 2 2 1 2 2 1 v m v m   =     = 9; 1 2 9 v v = = 3. Momentum ratio = 11 1 1 22 2 2 mv m v mv m v = × = 131 91 3 × = Choice (A) 10. referring figure 4T = Mg M = 4T g = 4 10 10 × M = 4 kg Choice (D) 11. s = ut + 2 2 gt u = 0, H1 = 2 2 gt H2 = 2 2 gt 2 1 1 2 2 t h t h    =       ; 1/2 1 1 2 2 t h t h   =     Choice (B) 12. let sphere M = body 1 sphere N = body 2 m1 and m2 are mass of sphere M and sphere N v1 , v2 are the velocities before impact 1 1 1 2 v v, are the velocities after impact ∴ m1 v1 = m1 . 1 1 1 22 v mv + As the balls are identical m1 = m2 v1 = 1 1 1 2 v v + Given that 0.5 = 11 11 21 21 12 1 vv vv vv V − − = − (\ V2 = 0) ∴ 0.5 = 1 1 2 1 1 1 1 2 v v v v − + 1 1 2 1 1 1 2 1 2 2 v v + =− v v 1 1 2 1 3 2 2 v v = 1 2 1 1 v v = 3 : 1 Choice (B) 13. Choice (D) 14. Choice (A) 15. Cosq = W PQ PQ = Cos W q = W Secq Choice (A) 16. Taking moments about A 10 3 Cos 10Sin 2 WW W ×− × = × qm q 3 91 3 5 10 10 10 −× = × × m From the figure Cosq = 3/10 , Sinq = 91 10 3 – 3/2 = m × 9.53 m = 1.5 9.53 = 0.157 ≈ 0.16 Choice (A)

3.10 | Engineering Mechanics Test 1 17. 3 2 Sin120 Sin P P q = 2P P 120° θ √3 P 3 2 sin120 Sin P q = 3 2 3/2 Sin P q = ⇒ Sinq = 1 q = 90° Choice (C) 18. As the lamina is symmetrical about y – y axis, bisecting the lamina, its centre of gravity lies on the axis. a = 30° = 6 p Centre of gravity of the lamina 2 Sin 3 r y a a = = 2 200 Sin 30 3 6 p × × = 400 0.5 3 6 p ×       = 128 mm Choice (C) 19. P θ RP RQ O α Q G R PR ^ PO and QR ^ QO therefore PR || QO QR || PO ∠PRQ = 90° PG = QG ∠GPO = ∠GOP q = a Choice (A) 20. Diameter of the wheel = 500 mm Weight of wheel = 5 kN Height of the block = 100 mm Let P = Least pull required just to turn the wheel in kN Sinq = 150 250 = 0.6 q = 36.86° AB = ( ) ( ) 2 2 250 150 − = 200 mm 5 kN B A R θ 250 mm 150 mm O P Taking moments about A P × 250 = 5 × 200 0 150 mm 100 mm A P = 1000 250 = 4 kN Choice (C) 21. D = 200 mm, R = 100 mm d = 150 mm, r = 75 mm IAB = 0.393 (R4 – r4 ) = 0.393 [1004 – 754 ] = 26.86 × 106 mm4 Choice (D) 22. Let R = normal reaction of the pulley on the beam at C R × 1 = 6 × 1.5 R = 9 kN Maximum force of friction at C = mR = 0.2 × 9 = 1.8 kN Taking moments about the centre of pulley W × 40 = 1.8 × 60 W = 60 1.8 40 × = 2.7 kN Choice (A) 23. Load (W) = 3 kN Mean diameter of the screw (d) = 60 mm Pitch (p) = 10 mm m = tanf = 0.075 tan(a) = 10 60 P p p d = × = 0.053 Efficiency h = ( ) tan tan a a j + = tan tan tan 1 tan tan a a j a j + − = ( ) 0.053 0.053 0.075 1 0.053 0.075 + − × 0.4124 = 41.24% Choice (B)

Engineering Mechanics Test 1 | 3.11 24. Let m1 = 20 kg, m2 = 70 kg, m3 = 5 kg From the system of pulleys and masses, we find that at pulley A, the 20 kg mass will come down with some acceleration as the total mass on the other side of the string is less than 20 kg At pulley B, the 7 kg mass will come down with some acceleration. Acceleration of 20 kg mass is = ( 1 2 ) 1 2 gm m m m − + = ( ( )) ( ) 9.81 20 7 5 20 7 5 − + + + = 2.45 m/s2 Choice (A) 25. Let m1 = 12 kg m2 = 6 kg g = 9.8 m/s2 Tension in the string, T = 1 2 1 2 12 6 9.8 12 6 mmg m m × × = + + = 39.2 N Choice (A) 26. Let q = Angle, which the chord will make with the vertical 3m/s2 θ T T ma mg Weight of the hob = mg = 2 × 9.8 = 19.6 N Inertia force acting on the hob (Opposite to the acceleration of the train) = ma = 2 × 3.2 = 6.4 N ma mg θ \ Tanq = 6.4 19.6 q = 18° Choice (D) 27. m = 0.5 kg, h = 300 m = 0.3 m O h G P  IG = 0.125 kg-m2 IG = m KG 2 0.125 = 0.6KG 2 KG 2 = 0.125 0.6 = 0.208 Length of equivalent simple pendulum L = 2 KG 0.208 0.3 0.3 h h + =+ = 0.993 m Choice (C) 28. L = 2 m N = 35 rpm r = 500 m = 0.5 m angular velocity of the bob w = 2 2 30 60 60 p p N × = w = 3.35rad/s tanq = ( ) 2 2 3.35 0.5 9.8 r g w × = = 0.572 q = tan–1(0.572) = 29.8° Choice (B) 29. A O B 55° I 2 m w = 5 rad/s VA = Velocity along the vertical VB = Velocity along the horizontal I A = OB = L cosq = 2 × cos55° I B = OA =L sinq = 2 × sin55° ∴ VB = w × IB = 5 × 2 × Sin55° = 8.19 m/s Choice (D) 30. T Cos(90 – q) = mg T Sinq = mg Tanq = 100 200 q = 26.56° T = ( ) 40 9.8 Sin Sin 26.56 mg q × = = 876.5 N Choice (A) 31. Velocity = acceleration × time 4 = a × 2 a = 2 m/sec2 Tension in Cable = ( ) W g a g + = ( ) 1000 11,810 9.81 2 9.81 9.81 + = ≈ 1200 kg Choice (B)

3.12 | Engineering Mechanics Test 1 32. For the minimum value of P, the body is at the point of sliding downwards. ∴ P = Sin ( ) Cos W a j j − × = Sin 25 20 ( ) 600 Cos 20 − × = 55.65 kN Choice (B) 33. For the maximum value of P, the body is at the point of sliding upwards. ∴ P = Sin ( ) Cos W a j j + × = Sin 25 20 ( ) 600 Cos 20 + × = 451.5 kN Choice (D) 34. µR µR R R 1500N 2mR = 1500 N R = 1500 2 0.1 × = 7500 N Choice (B) 35. F 200 100 R P Taking moments about Pin (R) × 100 = F × 200 F = 7500 100 200 × F = 3750 N Choice (C)

Directions for questions 1 to 35: Select the correct alternative from the given choices. 1. Two forces of 500 N and 600 N are acting simultaneously at a point. If the angle between them is 60o then the resultant of these two forces is (A) 781 N (B) 954 N (C) 1063 N (D) 881 N 2. A flywheel 400 mm in diameter is brought uniformly from rest to a speed of 240 rpm in 16 seconds. The tangential acceleration of a point on the rim (in m/s2 ) is (A) 1.57 (B) 0.628 (C) 0.314 (D) 0.419 3. A stone of mass 5 kg is tied to a spring of length 2 m and whirled in a horizontal circle at a constant angular speed of 10 rad/sec. The tension in the spring will be (A) 1000 N (B) 750 N (C) 500 N (D) 250 N 4. Ratio of moment of inertia of a sphere and that of a cylinder having same radius and mass about their centroidal axis is (A) 1 5 (B) 5 2 (C) 2 5 (D) 4 5 5. A pulley and rope arrangement is shown below 40 kg P (Hold by a person) If coefficient of friction between pulley and rope is 0.25 then the holding load by the person will be (A) 178.91 N (B) 860.64 N (C) 294.21 N (D) 741.23 N 6. The velocity-time graph of a body is shown in the figure. The acceleration at point A will be V (m/s) 10 A 2 t (seconds) (A) 5 m/s2 (B) 2.5 m/s2 (C) 10 m/s2 (D) Zero 7. If two bodies one light and other heavy have equal kinetic energies and equal mass then which one has a greater momentum? (A) Heavy body (B) Light body (C) Both have equal momentum (D) None of these 8. A ball of mass 9.81 kg is thrown with an angle a to the horizontal with a velocity of 9.905 m/s. What is the maximum range the ball reaches. (A) 96.2361 m (B) 9.81 m (C) 1 m (D) 10 m 9. A rescue airplane flying at a height of 500 m from ground for a flood affected area drops a rescue kit traveling at 200 m/s. How much distance does the airplane travel from the point of releasing the kit to the point of the kit hitting the ground. (Neglect air resistance) (A) 20.387 km (B) 20.387 m (C) 2.0192 m (D) 2.019 km 10. A thin solid circular disc of 10 kg is applied by a torque through a shaft connecting at the center of the disc. If the angular velocity reached is 5 rad/sec what is the amount of angular impulse acted upon the circular disc. (take r = 4 m and initially the disc is at rest). (A) 200 Nms (B) 400 Nms (C) 500 Nms (D) 1000 Nms 11. The angular speed of the seconds hand in a clock in rad/min is (A) 30 p (B) 120 p (C) 2 p (D) 60 p 12. Determine the point of action of the resultant of forces acting on the inclined plane as shown in figure. 30 N 50 N A B AB = 40 mm 30° (A) 20 mm from A (B) 20 mm from B (C) 25 mm from A (D) 25 mm from B 13. A car is moving along a straight road according to the equation x = 4t 3 + t + 7, where x is in meters and t is in seconds. What is the average acceleration during the fifth second? Engineering Mechanics Test 2 Number of Questions 35 Time:60 min.

3.14 | Engineering Mechanics Test 2 (A) 108 m/s2 (B) 109 m/s2 (C) 110 m/s2 (D) 112 m/s2 14. The magnitude of the force of friction between two bodies, one lying above the other depends upon the roughness of the (A) upper body (B) lower body (C) both the bodies (D) the body having more roughness 15. If the sum of all the forces acting on a body is zero, then the body may be in equilibrium provided the forces are (A) Parallel (B) Concurrent (C) Coplanar (D) Unlike parallel 16. Two sphere of same radius of 100 mm and same mass of 0.5 kg are in equilibrium within a smooth cup of radius 300 mm as shown in the figure. Reaction force between the cup and one sphere will be O A C R R Sphere cup D B (A) 5.14 N (B) 4.91 N (C) 6.41 N (D) 5.66 N 17. A A-Frame is shown in the given figure. Floor reaction at A and vertical pin reaction at D are respectively. 4 cm B C D 5 cm 60 45° ° 12 cm 9 cm 180 A E (A) 757 N and 1142 N (B) 612 N and 1013 N (C) 757 N and 1241 N (D) 612 N and 1142 N 18. A block of mass 50 kg is placed on an inclined surface, as shown in the figure. Coefficient of friction between block and surface is 0.3. Find the value of force ‘P’ required to be applied on the block to maintain uniform velocity of 5 m/s? 5 m/s 6 N 10° 20° P 30° (A) 612 N (B) 574 N (C) 438 N (D) 451 N 19. Particle A of mass ‘m’ is tied with 2 m cord at the instant shown in figure. At this instant angular velocity is 2.83 rad/sec. What will be the angular velocity (in rad/sec) when the angle turned by cord is 45o ? O A 30° Cord (2m) (A) 2.54 (B) 2.83 (C) 2.62 (D) 2.93 20. A 200 mm diameter pulley on a generator is being turned by a belt moving with 25 m/s and accelerating with 8 m/s2 . A fan with an outside diameter of 300 mm is attached to the pulley shaft. Linear acceleration of the tip of the fan (in m/s2 ) is (A) 12 (B) 3475 (C) 5671 (D) 9375 21. A homogenous cylinder of radius ‘R’ and mass ‘m’ is acted upon by a horizontal force ‘P’ applied at various positions along a vertical centre line as shown in the figure. Assume movement upon a horizontal plane. At what radius above the centre (h) should the force ‘P’ is applied so that the frictional force ‘F’ is zero? mg h y P R R F

Engineering Mechanics Test 2 | 3.15 (A) R (B) 3 R (C) 2 R (D) 4 R 22. Figure shows the line diagram of connecting rod AB of a slider crank mechanism. I is the instantaneous center of rotation of the rod. AI = 1.2 m BI = 1.6 m AB = 2 m 30° 60° VA = 12 m/s VB B I A Relative velocity of A and B is (A) 20 m/s (B) 18 m/s (C) 16 m/s (D) 14 m/s Statement for linked answer questions 23 and 24: 2 cm (radius) O 40 kg 730 N/m A 50 kg cylinder of radius 0.4 m rolls without slipping under the action of an 40 kg force. A spring is attached to a cord that is wound around the cylinder. The spring is streched when the 40 kg force is applied. 23. When the cylinder is moved by 0.15 m then the total work done will be (A) 33 N-m (B) 26 N-m (C) 92 N-m (D) 59 N-m 24. What is the speed of the center of the cylinder after it has moved 0.15 m? (A) 0.613 m/s (B) 0.921 m/s (C) 0.833 m/s (D) 0.731 m/s 25. A uniform chain of mass 10 kg and length 1 m lies on a smooth table such that one-fourth of its length is hanging vertically down over the edge of the table. Work done to pull the hanging part of the chain on the table is (A) 3.065 J (B) 12.625 J (C) 24.525 J (D) 6.131 J 26. A homogenous sphere of mass 1 kg is attached to the bar of negligible mass. In the horizontal position shown in the figure, the angular acceleration of the system (in rad/s2 ) is Bar O 0.5 m 0.2 m G sphere (A) Zero (B) 11.13 (C) 16.17 (D) 19.67 27. ABCD is a square which forms a plane truss with load P at point A. What is the axial force in the bar 1. B 1 D A P C 5 2 4 6 3 (A) 2 P (B) P (C) 0.707 P (D) 1.414 P 28. A stool rests on a smooth horizontal floor and is loaded with a load P. What is the value of a to have maximum shear force at the point E. α × a P A B a C E a/2 a/2 D (A) 0 (B) 1 (C) 0 or 1 (D) None of these 29. A body of mass m is suspended by a string of length L. The body traces a horizontal circle of radius 2 m when the semicone angle at the top is 30o . If the centrifugal force of the non-suspended body of same mass is 23 N while rotating in a circle of radius 4 m with the same angular velocity, what is the tension in the string. (A) 23 N (B) 11.328 N (C) 19.91 N (D) 11 N 30. During the replacement of machines in a workshop floor a nail was protruding 5 mm from the floor level. A hammer of 5 kg mass of the head is used to strike the nail to make it level with the floor. Consider the hammer as a free fall from a height of 100 mm and completes the job in single strike. What is the mass of

3.16 | Engineering Mechanics Test 2 the nail if the resistance offered by the floor is 1.032 N/mm (A) 45.87 gms (B) 98.77 gms (C) 198.77 gms (D) 99.385 gms Common data for questions 31 and 32: p R W φ α φ = 30° W = 30 N A block of 30 N weight is being pulled by a force P making an angle a with the horizontal. The reactive force R makes an angle of 30o with the vertical (angle of friction, f). 31. What is the minimum force Pmin required to impede the block to move (A) 15 3 N (B) 30 N (C) 45 N (D) 15 N 32. What is the value of a interms of the angle of friction when Pmin is acting on the block. (A) 2 f (B) f/2 (C) f (D) 90 – f 33. Two balls of weights 6 N and 2 N are made to collide with each other. The velocities of the balls before collision are 4 m/s and 8 m/s respectively and the 2 N ball is moving in opposite direction to 6 N ball. What is the ratio of velocities of the 6 N ball, after the collision, when the impact is considered to have a coefficient of restitution of 0.5 to when the impact is perfectly elastic. (A) 1 (B) 0.25 (C) 0.5 (D) 0.75 34. The wheel of a trolley bag which is being pulled by a force of 10 N (horizontal force) is of 50 mm radius. If the weight of the bag is 100 N what is the coefficient of rolling resistance in meters W P (A) 0.1 (B) 0.005 (C) 0.5 (D) 0.001 35. A plane truss is loaded as shown in the figure. What is magnitude of force in the member CD and is it in compression or tension 500 N 20√3 100 N B D 20 20 C A 20 E 100 N (A) 692.82 N, compression (B) 692.82 N, Tension (C) 1385.64 N, Compression (D) 1385.64 N, Tension Answer Keys 1. B 2. C 3. D 4. D 5. B 6. D 7. C 8. D 9. D 10. A 11. C 12. C 13. A 14. C 15. B 16. D 17. A 18. C 19. A 20. D 21. C 22. A 23. B 24. C 25. A 26. C 27. B 28. C 29. A 30. C 31. D 32. C 33. B 34. B 35. B Hints and Explanations 1. Resultant force, R = 2 2 2 Cos 1 2 12 F F FF + + q or = + +× × × 2 2 o R 500 600 2 500 600 Cos60 R = 953.94 N ~ 954 N Choice (B) 2. Angular acceleration, a = o t w w− or a = 2 240 0 60 2 1.57 rad sec 16 p × −       = now Tangential acceleration, at = r × a = 0.2 × 1.57 = 0.314 m/s2 Choice (C) 3. F = w × = 2 2 m 5 10 r 2 ⇒ F = 250 N Choice (D) 4. For sphere, I GS = 2 2 5 mR For cylinder, IGC = 2 2 mR

Engineering Mechanics Test 2 | 3.17 2 2 2 5 4 5 2 GS GC mR I I mR ∴= = Choice (D) 5. ( ) 0.25 40 9.81 P e × p = × ∴ P = 860.64 N Choice (B) 6. acceleration = dV dt ∴ Velocity is constant ∴ acceleration = 0 Choice (D) 7. 1 1 2 2 2 2 H L mV mV    =       VH = VL Now momentum = m × V ∴ Both has same mass and velocities ∴ Momentum is same for both. Choice (C) 8. Range of a projectile is given as 2 0 Sin2 V g a It is maximum when a = 45o 2 2 0 9.905 98.109025 max 9.81 9.81 V R g ∴ == = Rmax = 10.00091 ≃ 10 m Choice (D) 9. 200 m/s 500 m R Time taken for the kit to reach ground = 2 500 9.81 × = 10.096 sec Distance travelled by air plane in the time = 200 × 10.096 = 2019.2 m = 2.0192 km Choice (D) 10. Angular impulse = I(w2 – w1 ) I = 2 2 2 10 4 40 kg m 2 2 mr × = = Angular Impulse = 40 × (5 – 0) = 200 Nms Choice (A) 11. A seconds hand rotates 2p radians in 60 seconds i.e., in 1 minute. ∴ wsec = 2 rad rad 2 sec min 60 p = p Choice (C) 12. 30 50 30° A B x R P AB = 40 mm 30 2 √3 50 2 √3 Let the point P be the point of action of the resultant Taking moments around P. ΣMP = 0 ( ) ( ) 30 50 3 3 40 0 2 2 ×− × −= x x ( ) 30 50 3 3 40 80 200 25 mm 2 2 ⇒ = − ⇒ = ⇒= x xx x Choice (C) 13. V = 2 12 1 dx t dt = + 24 dV a t dt = = ∴ change in acceleration during 5th second is ( ) ( ) 5 4 12 25 1 12 16 1 1 1 V V V a t ∆ −    × +− × +    == = ∆ a = 108 m/s2 Choice (A) 14. Choice (C) 15. Choice (B) 16. Free body diagram of sphere A R RB (0.5 × 9.81) N 60° 90° 120° From geometry: OC = 300 mm, OA = 200 mm, OB = 200 mm, AB = 200 mm ∴ ∆OAB is equilateral triangle. ∴ Applying Lami’s theorem ( ) ( ) o o 0.5 9.81 Sin 90 Sin 120 R × = ⇒ R = 5.6638 N Choice (D)