Gravity and Gravitation | 201
4. What will be the time period of satellite moving around the earth if the radius of revolution is increased by
one and half time.
( )2 2/3 ( )2 1/3 ( )3 3/2 ( )3 2/3
a. 3 b. 3 c. 2 d. 2
5. If radius of earth shrinks by 1%, mass being unchanged; the value of g
a. decreases by 1% b. increases by 1%
c. increases by 0.5% d. increases by 2%
6. A satellite is moving round the earth in a circular orbit of radius 8000 km with a velocity of 800ms–1. The
acceleration due to gravity on the satellite is nearly
a. 0.04 ms–2 b. 0.06 ms–2
c. 0.08ms–2 d. 0.01ms–2
7. Two planets have radii r1 and r2 and density d1 and d2 respectively. The ratio of acceleration due to gravity
on them will be
a. r1d2 : r2d1 b. r1d1 : r2d2 c. r12d1 : r22d2 d. r1d12 : r2d22
8. let G be the gravitational constant and R be the radius of the earth. If the radius of earth becomes half,
then the new value of gravitational constant will be
a. G b. 2G c. RG d. G
2
9. The acceleration due to gravity at the equator is g. It's value at the pole is
a. less than at equator b. greater than g
c. less than g d. none of the above
10. The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is v. For a
satellite orbiting at an altitude of half the earth's radius, the orbital velocity is
3 3 2 2
a. 2 v b. 2 v c. 3 v d. 3 v
11. A small particle is held inside an isolated hollow sphere at distance x from the centre of sphere. The
gravitational force on that particle by the sphere is
a. directly proportional to x b. inversely proportional to x2
c. inversely proportional to x d. zero
12. Two planets of equal density having radii R and 2R. Then the ratio of acceleration due to gravity on their
surface will be
a. 1 : 1 b. 1 : 2 c. 2 : 1 d. 4 : 1
13. At what depth below earth surface, value of acceleration due to gravity becomes g . [If R = radius of the
4
earth]
2 R R 3
a. 3 R b. 2 c. 4 c. 4 R
RR
14. The ratio of values of 'g' at height 2 to that at depth 2 is (R = radius of earth)
a. 1 : 1 b. 3 : 4 c. 8 : 9 d. 9 : 8
15. Two satellites are orbiting around the earth in a circular orbit. One of them is 100 times heavier than the
other, than the ratio of their periods of revolution is
a. 1 : 100 b. 100 : 1 c. 1 : 10 d. 1 : 1
16. How much velocity should be given for a body if it is to be projected vertically upward from earth's surface
to reach a height of 10R?
GM 2GM 20GM 21GM
a. R2 b. R c. 11R d. 12R
202 | Essential Physics
R
17. A body of mass 'm' is taken to a height h = 5 from earth's surface. The increase in P.E. is
a. mgh 4 5 6
b. 5 mgh c. 6 mgh d. 7 mgh
18. A body falls on the surface of earth from infinite height. It will strike the earth with velocity of
a. infinity b. 0 ms–1 c. 8ms–1 d. 11.2 kms–1
19. If a missile launched with velocity less than escape velocity, then the sum of its KE and PE is always
a. positive b. zero c. negative d. arbitrary
20. The distance of the centres of moon and earth is D. The mass of the earth is 81 times the mass of the
moon. At what distance from the centre of the earth, the gravitational force will be zero?
D 2D 4D 9
a. 2 b. 3 c. 3 c. 10 D
21. A body is taken to a height equal to radius of the earth and is projected towards earth with 1 th of escape
4
velocity (ve), what will be the speed reaching the surface of earth?
a. ve 3 c. ve 1
2 b. 4 ve d. 4 ve
22. Two planets are revolving around the sun. Their mean orbital radii are r1 and r2. Then the ratio of their
time period T1 is equal to:
T2
( )a. r1 ( )b. r1 3/2 ( )c. r2 3 ( )d. r2 3/2
r2 r2 r1 r1
23. The tidal waves in the sea are due to
a. sun's attraction b. moon's attraction
c. earth's atmosphere d. attraction of venus
24. The force of gravitation of black hole is
b. repulsive b. electrostatic
c. conservative c. non–conservative
25. The escape velocity on earth (radius R, mass m) is 11.2 km/s. The escape velocity on another planet of mass
mR
4 and radius 2 in km/s is around
a. 11.2 b. 22.4 c. 15.8 d. 8
Answer Key
1. b 2. d 3. b 3. b 5. d 6. c 7. b 8. d 9. b 10. d
11. d 12. b 13. d 14. c 15. d 16. c 17. c 18. d 19. c 20. d
21. b 22. b 23. b 24. c 25. d
Ch apter Equilibrium
8
Teaching Manual Physics Grade – XI, Higher Secondary Education Board
Curriculum and Training Division Sanothimi, Bhaktapur
Syllabus:
Equilibrium– Moment of forces; Torque; Torque due to a couple; Center of mass; Center of
gravity; Conditions of equilibrium.
Objectives:
The objectives of this sub unit is to make the students able to understand the equilibrium of
forces.
Activities (Micro syllabus):
1. Define coplanar forces. State and prove Lami's theorem.
2. Define moment of force.
3. State and explain principle of moments
4. Define parallel force, couple and torque due to couple.
5. Define center of gravity and center of mass of a body.
6. Explain the conditions for an extended body to be in equilibrium.
7. Numerical problems: Focused numerical problems given:
(a) As exercise in the University Physics (Ref. 1) and
(b) In advanced level Physics (Ref.2) and also in Physics for XI (Ref. 3).
204 | Essential Physics
8.1 Introduction
When a number of forces are acting on a rigid body and the net force on the body is zero, then the body is
said to be in linear equilibrium. If the body is at rest under the application of a number of forces, then
the body is said to be in static equilibrium. If the body is moving with uniform velocity under the action
of a number of forces, then the body is said to be in dynamic equilibrium.
8.2 Lami's Theorem
This theorem states that if a body is in equilibrium under the action of three forces, then each of the
forces is proportional to the sine of the angle between the other two forces.
Q αC
α R
γ Q
β P A π–β π–γ γ
R
P B
(a) β
(b)
[Fig.8.1, Lami's theorem]
Let P, Q and R be the three forces acting on a body as shown in figure 8.1 (a). Let α, β and γ be the angle,
between Q and R, R and P, and P and Q respectively. Then from Lami's theorem,
P QR
sin α = sin β = sin γ
Proof: Let us draw a triangle ABC in such a way that the sides AB, BC and CA represent the forces P, Q
and R respectively as shown in figure 8.1(b). From figure,
∠ BAC = π – β, ∠ ABC = π – γ and ∠ ACB = π – α
Now, using sine law in ∆ ABC,
abc
sin A = sin B = sin C
BC CA AB
or, sin (π – β) = sin (π – γ) = sin (π – α)
QR P
or, sin β = sin γ = sin α
∴ P QR
sin α = sin β = sin γ
This proves the Lami's theorem
8.3 Moment of a Force
The point on a body at which the force is applied is called the point of action of the force. A line passing
through the point of action of the force and parallel to the direction of force is called line of action of the
force. A line about which the body rotates is called axis of rotation of the body. The perpendicular
distance between the line of action the force and the axis of rotation is called moment arm.
Equilibrium | 205
The turning effect of a force is called moment of the force and it is measured as the product of force
applied and perpendicular distance of the line of action of force from the Point of rotation
axis of rotation (moment arm). It is also called torque and denoted by τ.
Mathematically,
Torque = Force applied × Moment Arm
or, τ = F × r = rF . . . .(1) O
Torque is a vector quantity. In vector form, the torque can be expressed as, rF
P
→τ = →r × → = rF sinθ n^ . . . (2)
[Fig 8.2, Moment of a force]
F
Where θ is angle between →r and → , n^ is a unit vector along the direction
F
of →τ . The direction of torque (→τ ) is determined by the right hand rule.
When θ = 90°, τ = rF = maximum torque A x1 O x2 B
When θ = 0°, τ = 0 = minimum torque
Its S.I. unit is Nm and its dimensional formula is [ML2T–2]. F1 F2
The moment or torque on a body is said to be clockwise if it rotates the Anticlockwise Clockwise
body in clockwise direction and anticlockwise if it rotates the body in
anticlockwise direction. [Fig. 8.3, Clockwise and
Anticlockwise moments]
Principle of Moments
It states that if a body is in equilibrium under the action of a A BC OD E
number of forces, then the sum of clockwise moments is
equal to the sum of anticlockwise moments. That is,
Sum of clockwise moments = Sum of anticlockwise F2 F4 F5
moments, F1 F3
Suppose a rigid rod or meter scale AE is equilibrium under [Fig. 8.4, Moments of forces acting at
the action of forces F1, F2, F3, F4 and F5 acting at points A, B, different points of a meter scale]
C, D and E respectively as shown in figure 8.4. Here, the rod
is supported at point O. The forces F1, F2 and F3 produce
anticlockwise moments about point O and F4 and F5 produce
clockwise moments about the point O. Thus,
Sum of anticlockwise moments = F1× AO + F2 × BO + F3 × CO
Sum of clockwise moments = F4 × DO + F5 × EO
Now, from principle of moments,
Sum of anticlockwise moments = Sum of clockwise moments
or, F1 × AO + F2 × BO + F3× CO = F4 × DO + F5 × EO
or, F1× AO + F2 × BO + F3 × CO – F4 × DO – F5× EO = 0
This shows that if a body is in equilibrium under the action of a number of forces, then the algebraic sum
of moments of all these forces about any point is zero. It should be noted that if anticlockwise moments
are taken as positive, then the clockwise moments should be taken as negative and vice–versa
206 | Essential Physics
8.4 Coplanar and Parallel Forces
The forces are said to be coplanar if the lines of F4
action of all the forces lie on a same plane. If the
lines of action of forces acting on a body are F1
parallel, then forces are called parallel forces. If the F2
parallel forces are directed in same direction, then
they are called like parallel forces. Two parallel (a) Like parallel forces F3
forces are said to be unlike parallel forces, if they (a) Unlike parallel forces
are directed in opposite direction.
[Fig. 8.5, Parallel forces]
8.5 Couple and Torque Due to a couple
Two equal and unlike parallel forces acting at two different points of a rigid body form a couple. A
couple always produces rotational motion on a body. The moment or torque due to a couple is defined as
the turning effect produced by the couple and it is measured as the sum F
of moments of the two forces of the couple.
Let both the forces of a couple produce clockwise moments on the rod P Q
PQ about a point O as shown in figure 8.6. Then, the torque of the O
couple or moment of the couple about point O is
F
τ =moment of force F at P about O + moment of force F at Q about O
= F × PO + F × OQ =F × (PO + OQ) =F × PQ [Fig. 8.6, Two equal and unlike
∴ τ =F × PQ = Force × Arm of the couple parallel forces]
Here, PQ is the perpendicular distance between the two forces of the couple called arm of the couple.
Hence, torque due to a couple is equal to the product of one force of the couple and arm of the couple.
Some examples of couple are the opening of a door or a window, turning a cork crew, opening the cap of
a medicine bottle, turning on or off the cap of a pen, turning on or off a water tap etc.
8.6 Centre of Gravity (C.G.)
A point with respect to a body where the whole weight of the body (i.e. sum of weights of each particles
that constitute the body) is supposed to be concentrated and the total gravitational torque on the body
about the point is zero, is called centre of gravity (C.G.) of the body.
A rigid body can be supposed to be made up of large number of C.G.
particles each having certain mass. All the particles of the body are m1g
attracted by the earth towards the centre. Thus, the weight of each
particle is directed towards, the centre of earth. All these forces m2g
(weights) are parallel forces and resultant force (weight) can be
obtained by algebraic sum of the individual forces (weights). Therefore, m3g mng
the C.G. of a body is a point through which the resultant of individual W= Mg
forces (weights) of all the particles of the body passes.
[Fig.8.7, Centre of gravity]
Let us consider a rigid body of mass M consists of n particles of masses
m1, m2, m3, . . ., mn respectively as shown in figure 8.7. The forces due
to gravity on these masses (weights) are m1g, m2g, m3g, …….., mng
respectively, all acting vertically downward. The resultant downward
force (resultant weight) on the particle is,
Equilibrium | 207
W = m1g + m2g + m3g + . . . + mng
or, W = (m1 + m2 + m3 + . . . + mn)g
∴ W = Mg, where M = m1 + m2 + m3 +. . . + mn = mass of rigid body.
Characteristics of C.G. of a Body
i. A rigid body remains in equilibrium when supported through its C.G.
ii. The position of C.G. of a body depends upon its size and shape.
iii. The C.G. of a body is independent on its position.
iv. The C.G. of a body may or may not lie within the material of the body. For example, the C.G. of a
solid sphere lies at its geometric centre where there is matter of the sphere. The C.G. of a hollow
sphere also likes at its geometric centre where there is no matter of the sphere.
v. For regular shaped bodies, the C.G. lies at their geometric centres.
vi. The position of C.G. of a body changes with the change in shape of the body.
The following table shows the position of C.G. of different shape of bodies.
S.N. Shape of Body Position of C.G.
1. Circular ring. At the geometric centre.
2. Circular disc. At geometric centre.
3. Solid sphere, hollow sphere and angular disc. At the centre.
4. Uniform rod or bar At the middle point.
5. Hollow or solid cylinder. At the middle point of its axis.
6. Rectangular or cubical block At the point of intersection of diagonals
7. Parallelogram At the point of intersection of diagonals.
8. Triangular lamina At the point of intersection of medians.
9. Square lamina or rectangular lamina At the point of intersection of diagonals.
10. Pyramid or cone
( )1 th
At a distance equal to 4 of the line joining the
centre of the base with its apex from the base.
8.7 Centre of Mass (C.M.)
A point with respect to a body at which the whole mass of the body is supposed to be concentrated, is
called its centre of mass. The applied force on centre of mass of a body produces linear acceleration but
not rotation.
In translational motion, when a force is applied, each point on a rigid body experiences the equal linear
displacements so that the motion of the whole body can be represented by a single particle. But, in case of
rotational motion, the different points of a body have different linear displacements. These all points of
the body rotate with the same angular velocity except one point on the body which does not rotate and
possesses linear motion only. This point is called the centre of mass of the body.
Calculation of Centre of Mass
Y
x1 x2 x2 c.m. x
c.m. x1 M
m1 O O m2
m1
m2
xcm
(a) (b)
[Fig. 8.8, Centre of mass between two bodies]
208 | Essential Physics
For two body system, let m1 and m2 be the masses of two bodies and c.m. be their centre of mass. Let
origin O also lies at C.M. as shown in figure 8.8 (a). Then, in equilibrium,
F1x1 = F2x2
Where x1 and x2 are the distances of m1 and m2 respectively from C.M. (i.e. origin).
or, m1gx1 = m2gx2
∴ m1x1 =m2x2 . . . (1)
This shows that greater the mass of the body, nearer is their C.M. and vice–versa.
Again, suppose m1 and m2 lie along x–axis, and c.m. and origin 'o' don't consider at a point as shown in
figure 8.8(b). If x1, x2 and xcm be the respective distances of m1, m2 and C.M. from the origin O, then
m1x1 + m2x2 = (m1 + m2)xcm
∴ xcm = m1x1 + m2x2 . . . (2)
m1 + m2
Similarly, if m1 and m2 lie along y– axis, then we can write Y
ycm = m1y1 + m2y2 . . . (3) x2 m2
m1 + m2
For a system consisting of n particles, let m1, m2, m3, ….., mn xn mn
be the masses of n – particles. Let (x1, y1), (x2, y2), ……....., c.m.
(xn, yn) be the co–ordinates of masses m1, m2, …..........…., mn xcm
respectively as shown in figure 8.9. Now, if (xcm, ycm) be the m1
x1
co–ordinate of C.M., then
m1x1 + m2x2 + ..... + mnxn Σi mixi yn
m1 + m2 + ......... + mn M
xcm = = y1 y2 ycm
m1y1 + m2y2 + ..... + mnxn Σmiyi O X
m1 + m2 + ......... + mn [Fig. 8.9, Calculation of centre of mass]
And, ycm = = i
M
Where, M = m1 + m2 + . . . + mn is the total mass of the body
For a small sized body and having uniform density, the C.G. and C.M. of the body coincide. For a
homogeneous sphere, both C.G. and C.M. lie at its geometric centre. But, for a body having very large
dimensions and non–uniform density the C.G. and C.M. don't coincide at a point.
8.8 Conditions of Equilibrium of Rigid Bodies
A rigid body is said to be in equilibrium under the action of a number of forces if no acceleration of
translation and no acceleration of rotation are produced on the body. Thus, there are two types of
equilibrium of a rigid body–translational equilibrium and rotational equilibrium.
Translational Equilibrium
A rigid body is said to be in translational equilibrium if the vector sum of all the forces acting on the body
is zero. That is, ΣF = 0
In terms of components, ΣFx = 0, ΣFy = 0 and Σ FZ = 0
Here, ΣFx = 0 means that all the components of forces, along the x–axis add to zero. Same is force along
Y and Z axes. Now, from Newton's second law, ΣF= ma
For translational equilibrium, ΣF = 0
∴ 0 = ma
or, a = 0 [since, m ≠0]
or, v = 0 or constant
Therefore, in translational equilibrium, the body should be either at rest or moving with uniform linear
velocity.
Equilibrium | 209
Rotational Equilibrium
A rigid body is said to be in rotational equilibrium if the vector sum of all the external torques about an
axis is zero. That is
Στ=0
Since, Στ = Iα
∴ 0 = Iα
or, α = 0
or, ω =0 or constant
Therefore, in rotational equilibrium, the body should be either at rest or rotating with uniform angular
velocity.
8.9 States of Equilibrium
There are three types of states of translational equilibrium of a rigid body.
1. Stable Equilibrium : A rigid body is said to be in stable equilibrium if it returns to its original
equilibrium position when it is slightly displaced and released as shown in figure 8.10(a). In stable
equilibrium position, the C.G. lies low and C.G. at the displaced position is at higher position than
the original equilibrium position.
C.G. C.G. C.G.
(a) Stable Equilibrium (b) Unstable Equilibrium (c) Neutral Equilibrium
[Fig. 8.10, States of Equilibrium]
2. Unstable Equilibrium : A rigid body is said to be in unstable equilibrium position if it doesn't
return to its original equilibrium position when it is slightly displaced and released as shown in
figure 8.10(b). In unstable equilibrium position, the C.G. of the body lies at its upper position. But,
when it is slightly displaced from equilibrium position, then the C.G. lies at lower position than
before.
3. Neutral Equilibrium : A body is said to be in neutral equilibrium position if it doesn't return to
the original equilibrium position but comes to a new equilibrium position which is similar to the
original equilibrium position when it is slightly displaced and released as shown in figure 8.10 (c).
In neutral equilibrium position, the height of C.G. of body remains at the same height from the
base in all displaced positions.
Conditions of a Body to be in Stable Equilibrium
[Fig. 8.11, Conditions of stable equilibrium]
210 | Essential Physics
1. The C.G. of the body should lie as low as possible
When the C.G. of a body lies close to its base, then the body is more stable. Due to this reason, the
bottom of a ship is made heavy and the cargo is loaded at the base. This lowers the C.G. in the ship
and becomes more stable. Similarly, a truck loaded with heavy materials like iron rod is more
stable than a truck loaded with light materials like straw or cotton clothes. To lower the C.G. in a
loaded truck, the heavy goods are loaded at the bottom and lighter goods are loaded at the top.
2. The base area of the body should be as large as possible
When the base area of a body is large, then the body is more stable. Due to this reason, a cow is
more stable than a man. This is because a cow with four legs has more base area than a man with
two legs. A man standing on two legs is more stable than a man standing on one leg, a four
wheeler vehicle (bus or car) is more stable than a two wheeler vehicle (cycle), etc.
3. Vertical line passing through the C.G. should lie within the base of the body in displaced
position
When a vertical line passing through C.G. of a body on displaced position lies within its base, the
body is more stable. Due to this reason, a man carrying a load on his back leans forward so that a
vertical line passing through its C.G. lies within the base area of the body provided by his feet.
Similarly, a man carrying a bucket of water in its hand leans opposite to the bucket, a pregnant
woman leans backward etc. to keep a vertical line passing through C.G. within the base area.
Boost for Objectives
• The centre of mass of a body may or may not lie within the material of the body.
• If a body falling vertically downward under gravity breaks into pieces, then the C.M. of particles lies on the
same vertical line.
• Location of C.M. remains unchanged in rotatory motion but changes in translatory motion.
• The nature of motion of C.M. depends on external force and is independent of internal forces.
• Sum of moments of mass of particles about C.M. is zero.
• In the absence of external force, velocity of C.M. remains constant.
• If a shell moving on parabolic path explodes, the pieces fly in such a way that C.M. of all pieces continue
moving on the same parabola on which that of the shell was moving.
• The position of C.M. of a system of particles depends on mass, position and relative distance between the
particles and is independent of forces on particles.
• For stable equilibrium, the centre of gravity should be at lowest position.
• The centre of mass of a thin triangular plate lies on its centroid.
• When CG rises, body becomes unstable.
• Centre of mass and geometrical center of the body may not coincide.
→→
• Velocity of centre of mass of particles of masses m1 and m2 moving with speeds v1 and v2 is given by ,
→ = → + → . Acceleration of centre of mass is, → = → + →
vcm .m1mv11 + m2 v2 acm m1 a1 + m2 a2
m2 m2
m1
Short Questions with Answers
1. Does the center of gravity of a solid body always lie within the material of the body? If not, give a
Ans:
counter example. [HSEB Sample Question, 2072, C, 2070, 2069, 2073 C]
2.
Ans: No, it is not necessary. The centre of gravity of a solid body may or may not lie within in the material of
the body. For example, the centre of gravity of a circular ring, hollow sphere, football etc. lies at their
geometrical centre where there is no matter.
Why a wrench of longer arm is preferred more? [HSEB Sample Question]
The turning effect of force is called moment of the force or torque and it is measured as the product of
force applied and perpendicular distance of the line of action of force from the axis of rotation
Mathematically, Torque = Force applied × Moment arm
Equilibrium | 211
or, τ = F×r
So, if a force is applied to a wrench of longer arm, then torque is large, which makes easier to loosen or
tighten the knot. Hence, wrench of longer arm is preferred more.
3. A handle or a knob is fixed at the free end of the door. Explain why? [HSEB 2072 D]
Ans:
The torque or moment of a force is, Torque = Applied force × Moment arm
4.
Ans: or, τ = F × r
5. The moment arm is the perpendicular distance between the line of action of applied force and the axis
Ans:
6. of rotation. Moment arm (r) is maximum when a handle or a knob is fixed at the free end of the door
Ans:
and hence even a small applied force can produce large turning effect (torque) and it becomes easy to
7.
Ans: open the door.
8.
State the conditions of equilibrium of a system of coplanar forces. [HSEB Model question]
Ans:
9. The forces are said to be coplanar if the lines of action of all the forces lie on a same plane. A particle is
said to be in the conditions of equilibrium, when
→→ → →
i. The vector sum of all the forces acting on the particle is zero i.e., F1 + F2 + F3 + ....... Fn = 0
and
ii. The vector sum of moments of all the forces acting on the particle is zero, i.e.
→→→ →
τ1+ τ 2 + τ 3 + ........τ n = 0
Mention the conditions of stable equilibrium. [HSEB 2071 C]
The followings are the conditions for a body to be in stable equilibrium.
a. The C.G. of the body should lie as low as possible.
b. The base of the body should be as large as possible.
c.. A vertical line passing through C.G. should lie within the base of the body on displaced position.
What is the difference between centre of mass and centre of gravity of a system? [HSEB 2071 S, 2051]
The centre of gravity (C.G.) of a body is defined as a point where the whole weight of the body is
supposed to be concentrated and the total gravitational torque on the body about the point is zero. The
centre of mass (C.M.) of a body is defined as a point where whole mass of the body is supposed to be
concentrated. The applied force on C.M. of a body produces linear acceleration but no rotation.
For a small sized body and having uniform density, the C.G,. and C.M. of the body coincide. For a
homogenous sphere, both C.G. and C.M. lie at its geometrical centre. But, for a body having very large
dimensions and non–uniform density, the C.G. and C.M. don't coincide at a point. For a gravity free
space, C.G. loses its identity but C.M. has still its meaning.
When four–legged animals walk, they always have three of their legs on the ground at any
instant. Explain, why? [HSEB 2070 D]
For a body to be in stable equilibrium the base area of the body should be as large as possible and a
vertical line passing through C.G. of the body should lie within the base area of the body. While
walking, when three legs of animals lie on the gerund, base area increases and animal becomes more
stable. Thus, when four legged animals walk, they always have three legs on the ground at any instant.
A man carrying load on his back leans forward. Why should he do so? [HSEB 2069 S]
OR
Explain why a man carrying a load on his back leans forward. [HSEB 2068, 2053]
OR
A coolie carrying a load on his back leans forward. Why? [HSEB 2069]
For a body to be in stable equilibrium, the vertical line passing through C.G. must pass within the base
area of the body. The C.G. shifts backwards while carrying a load on his back. If he doesn't lean
forward, then a vertical line passing through C.G. falls outside (backwards) and it will be difficult to
carry the load. Therefore, he leans forward to be in stable equilibrium.
A man carrying a bucket full of water on his right hand always leans towards his left hand. Why?
[HSEB 2068 old]
OR
A man carrying a bucket of water on his hand always leans to the opposite side. Exp[lain why?
[HSEB 2058]
212 | Essential Physics
Ans: For a body to be in stable equilibrium, the vertical line passing through C.G. must pass within the base
area of the body. So, to keep vertical line passing through C.G. within the base area provided by his
feet, the man carrying a bucket of water leans to the opposite side.
10. During pregnancy, women often develop back pains from learning backward while walking. Why
do they have to walk this way? [HSEB 2067]
Ans: During pregnancy, when women walk leaning backward, a vertical line passing through C.G. lies within
the base area provided by her feet and she remains in stable equilibrium. If she doesn't lean backward,
then a vertical line passing through C.G. lies in front of her body and she becomes in unstable
condition.
11. Define the terms; couple and moment of couple. [HSEB 2064]
OR
What is meant by the moment of a couple? [HSEB 2053]
Ans: Two equal and unlike parallel forces acting at two different points of a rigid F
body forms a couple. A couple always produces rotational motion on a
body. The torque or moment due to a couple is defined as the turning effect P Q
produced by the couple and it is measured as the product of one force and O
perpendicular distance between the two forces of the couple. F
∴ Moment of couple = One force × Arm of Couple
or, τ = F × PQ
12. Why is horse more stable than a man? [HSEB 2055]
OR
Why is cow more stable than a man? [HSEB 2051]
Ans: For a body to be in stable equilibrium, the base area of the body should be as large as possible. In case
of a horse (or a cow), it has four legs which cover more base area than the small base area covered by
two legs of a man. Therefore, the horse (or a cow) is more stable than a man.
13. Can a body be in equilibrium if it is in motion? Explain. [HSEB 2054]
Ans: Yes. A body will be in translational equilibrium if either it is at rest or moving with uniform velocity.
Similarly, a body will be in rotational equilibrium if either it is at rest or moving with uniform angular
velocity. In other words, a body in equilibrium may be at rest or may have zero translational and
angular accelerations.
14. State principle of moments and give an example [HSEB 2052]
Ans: It states that if a rigid body is in equilibrium under the action of a number of forces, then the algebraic
sum of the moments of all the forces about any point is zero i.e. sum of clockwise moments must be
equal to the sum of anticlockwise moments. For example: a ladder resting on a wall is in equilibrium
under the action of weight of ladder, reactions of ground and wall.
15. Why is the bottom of a ship made heavy? Explain. [NEB 2074, HSEB 2051]
Ans: For a body to be in stable equilibrium, the C.G. of the body should lie as low as possible. So, to make
the C.G. at lower position, the bottom of the ship is made heavy.
16. Is it possible for a solid body to have no matter at its centre of mass?
Ans: Yes. The centre of mass (C.M.) of a body may or may not lie within the material of the body. For
example; the C.M. of a uniform hollow sphere, football, circular ring etc lie at their geometrical centres
where there is no matter.
17. A boat is likely to capsize if persons in the boat stand up explain. [HSEB 2063 S]
Ans: For a floating body to be in stable equilibrium, the metacentre of the body should lie above its C.G. For
this, the C.G. should lie as low as possible. If persons in the boat stand up, then C.G. is raised up and
becomes above the metacentre. In this condition, the boat becomes unstable and hence the boat is likely
to capsize (overturn).
18. Is a body in uniform circular motion in equilibrium?
Ans: No. A body in equilibrium should have zero acceleration. But in uniform circular motion, the body is
under the action of centripetal acceleration towards the centre of circular path. So, the body in uniform
circular motion is not in equilibrium. But, the body is in rotational equilibrium.
19. Can the couple acting on a rigid body produce translatory motion?
Ans: No. A couple always produces rotational motion on a body.
Equilibrium | 213
Numerical Examples
1. A roller whose diameter is 1.0m weighs 360N. What horizontal force is necessary to pull the roller
over a brick 0.1m high when the force is applied at the centre? [HSEB 2050]
Solution:
Diameter of roller (d) = 1.0m
d 1.0 m OB F
Radius of roller (r) = OP = 2 = 0.5m
Weight of roller (W) = 360N.
Height of brick = 0.1m AP
Let F be the horizontal force necessary to pull the roller over 0.1m
the brick.
Now, taking moment about point P,
Clockwise moment about p = Anticlockwise moment about P
or, F × PB = W × PA W
or, F×(0.5 – 0.1) = W × OP2 – OA2
or, F × 0.4 = 360 × 0.52 – 0.42
or, F × 0.4 = 360 × 0.3
∴ 360 × 0.3
F = 0.4 = 270N
2. Two people are carrying a uniform wooden board that is 3.00m long and weighs 160N. If one person
applies an upward force equal to 60N at one end, at what point does the other person lift?
[HSEB Sample question]
Solution:
Given, length of wooden board (l) = 3m
Weight of wooden board (W) = 160N
Upward force at one end (F1) = 60N
Distance of F1 from C.G. = 1.5m
Let another force F2 should be applied at a distance x from the C.G. of the board.
Since, the board is in equilibrium, C.G.
∴ F1 × 1.5 = F2 × x
or, F1 × 1.5 = (W – F1) × x 1.5 m x
or, 60×1.5 = (160 – 60) × x
∴ 60 × 1.5 90 F1= 60 N W= 160 N F2
x = 100 = 100 = 0.9m
Hence, the other person should lift at 0.9m from C.G. or 1.5 + 0.9 = 2.4m from first person.
3. Three odd shaped blocks of chocolate have the following masses and centre of mass co–ordinates:
(a) 0.300 kg, (0.200m, 0.300 m)
(b) 0.400 kg, (0.100m, –0.400m)
(c) 0.200 kg, (–0.300m, 0.600m)
Find the co–ordinates of the center of mass of the system of three chocolate blocks.
[HSEB Sample Question]
Solution:
Given, m1(x1, y1) = 0.300 kg(0.200 m, 0.300 m)
m2(x2, y2) = 0.400kg (0.100 m, –0.400m)
m3(x3, y3) = 0.200kg (–0.300m, 0.600m)
214 | Essential Physics
Position of C.M. = (xcm, ycm) = ?
xcm = m1x1 + m2x2 + m3x3
m1 + m2 + m3
0.300 × 0.200 + 0.400 × 0.100 + 0.200 × (–0.300)
= 0.300 + 0.400 + 0.200
0.060 + 0.040 – 0.060
= 0.900
0.040
= 0.900
= 0.044m
And,
ycm = m1y1 + m2y2 + m3y3
m1 + m2 + m3
0.300 × 0.300 + 0.400 × (–0.400) + 0.200 × 0.600
= 0.300 + 0.400 + 0.200
0.090 – 0.160 + 0.120
= 0.900
0.050
= 0. 900
= 0.056m
∴ Co–ordinates of center of mass of the system = (xcm, ycm) = (0.044m, 0.056m)
Important Numerical Problems
1. Two parallel forces of 100N and 40N act on a body in the same direction from two points A and B at
a distance of 0.4m from each other. Find the magnitude and the point of application of the resultant
of two forces.
Solution:
Let, C be the point at which the resultant force acts and it is at a distance x from point A.
Here, P = 100N, Q = 40N
AB = 0.4m, AC = x, BC = 0.4 – x
The resultant R of two like parallel force is,
R = P + Q = 100 + 40 = 140N P = 100 N R
In case of two like parallel forces, the point C divides, AB A C Q = 40 N
internally in the inverse ratio of the forces. That is, x 0.4 – x B
P BC 0.4 m
Q = AC
or, 100 = 0.4 – x
40 x
or, 100x = 16 – 40x
∴ 16
x = 140 = 0.1143 m
So, the magnitude of the resultant of two forces is 140N which acts at a distance 0.1143 m from force
100N.
2. The meteorites in free space have masses of 10kg and 15 kg respectively. Find the position of their
centre of mass if they are 12m apart. [T.U. 2039]
Solution:
Given, m1 = 10kg, m2 = 15kg
Equilibrium | 215
Distance between m1 and m2 = 12m
Let, x be the distance of C.M. from m1. Then distance of m2 from C.M. is (12–x)
We have, m1x1 = m2x2 m1 C.M. m2
or, 10x = 15(12 –x ) x
or, 10x + 15x = 180 12 – x
∴ 180 12 m
x = 25 = 7.2 m.
Hence, their C.M. lies at 7.2m from 10kg meteorite.
3. The foot of a uniform ladder is on a rough horizontal ground, and the top rests against a smooth
vertical wall. The weight of the ladder is 400N, and a man weighing 800N stands on the ladder one–
quarter of this length from the bottom. If the inclination of the ladder to the horizontal is 30°, find
the reaction at the wall and the total force at the ground. [T.U. 2044, 2055]
Solution: F
Given, weight of ladder (W1) = 400N R B
Weight of man (W2) = 800N 30°
Inclination of ladder (θ) = 30° D
Let, AB = l = length of the ladder
R = reaction at the wall = ? θC
F = total force on ground = ?
Since the ladder is in equilibrium, from principle of moments, A 30°
Sum of clockwise moments = sum of anticlockwise moments [about point A] w1 = 400 N
or, W2 × AC cos30° + W1 × AD cos30° = R × AB sin30°
w2 = 800 N
l3 l3 1
or, 800 × 4 × 2 + 400 × 2 × 2 = R × l × 2
800 3 400 3
or, R = 4 + 2 = 200 3 + 200 3
∴ R = 400 3 N = 692.8 N
Again, let θ be the angle made by F with horizontal. Then,
ΣFx = 0 . . . (1)
or, Fcosθ – R = 0
or, Fcosθ = R
Also, ΣFy = 0
or, Fsinθ – W1 – W2 = 0
or, Fsinθ = W1 + W2 . . . (2)
Squaring and adding equations (1) and (2), we get
F2(cos2θ + sin2θ) = R2 + (W1+ W2)2
or, F2 = (692.8)2 + (400 + 800)2
∴ F = (692.8)2 + (1200)2 = 1385.63 N.
So, the reaction at the wall is R= 692.8N and the total force at the ground is
1385.63N. A 3m B
D 100 N
4. A rectangle plate ABCD has two forces of 100N acting along AB and CD in 5m
opposite directions. If AB = 3m, BC = 5m, what is the moment of the couple or 100 N C
torque acting on the plate ? What forces acting along BC and AD respectively
are required to keep the plate in equilibrium?
Solution:
Given, AB = DC = 3m, BC = AD = 5m
216 | Essential Physics
The two equal and opposite forces (F = 100N) acting on either sides AB and CD form a couple.
∴ Couple or torque acting on the plate is,
τ = F × BC = 100 × 5 = 500 Nm.
Let two equal and opposite forces of magnitude T, are acting along sides BC and DA, to keep the plate
in equilibrium. Then,
τ = T × AB
or, 500 = T × 3
∴ 500
T = 3 = 166.67N
5. Two particles of masses 5kg and 4kg are located at (–6^i + 4^j + 2^k ) and (2^i + 5^j + 13^k )
respectively. Locate the position of centre of mass of the system.
Solution:
→
Let R = position of C.M. of the system
Given, m1 = 5 kg, m2 = 4kg
→ = – 6^i + 4^j + 2^k
r1
→ = 2^i + 5^j + 13^k
r2
Total mass, M = m1 + m2 = 9kg
→→→
We know, (m1 + m2) R = m1 r1 + m2 r2
or, → ∧ + 4^j + 2^k ) + 4 (2^i + 5^j + 13^k )
9R = 5(–6 i
or, → = –22^i + 40^j + 62^k
9R
∴ → = – 292 ^i + 490 ^j + 692 ^k .
R
Practice Short Questions
1. Does the centre of gravity of a solid body always lie within the material of the body? Explain.[HSEB 2070 C]
2. Is it possible for a solid body to have no matter at it centre of gravity? [HEB 2059]
3. Why a wrench of longer arm is preferred in comparison to a wrench of short arm? [HSEB 2065]
4. What is equilibrium?
5. Why does a table lamp have a heavy and large base?
6. The cargo is loaded at the bottom of a ship, why?
7. Why we cannot rise from a chair without bending a little forward?
8. Standing is not allowed in a double decker bus. Why?
9. Why is it difficult to open the door by pulling or pushing it at the hinge?
10. Can three unequal coplanar forces bring the body in equilibrium?
11. What are the factors on which position of centre of mass depends?
12. Can two unequal coplanar forces acting together produce condition of equilibrium?
13. Define like parallel forces and unlike parallel forces.
14. A truck loaded with iron is more stable than a truck loaded with straw, Why?
Equilibrium | 217
Practice Long Questions
1. What do you mean by centre of mass? Derive an expression for centre of mass of a two particle system.
[HSEB 2053]
2. State principle of moments. Explain how you verify the principle of moment in the laboratory. [HSEB 2059]
3. State and prove Lami's Theorem.
4. State parallelogram law of forces. Obtain the expressions for magnitude and direction of the resultant force.
5. State and explain the conditions under which a rigid body remains in equilibrium under the action of a set of
coplanar forces.
Practice Numerical Questions
1. Three masses 1kg, 2kg and 3kg are located at the corners of an equilateral triangle of side 1m. Locate the
centre of mass of the system. [Ans: (0.58m, 0.44m)]
2. Find the torque of a force 2^i + 3^j – 4k^ about the origin which acts on a particle whose position vector is ^i +
2^j – k^. [Ans: –5^i + 2^j – ^k]
3. A 12m long uniform ladder of 200N weight is placed at angle 60° to horizontal with one end leaning against a
smooth wall and the other end on the ground. Find the reactions of the wall and the ground.
[Ans: 57.74N, 2108.17N.]
4. A ball with radius with radius r1 = 0.080 m and mass 1.00 kg in attached by a light rod 0.400 m in length to a
second ball with radius r2 = 0.100 m and mass 2.00 kg. Where is the centre of gravity of this system
[Ans: 0.387 m]
5. Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts one end
with a force of 400 N, and the other lifts the opposite end with a force of 600 N. What is the weight of the
motor and where along the board is its centre of gravity located? [Ans:100N,1.20 m]
6. A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall.
The ladder weights 160 N. The coefficient of static friction between the foot of the ladder and the ground is
0.40. A man weighing 740 N climbs slowly up the ladder. (a) What is the maximum frictional force that the
ground can exert on the ladder at its lower end? (b) What is the actual frictional force when the man is
climbed 1.0 m along the ladder? How far along the ladder can the man climb before the ladder starts to slip?
[Ans: 360 N, 170 N, 2.7 m]
7. A metal bar 70 cm long and 4.00 kg in mass is supported on knife–edges placed 10 cm from each end. A 6.00
kg weight is suspended at 30 cm from one end. Find the reactions at the knife–edges. (Assume the bar to be
of uniform cross section and homogeneous). Given g = 9.8 ms–2. [Ans: 54.88 N, 43.12 N]
MBBS, B.E., B.Sc. Nursing Entrance Preparation Questions
1. The torque due to gravitational force on a body about its centre of mass is:
a. infinite b. finite
c. zero d. can not be measured
2. Two point masses of 1 kg and 2 kg separated by 0.5m constitute a system. The distance of the centre of
mass of the system from 1 kg mass is:
a. 0.15m b. 0.25m
c. 0.33m d. 0.4m
3. If no internal force is applied in a body the velocity of the centre of mass:
a. Zero b. Increases
c. Decreases d. Remains constant
4. The product of moment of inertia and angular acceleration gives,
a. Linear momentum b. Angular momentum
c. Torque d. Force
218 | Essential Physics
5. Two particles of masses m1 and m2 are at distance x apart. Then, their centre of mass lies at distance..........
from m1.
( )a.m1 m1
m2 m1 + m2
x ( )b. x
( )c.m2 m2
m1 m1 + m2
x ( )d. x
6. Out of two particles of masses m1 and m2, the particle m1 is moved through a distance d towards their
centre of mass. What is the displacement of centre of mass?
m1 m2
m1 + m2 m1 + m2
( )a. d ( )b. d
m1 + m2
m1
( )c. d d. None
7. Two masses of 1kg and 2kg are 9m apart and make a two body system. Their centre of mass from 1kg mass
will be at
a. 6m b. 4m
c. 3m d. 2m
Answer Key 5. d 6. a 7. a
1. c 2. c 3. d 4. c
Ch apter Rotational
Dynamics
9
Teaching Manual Physics Grade – XI, Higher Secondary Education Board
Curriculum and Training Division Sanothimi, Bhaktapur
Syllabus :
Rotational Dynamics– Rotation of rigid bodies; Equation of angular motion; Relation between linear and
angular kinematics; Kinetic energy of rotation of rigid bodies; moment of inertia: Radius of gyration.
Moment of inertia of a uniform rod; Torque and angular acceleration for a rigid body; Work and power in
rotational motion; angular momentum; Conservation of angular momentum.
Objectives:
The objective of this sub unit is to make the students able to identify and describe simple rotational
motion.
Activities (Micro syllabus) :
1. Define angular displacement, angular velocity and acceleration of a rigid body rotating about a
stationary axis
2. Derive the kinematics equations for a body rotating about a stationary axis with a uniform angular
acceleration.
3. Derive a relation for kinetic energy of a rotating and rolling rigid body.
4. Define and calculate moment of inertia of a rigid body. Define and explain the meaning of radius of
gyration.
5. Calculate the moment of inertia for a thin rod about an axis passing through the center and
perpendicular to it.
6. Define torque on a body and establish its relation with moment of inertia.
7. Define angular momentum of a body and establish its relation with moment of inertia.
8. Establish a relation between angular momentum and torque.
9 . State and explain principle of conservation of angular momentum.
10. Calculate work done by a torque and power associated with it.
11. Numerical problems: Focused numerical problems given:
(a) As exercise in the University Physics (Ref. 1) and
(b) In advanced level Physics (Ref. 2) and also in Physics for XI (Ref. 3).
220 | Essential Physics
9.1 Introduction
In translational of motion, a body as a whole moves along a straight line or curved line. In this chapter,
we focus our study to another type of motion in which a body turns about an axis, called rotational
motion. A solid body is said to be rigid if the particles in the body are compactly arranged and the
interparticle distance is not disturbed by an external force applied on it. The shape of a rigid body remains
unaltered under the application of any external force. No real body is perfectly rigid. But, for all practical
purposes, all the solid bodies are regarded as rigid bodies. Rigidity is the property of a rigid body by
virtue of which it offers resistance to the external deforming force.
9.2 Rotation of Rigid Bodies
A rigid body can undergo both translational and rotational motion. A rigid body is said to be in
translational motion if each particle of the body has same linear displacement in the equal interval of
time. For example; when a bus is moving, then the passengers and the bus itself are in translational
motion. A rigid body is said to be in rotational motion about a given axis if each particle of the body has
same angular displacement in the equal interval of time. In rotational motion, the different particles of the
rigid body have same angular velocity but different linear velocities. For example; the motion of a wheel
of a moving bus about its axle is rotational motion.
9.3 Equations of Angular Motion
When a rigid body is rotating about an axis, then its each particle moves in a circular path.
Angular Displacement
The angle through which a particle rotates in a certain interval of time is called angular displacement. It is
denoted by θ and is measured in radian in S.I. Unit.
Angular Velocity
The rate of change of angular displacement is called angular velocity.
That is, ds
dθ
Angular velocity = Angular displacement
r
time taken
[Fig. 9.1, Angular displacement]
or, ω = θ2 – θ1 = dθ . . . (1)
t2 – t1 dt
Again, from figure 9.1,
dθ = ds
r
or, ds = rdθ . . . (2)
Where, r = radius of the circular path,
ds= linear displacement in time interval dt
and, dθ = angular displacement in time interval dt.
Now, dividing both sides of equation (2) by dt, we get
ds = r dθ
dt dt
or, v = rω . . . (3)
where, ds = v = linear velocity, and dθ = ω = angular velocity
dt dt
Angular velocity (ω) is measured in rad/sec.
Rotational Dynamics | 221
Angular Acceleration
The rate of change of angular velocity is called angular acceleration.
That is, Angular acceleration = change in angular velocity
time taken
or, α = ω2 – ω1
t2– t1
dω . . . (4)
= dt
or, α = ω2– ω1, where t2 – t1 = t
t
∴ ω2 = ω1 + αt . . . (5)
Which is similar to, v = u + at, in linear motion.
Again,
average angular velocity is, ωav = ω1 + ω2
2
Also,
ωav = θ
t
∴ θ = ω1 + ω2
t 2
or, θ = ω1 + ω2t . . . (6)
2
From equations (5) and (6), we have
θ = ω1 + ω1 + αt t = 2ω12+ αt t
2
or, θ = ω1 + 1 αtt
2
∴ θ = ω1t + 1 αt2 . . . (7)
2
This equation is similar to, s = ut + 1 at2, in linear motion.
2
Now, squaring equation (5), we get
ω22 = ω12 + 2ω1αt + α2t2
or, ω22 = ω12 + 2α ω1t + 21α t2
∴ ω22 = ω21 + 2αθ . . . (8)
This equation is similar to, v2 = u2 + 2as, in linear motion.
222 | Essential Physics
9.4 Moment of Inertia (Rotational Inertia)
The inertia of a body in rotational motion is called moment of inertia. It is the inability of a body to
change its state of rest or state of uniform rotational motion by itself. A body rotating about an axis has a
tendency to be rotating even if a stopping torque is applied to it. Such a
property of a rotating body is called rotational inertia or moment of
inertia. For example, a rotating fan doesn't stop immediately even if the
switch is put off due to rotational inertia. m
Suppose, a particle having mass 'm' is rotating about an axis passing
through a point 'o' which is at a distance r from the particle as shown in F
figure 9.2. If F be the force applied on the particle, then r
F = ma
Where 'a' is the tangential acceleration of the particle O
Also, a = rα
Where, α is the angular acceleration of the particle. Now, torque on the [Fig. 9.2 : Moment of inertia]
particle due to this force F is
τ = r × F = r × ma = r × m (rα)
∴ τ = (mr2)α . . . (1)
For a linear motion, we have,
Force = linear inertia (mass) × linear accleariton.
For a rotational motion, we must have
Torque = Roational inertia (moment of inertia) × angular acceleration
∴ τ=Iα . . . (2)
Comparing equations (1) and (2), we get . . . (3)
I = mr2
Hence, the moment of inertia (I) of a particle about an axis is measured as the product of its mass and
square of its distance from the axis of rotation. In SI unit, I is measured in kgm2 and in CGS unit, it is
expressed in gm cm2. The dimensional formula, of I is, I = M × L2 = [ML2 T0].
9.5 Moment of Inertia of a Rigid Body
Consider a rigid body having mass M is rotating about an axis YOY' with constant angular velocity ω as
shown in figure 9.3. Let the rigid body consists of n particles of masses m1, m2, . . ., mn which are at
perpendicular distances r1, r2, . . ., rn respectively from Y
the axis YOY'. Then the moment of inertia of these n
particles about the axis YOY' are given by I1 = m1r21, m2 r2 m3 M
I2 = m2 r22, . . . In = mnrn2 respectively. Now, the O r3
moment of inertia (I) of the rigid body about the axis
m1 r1 mn
YOY' is equal to the sum of moments of inertia of rn
these all the particles. Therefore, Y'
I = I1 + I2 + . . . + In
or, I = m1r21 + m2 r22 + . . . + mnrn2
n
Σ∴ I = mir2i [Fig. 9.3, Moment of inertia of a rigid body]
i=1
Where mi is the mass of ith particle of the body which is at a distance ri from the axis YOY'. Thus, the
moment of inertia (I) of a rigid body about an axis is the sum of the products of the masses of its particles
Rotational Dynamics | 223
and the squares of their respective perpendicular distances from the axis of rotation. The moment of
inertia (I) of a body depends on the mass of the body and the distribution of its particles. The moment of
inertia of a body is not a fixed quantity. It has different values for different axis of rotation for a body.
The moment of inertia (I) plays the same role in rotational motion as the mass (m) plays in the linear
motion.
9.6 Moment of Inertia of a Uniform Rod
A. About an axis through its centre and perpendicular to its length
Suppose AB be a thin uniform rod having mass M and length l as shown in fig 9.4. Let YOY' be
an axis passing through centre O of the rod and perpendicular to its length about which the M.I.
has to be determined. Consider a small length dx of the rod which lies at a distance x from the
centre O of the rod and has mass dm.
M Y
Here, mass per unit length of the rod = l .
Mass of element dx of the rod, dm = Ml dx l/2 l/2
The small moment of inertia of mass dm about the A O B
axis YOY' is x dx
dI = (dm) x2 = Ml dx x2
∴ dI = Ml x2 dx . . . (1) Y'
[Fig. 9.4, M.I. of a uniform rod.]
The total moment of inertia of the rod AB about the axis YOY' is obtained by integrating equation
ll
(1) from the limit x = – 2 to x = 2 i.e.,
l/2 l/2
∫ ∫I
= dI = –l/2Ml x2dx
–l/2
∫or, l/2 x33–ll/2/2 l83 l83
I = M x2 dx = M = M 2l 3 – – 2l 3 = M +
l l 3l 3l
–l/2
∴ Ml2 . . . (2)
I = 12
B. About an axis through its one end and perpendicular to its length
In this case, the axis lies at one end, say A, and is perpendicular to its length as shown in figure
9.5. Now, the M.I. can be obtained by integrating equation (1) from limit x = 0 to x = l, i.e.
ll
∫ ∫I =
0Ml x2dx Y
dI =
0
∫or, l x330l =
M x2 dx = M M [l3 – 03] AB
I= l l 3l x dx
0
∴ Ml2 . . . (3)
I= 3
Equations (2) and (3) show that the M.I. of a thin Y'
uniform rod about an axis passing through its one
end and perpendicular to its length is four times [Fig. 9.5, M.I. of a uniform rod.]
greater than that of the rod about an axis passing through its centre and perpendicular to its length.
224 | Essential Physics
9.7 Moment of Inertia of Various Bodies
S.N. Body Moment of Inertia Axis
1. Rectangular plate M(l2 + b2) Through its centre and perpendicular to its plane
2. Circular ring I = 12 Through its centre and perpendicular to its plane
3. Circular disc Through its centre and perpendicular to its plane
I = MR2
4. Solid cylinder MR2 Through its axis of symmetry
5. Hollow cylinder I= 2 Through its axis of symmetry
6. Solid sphere About its any diameter
MR2
7. Hollow sphere I= 2 About its any diameter
I = MR2
I = 2 MR2
5
I = 2 MR2
3
9.8 Radius of Gyration
The radius of gyration of a body about a given axis of rotation is defined as the distance from the axis at
which the whole mass of the body is supposed to be concentrated so that the moment of inertia of the
body about the axis of rotation remains the same. It is denoted by K.
Suppose, a body consists of n particles, each of mass m. Thus, by definition of moment of inertia
I=Σ mr2
222 2
or I= m [r + r + r + . . . + r ]
123 n
or, I = mn r12 + 2 + 2 . . . + r2n
r r+
2 3
n
∴ I = MK2 . . . (1)
Where, mn = M = Total mass of the body
222 2
r +r +r +...r
and, K2 = 1 23 n
n
222 2
r +r +r + ...r
∴ K= 123 n . . . (2)
n
Thus, the radius of gyration of a body about a given axis can also be defined as the root mean square
distances of all the particles of the body from the axis of rotation.
Radius of Gyration of a Thin Rod
Ml2
The M.I. of a thin rod about an axis through its centre and perpendicular to its length is, I = 12 . . . (3)
If K be the radius of gyration of the rod about the axis, then, I = MK2 . . . (4)
From equations (3) and (4), we get
MK2 = Ml2
12
∴ l
K=
12
If the axis of rotation passes through one end and perpendicular to the length of the rod, then
MK2 = Ml2
3
Rotational Dynamics | 225
∴ l 3
K=
Similarly, Radius of Gyration of various bodies are given in the following table:
S.N Body Moment of Inertia(I) Radius of Gyration(K) Axis
. l2 + b2
MK2 = M(l2 + b2) Through its centre and
1. Rectangular plate 12 K = 12 perpendicular to
K=R its plane.
2. Circular ring MK2 = MR2
R Through its centre and
3. Circular disc. MK2 = MR2 K= perpendicular to
2 its plane.
2
4. Solid cylinder MK2 = MR2 R Through its centre and
2 K= perpendicular to
2 its plane.
5. Hollow culinder MK2 = MR2 K=R
Through its axis of
6. Solid sphere MK2 = 2 MR2 2 symmetry
7. Hollow sphere 5 K= 5 R
Through its axis of
MK2 = 2 MR2 2 symmetry
3 K= 3 R
About its any diameter
About its any diameter
9.9 Torque or Moment of a Force
The turning effect of a force on a body is called torque or moment of the force. It is denoted by τ. The
torque acting on a rotating body is measured as the product of the force and perpendicular distance of the
force from the axis of rotation. Mathematically,
Torque =Force × perpendicular distance of the force from the axis of rotation.
or, τ = Fr
∴ τ = rF . . . (1)
It is a vector quantity. Its SI unit is Newton meter (Nm). In vector from, torque is expressed as,
→τ = →r × →F . . . (2) Y
← axis
The direction of →τ is perpendicular to the plane containing →r and
→F . The torque may be clockwise torque or anticlockwise torque. Its →τ F
dimensional formula is,
τ = rF = L × MLT–2 = [ML2Τ–2].
Equation (2) can be written as, O θ
rp
→τ = →r × → = rF sinθ n^ . . . (3)
F
Where, θ is the angle between →r and → and n^ is unit vector along
F,
the direction of →τ .
When θ = 0°, then τ = Fr sin 0° = 0 = minimum [Fig. 9.6 , Torque or moment of
When θ = 90° , then τ = Fr sin90° = Fr = maximum force]
226 | Essential Physics
Relation between torque, angular acceleration and moment of inertia
Suppose a rigid body is rotating about an axis YY' with angular velocity ω as shown figure 9.7. Suppose
the body consists of n particles having masses m1, m2, m3 …………., mn which are at a respective
distances r1, r2, r3, ……, rn from the axis of rotation. Let τ be the external torque applied to the body. The
torque τ will produce constant angular acceleration α on each particle but different linear accelerations. If
a1, a2, a3, ……, an be the respective linear accelerations produced on n Yω
particles, then
a1 = r1α, a2 = r2α , a3 = r3 α, . . ., an = rnα rn mn
The forces acting on n particles are,
r3
F1 = m1 a1, F2 = m2a2, . . ., Fn = mnan respectively. m3
Similarly, the torques due to these n forces are
r2 m2
τ1 = F1r1, τ2 = F2r2, . . ., τn = Fnrn respectively.
Since, total torque acting on the body is equal to the sum of the individual m1 r1
torques acting on the n particles of the body. Therefore,
τ = τ1 + τ2 + τ3 + . . . + τn
= F1r1 + F2r2 + F3r3 + . . . + Fnrn Y'
= (m1a1)r1 + (m2a2) r2 + (m3a3) r3 + . . . + (mnan) rn [Fig. 9.7, Rotating
= (m1r1α)r1 + (m2r2α)r2 + (m3r3α)r3 + . . . + (mnrnα)rn body]
= 2 + 2 + 2 + . . . + mnrn2 )α= (Σmr2)α
(m1r1 m2r2 m3r3
∴ τ=Iα ...(1)
Where, I = Σmr2 = 2 + 2 + 2 + . . . + mnrn2
m1r1 m2r2 m3r3
is the moment of inertia of the body about the axis of rotation YY'. Equation (1) gives the relation
between torque (τ), angular acceleration (α) and moment of inertia (I).
If α = 1, then τ = I × 1
∴ I=τ
So, the M.I. of a body about a given axis is equal to the external torque required to produce unit angular
acceleration in the body about the axis.
9.10 Work Done by a Couple and Power in Rotational Motion
Two equal and unlike parallel forces acting at two different points of a rigid AF F
body forms a couple. A couple always produces the rotational motion on a A'
body. The torque due to a couple is,
τ =Either of the forces of a couple × perpendicular distance between the rθ
two forces of the couple FO
Let two equal and unlike parallel forces, each of magnitude F act an a wheel θ
at two points A and B tangentially such that AB is the diameter of the r
wheel. Let the wheel rotates through an angle θ and the points A and B get B' B
shifted to new positions A' and B' respectively. The torque due to the couple F
is,
τ = F × 2r . . . (1) [Fig. 9.8 : Rotating wheel]
Where, r is the radius of the wheel
Now, work done by each force of the couple is,
Rotational Dynamics | 227
= Force × distance moved
= F × AA' (or BB')
= F × rθ; [since AA' = BB' = rθ and θ is in radian.]
The total work done by the two forces of the couple is given by
W = F × rθ + F × rθ
or, W = 2F × rθ
or, W = (F × 2r)θ
∴ W= τθ . . . (2),[using equation (1)]
Which is similar to, W = FS, in translational motion.
Again, differentiating equation (2) with respect to time t, we get.
dW = d (τθ) = τ dθ
dt dt dt
∴ P = τω . . . (3)
dW and dθ = ω = angular velocity
Where, dt = P = power developed dt
Equation (3) is similar to, P = F.v, in translational motion.
9.11 Angular Momentum
The moment of linear momentum of a body is called angular momentum. It is denoted by L. A
rotating body possesses angular momentum. It is measured as →L
the product of the linear momentum of a body and the →p = m→v
perpendicular distance between the body and the axis of
rotation.
Suppose a body having mass m is revolving around a circle of O →r m
radius r with speed v about an axis passing through the centre
0 as shown in figure 9.9. Then angular momentum of the body
is,
L =Linear momentum × perpendicular distance between [Fig. 9.9, Body revolving around a circle.]
the body and axis of rotation
or, L = Pr
or, L= mvr . . . (1)
If ω be the angular velocity of the body, then, v = rω
So, L = m (rω)r
or, L = mr2ω . . . (2)
Equations (1) and (2) are the expressions for the angular momentum of the body. The S.I. unit of angular
momentum is kgm2s–1 and C.G.S. unit is gmcm2s–1. The dimensional formula is [ML2 T–1]. It is a vector
quantity. In vector form,
→ →r × → = rp sinθ n^ . . . (3)
L= P
Where θ is the angle between →r and → and n^ is the unit vector along direction of → The direction of →
P, L. L
is perpendicular to both →r and → as shown in figure 9.9. The magnitude of angular momentum is,
P
L = rP sinθ . . . (4)
When θ = 0°, then L = rpsin0° = 0. In this case, there is no rotational effect on the body.
When θ = 90°, then L = rpsin90° = rP. In this case, there is maximum rotational effect on the body.
228 | Essential Physics
Relation between angular momentum and moment of inertia
Consider a rigid body having mass M is rotating about an axis YY' with constant angular velocity ω as
shown in figure 9.10. Let the body consists of n particles having masses m1, m2, m3, . . ., mn which are at
respective perpendicular distances r1, r2, r3, . . ., rn from the axis of rotation. Here, all the particles of the
body have same angular velocity ω but different linear velocities. Let v1, v2, v3, . . ., vn be their respective
linear velocities. Then v1 = r1ω, v2 = r2 ω, v3 = r3ω, . . ., vn = rnω.
The magnitude of angular momentum of particle of mass m1 is Y
given by, L1 = m1v1r1 = m1 (r1ω) r1 = m1r12 ω. Similarly, if L2, L3, . . ω
., Ln be the magnitudes of angular momentum of particles of m1
masses m2, m3, . . . , mn respectively, then, L2 = m2r22 ω, L3 = m3r23 r1
ω, . . ., Ln = mnr2n ω. Now, the total angular momentum of the body
r3 m3 M
about the axis YY' is equal to the sum of the angular momenta of
m2 r2
all the n particles of the body about that axis. Thus,
rn mn
L = L1 + L2 + L3 + . . .+ Ln
= m1r12ω + m2r22 ω + m3r23 ω + . . . + mnr2n ω Y'
= (m1r12 + m2r22 + m3r32 + . . . + mn r2n) ω
= (Σmr2) ω [Fig. 9.10, Rotating rigid body]
∴ L =Iω
Where, Σ mr2 = I, is the moment of inertia of the rigid body about
the axis of rotation YY'. Above equation L = Iω is similar to P =
mv is translational motion. Thus, the angular momentum in
rotational motion is analogous to linear momentum in translational
momentum.
Relation between angular momentum and torque
The angular momentum of a rigid body having moment of inertia I and rotating with angular velocity ω
about an axis of rotation is,
L = Iω . . . (1)
Differentiating both sides w.r.t. time t, we get
dL = d (Iω)
dt dt
The moment of inertia I of a body about a given axis of rotation is constant.
∴ dL = I dω
dt dt
or, dL = Iα . . . (2)
dt
Where, dω = α is the angular acceleration of the body. Also, the torque acting on the body is
dt
τ = Iα . . . (3)
From equations (2) and (3), we get
τ= dL . . . (4)
dt
Hence, the torque acting on a body is equal to the rate of change of angular momentum of the body.
dP
Equation (4) in rotational motion is analogous to F = dt in translational motion.
Rotational Dynamics | 229
9.12 Principle of Conservation of Angular Momentum
It states that if no external torque acts on a system, then total angular momentum of the system
remains conserved. That is, Iω = constant
Where, I is the moment of inertia of a body about a given axis of rotation and ω is its angular velocity.
Proof: We have, the torque acting on a body is,
τ = dL
dt
If external torque acting is zero i.e. τ = 0, then,
dL
dt = 0
or, dL = 0
Integrating this equation, we get
∫dL = constant
or, L = constant
∴ Iω = constant [ L = Iω ]
This is the principle of conservation of angular momentum. If I1 and ω1 be the initial values of moment of
inertia and angular velocity respectively, and I2 and ω2 be their respective final values, then in general
I1ω1 = I2ω2
Examples of conservation of angular momentum
i. Motion of a Planet around the Sun: The motion of a planet around the sun in elliptical orbit
follows the angular momentum conservation principle. When the planet is far from the sun, its
moment of inertia about the axis of rotation increases and its angular velocity ω decreases keeping
Iω = constant. Again, when the planet approaches near to the sun, then I decreases and ω increases
keeping Iω = constant.
ii. Earth – Meteorite system: The effective mass of earth increases when a meteorite falls on its
surface. This increases moment of inertia of the earth. From angular momentum conservation
principle, Iω = constant. So, ω should decrease. So, the earth should slightly slow down in its
speed of rotation. But, due to very small change in speed of rotation, we don't experience it.
iii. Ballet dancing: A ballet dancer or ice–skater uses angular momentum conservation principle to
change rate of spin during the performance as shown in fig 9.11. When she stretches her arms
(hands and legs), the moment of inertia increases and hence angular velocity decreases keeping Iω
= constant. Then the dancer decreases the spinning rate. To increase the rate of spin, she folds her
arms (hands and legs) decreasing moment of inertia and hence angular velocity increases, keeping
Iω = constant.
[Fig. 9.11, Ballet dancing] [Fig.9.12, Rotating turntable]
230 | Essential Physics
iv. Motion on a Rotating Turntable: By using angular momentum conservation principle, a man
standing on a rotating turntable and holding weights on his hands, can rotate with a desired angular
velocity. When he draws his hands close to the chest, moment of inertia decreases and hence
angular velocity increases keeping Iω = constant. On the other hand, when he stretches his hands,
the moment of inertia increases and hence angular velocity
decreases, keeping Iω = constant as shown in figure 9.12.
v. A Diver diving into a Swimming Pool: A diver diving into a
swimming pool can increase the number of loops in air by using
angular momentum conservation principle as shown in figure 9.13.
During jumping, from a springboard, the diver curls his body by
rolling his arms (hands and legs) as shown in figure. This decreases
the moment of inertia and hence the angular velocity increases,
keeping Iω = constant. This increase in angular velocity causes the
diver to have more number of loops in air. [Fig. 9.13, Diving]
9.13 Kinetic Energy of a Rotating Body
Let us consider a rigid body having mass M is rotating about an Y
ω
axis YY' with constant angular velocity ω as shown in figure
9.14. Let the body consists of a number of particles having
masses m1, m2, m3, . . ., mn which are at perpendicular distances r1
r1, r2, r3, . . ., rn respectively from the axis of rotation. All of m1 r2 m2
these n particles have same angular velocity but different linear
velocities. Let their respective linear velocities be v1, v2, v3,...vn. rn
mn
Then, v1 = r1ω, v2 = r2ω, v3 = r3ω, . . ., vn = rnω M
The rotational kinetic energy of particle m1 is, r3 m3
K.E1 = 1 m1 v12 = 1 m1 (r1 ω)2 = 1 m1 r12 ω2
2 2 2
Similarly, the rotational kinetic energy of particles of masses Y'
m2, m3, …., mn are [Fig. 9.14, Kinetic Energy of rotation]
K.E2 = 1 m2 r22 ω 2, K.E3 = 1 m3r23 ω 2, . . ., K.En = 1 mnrn2 ω 2
2 2 2
respectively.
Now, the rotational kinetic energy of a rigid body is equal to the sum of the kinetic energies of the
particles of the body. Therefore,
K.Erot = K.E1 + K.E2 + K.E3 + . . . + K.En
or, K.Erot = 1 m1r12 ω 2 + 1 m2 r22 ω 2 + 1 m3r32 ω2 + . . . + 1 mnrn2 ω2
2 2 2 2
or, K.Erot = 1 (m1r21 + m2 r22 + m3r23 + . . . + mnrn2) ω2
2
or, K.Erot = 1 (Σmr2) ω2
2
∴ K.Erot = 1 I ω2
2
Where, m1r12 + m2r22 + m3r32 + . . . + mnrn2 = Σ mr2 = I, is the moment of inertia of the body about the axis of
rotation YY'. The above equation is analogous to K.E. = 1 mv2 is translational motion.
2
Rotational Dynamics | 231
9.14 Kinetic Energy of a Rolling Body
When a body rolls, it rotates about a horizontal axis and its centre of mass moves linearly. Therefore, a
rolling body possesses two types of kinetic energy translational kinetic energy and rotational kinetic
energy. Suppose, a body having mass M and radius R is rolling without slipping on a horizontal plane as
shown in figure 9.15. Let ω be the constant angular velocity of the body and V be the linear velocity of
the centre of mass of the body. If I be the moment of inertia of the body about the axis of rotation, then,
K.E. of rolling body = K.E. of rotation + K.E. of translation.
or, K.ET = K.Erot + K.Etrans
or, K.ET = 1 Iω2 + 1 MV2 . . . (1) w
2 2 R
If K be the radius of gyration of the rolling body about the axis OV
of rotation, then
I =MK2
∴ K.ET= 1 (MK2)ω2 + 1 MV2
2 2
or, K.ET = 1 MK2 VR2+ 1 MV2 [Fig. 9.15, K.E. of rolling body]
2 2
∴ K.ET = 21.MV2 KR22 + 1 . . . (2)
Also, K.ET 1 (Rω)2 KR22 + 1
=2.M
∴ K.Et = 1 Mω2 (K2 + R2) . . . (3)
2
Equations (1), (2) and (3) are the expressions for kinetic energy of a rigid body rolling on a horizontal
plane without slipping.
For a circular ring, I = MR2. Also, I = MK2
∴ MK2 = MR2 ⇒ K2 = 1
R2
∴ K.ET = 1 MV2 (1 + 1) = MV2
2
Similarly, the K.E. of different shaped rolling bodies are as shown in table below:
S.N. Body K2 1 K2
Value of R2 2 R2 +1
( )K.ET
= MV2
1. Circular disc 1 3 MV2
2 4
2. Solid cylinder 1 3 MV2
2 4
3. Hollow Sphere 1 MV2
4. Solid Sphere 2 7 MV2
5 10
5. Hollow sphere 2 5 MV2
3 6
6. Circular ring 1 MV2
232 | Essential Physics
9.15 Acceleration of a Body Rolling Down an Inclined Plane
Let us consider a rigid body having circular symmetry (sphere, disc, cylinder etc.) of mass M and radius
R is rolling down along an inclined plane having angle of inclination θ to the horizontal as shown in
figure 9.16. Let V be the velocity acquired by the body after rolling down the inclined plane of height h,
then the kinetic energy gained by the body is
K.ET = 1 MV2 KR22 + 1 . . . (1)
2
Where, K is the radius of gyration of the body. Also, the rigid body loses it potential energy as it rolls
down the inclined plane. So,
Loss in P.E. of the body = Mgh . . . (2) ω0 = 0 M
u=0 R
Since total mechanical energy remains conserved. So,
Loss in P.E. = Gain in K.E.
or, Mgh = 1 MV2 KR22 + 1 S Mg
2 h
ω
or, 2gh = V2 KR22 + 1 θ
2gh V
θ
= KR22 + 1
∴ V2 . . . (3) [Fig. 9.16, Acceleration of a body rolling
down an inclined plane]
From figure 9.16, sinθ = h ⇒ h = S sinθ . . . (4)
s
Where, S is the distance moved along the inclined plane.
From equations (3) and (4), we have
V2 = 2g S sinθ . . . (5)
KR22 + 1
If 'a' be the linear acceleration of the rolling body down the inclined plane, then using,
V2 = u2 + 2as.
Since, initial velocity, u = 0 . . . (6)
∴ V2 = 2as
From equations (5) and (6), we get
2g S sinθ
2as = KR22 + 1
∴ g sinθ . . . (7)
a = KR22 + 1
This is the general expression for acceleration of a rigid body rolling down an inclined plane. Its value
depends on the ratio of K2 and the angle of inclination θ of the inclined plane. For a circular ring,
R2
I = MK2 = MR2 ⇒ K2
R2 = 1
∴ a = g sinθ = 1 g sinθ
1+1 2
Similarly, the acceleration of different summetrical bodies rolling down an inclined plane are as
shown in following table:
Rotational Dynamics | 233
S.N. Body K2 ( )Acceleration, a= g sinθ
Value of R2 K2
R2 +1
1. Circular disc 1 2 g sinθ
2 3
2. Solid cylinder 1 2 g sinθ
2 3
3. Hollow cylinder 1 1 g sinθ
2
4. Solid sphere 2 5 g sinθ
5 7
5. Hollow sphere 2 3 g sinθ
5
3
6. Circular ring 1 1 g sinθ
2
9.16 Relation Between Translational and Rotational Quantities
The translational quantities and rotational quantities of a rotating body are related to each other. The
relationship between the various parameters appearing in translational and rotational motions can be
summarized as in the table shown below.
S.N. Translational Motion S.N. Rotational Motion Relation
1. Linear displacement = S 1. Angular displacement = θ S = rθ
2. dS 2. Angular velocity; ω = dθ V = rω
Linear velocity, V = dt dt
3. Linear acceleration 3. Angular acceleration a = rα
dv d2S I = mr2
α = dω = d2θ
a = dt = dt2 dt dt2
4. Mass = m 4. Moment of inertia = I
5. Linear momentum, P = mV 5. Angular momentum L = Iω L = rP
6. dP 6. Torque, τ= dL = I dω = Iα τ = rF
Force, F = dt = ma dt dt
7. Work done by a force, W = FS 7. Work done by a torque, W = τθ −
8. Power, P = FV 8. Power, P = τω −
9. Translational K.E. = 12mV2 9. Rotational K.E. = 1 Iω2 −
2
10. Initial linear velocity u 10. Initial angular velocity ω0 u = rω0
–
11. Equation of translational motion 11. Equations of rotation motion
(i) S = ut (i) θ = ωt
(ii) V = u + at (ii) ω = ω0 + αt
(iii) s = ut + 1 at2 (iii) θ = ω0t + 1 αt2
2 2
(iv) V2 = u2 + 2as (iv) ω2 = ω02 + 2αθ
234 | Essential Physics
9.17 Theorems of Parallel and Perpendicular Axes
A. Parallel Axis Theorem A I C Icm
This theorem states that the moment of inertia of a body rO
about any axis is equal to its moment of inertia about a M
parallel axis through its centre of mass, plus the product of
the mass of the body and the square of the distance between BD
the two axes.
[Fig. 9.17, Parallel axis
Let AB be the axis in the plane of paper about which the theorem]
moment of inertia (I) of the rigid body has to be determined.
Let CD be an axis parallel to AB through the centre of mass Z
O of the body, at a distance r from AB, then
Iz
I = Icm + Mr2
O Iy Y
Where Icm is the M.I. of the body about the parallel axis CD
through centre of mass O and M is the mass of the body Ix
X
B. Theorem of Perpendicular Axes
[Fig. 9.18, Theorem of
This theorem states that the moment of inertia of a plane perpendicular axes]
lamina (a plane body) about an axis perpendicular to the
plane of the lamina is equal to the sum of the moments of
inertia of the lamina about two axes perpendicular to each
other, in its own plane, and intersecting at the point where
the perpendicular axis passes through it.
If Ix, Iy and Iz be the moments of inertia of the plane lamina
about three mutually perpendicular axes through the point
O, then
Iz = Ix + Iy
Here, OX and OY are the axes in the plane of the lamina,
and OZ is an axis perpendicular to the plane of the lamina as
shown in figure 9.18.
Boost for Objectives
• A rigid body is a system of particles in which distance between any two particles does not change under the
influence of external forces.
• Moment of inertia depends on mass and distribution of mass of body about the axis of rotation.
• The spokes are fitted in cycle wheel to increase moment of inertia so that the cycle runs smoother and
steadier.
• Fly wheel is an important part of engine. It helps the engine in keeping the speed uniform.
• A couple produces purely rotational motion.
• Ice skaters use the principle of conservation of angular momentum.
• If body shrinks, moment of inertia decreases.
• A dancer on ice spins faster, when she folds her arms, then M.I decreases, angular velocity is increased, K.E
increases and angular momentum is conserved (I1ω1 = I2ω2)
• When a mass is rotating in plane about a fixed point, its angular momentum is directed along the axis of
rotation
• If polar ice on the earth gets melted then length of the day on the earth gets increased.
• When a solid cylinder, a solid sphere and a hollow sphere are each released from same height on an inclined
plane, then solid sphere reach the bottom at first.
Rotational Dynamics | 235
• A body can't roll on an inclined plane in the absence of friction i.e. on smooth inclined plane.
• When a body starts to roll on an inclined plane, its potential energy is converted into translation and rotational
kinetic energy.
• Position of centre of mass is independent of the reference frame. It depends only on masses of the particles
and their relative positions.
Short Questions with Answers
1. Does the angular momentum of a body moving in a circular path change? Give explanation to
Ans:
your answer. [NEB 2074]
→ →
The angular momentum L of a body having mass m moving in a
L
→→ →p = m→v
circle of radius vector r and having linear momentum P is,
→ →→ = rp sinθ ∧n
L = r ×P
Where → → and ∧n is the unit vector
θ is angle between r and P
→→ O →r m
perpendicular to both r and P i.e. along the axis of rotation.
In uniform circular notion, θ = 90°
∴ → = rp sin90° ∧n = rp ∧n .
L
Here r and P are also constants. Hence, there is maximum rotational effect on the body and the angular
momentum (L) remains constants.
2. One end of a solid rod of length L (in m ) is fixed at one end. Can the magnitude and direction of
→
torque be estimated if a force F (in N) act at a point l (in m) from the fixed end making an angle
150° with the horizontal ? Explain. [NEB 2074]
Ans:
→
Since, torque τ is,
→ → → = rF sinθ ∧n
τ =r ×F
And, →
τ = rF sinθ (in magnitude).
→
or, τ = lF sin (180 - 150°)
or, → 1
τ = lF sin 30° = lF ×2
∴ → lF
τ =2
Here, the magnitude of torque is lF → → → →
2 which is perpendicular to the plane of l and F i.e. if l and F
→
are in the plane of page, then τ is out of page. The direction of rotation is clockwise.
3. The cap of the bottle can be easily opened with the help of two fingers than with one finger, why?
[HSEB 2069]
Ans: When the cap of the bottle is opened with the help of two fingers, then a couple is formed. But, when it
is opened with the help of a single finger, then a torque is formed. We know, the torque due to a couple
is double of the torque due to a single force. Hence, the cap is opened easily with the help of two
fingers.
236 | Essential Physics
4. A solid sphere and a hollow cylinder of same mass and same size are rolling down on an inclined
Ans:
plane from rest. Which one reaches the ground first? Why? [HSEB 2069 old, B]
5.
Ans. The acceleration attained by a body rolling down on an inclined plane is a = g sin θ
6. I
Ans: 1 + mr2
7.
Ans: For a hollow cylinder, I = mr2, ac = g sin θ = g sin θ
8. mr2 2
Ans: 1 + mr2
9. For a solid sphere, I = 2 mr2, as = g sin θ = 5 g sin θ
Ans: 5 2 7
10. 1+5
Ans:
Here, as > ac. So, the solid sphere reaches the ground first than the hollow cylinder.
11. A ballet dancer can increase or decrease her spinning rate by using the principle of conservation
Ans:
of angular momentum, how? [HSEB 2068 old]
From principle of conservation of angular momentum, I1ω1 = I2ω2
Also, I = mr2
i. When she lowers her hands, then her moment of inertia (I2 = mr2) decreases and hence her
angular velocity (ω2) increases. As w2 increases, the spinning rate is also increased.
ii. When she stretches her hands outward, then her moment of inertia (I2 = mr2) increases and hence
ω2 decreases. When ω2 decreases, the spinning rate is also decreased.
A fan with blades takes longer time to come to rest than without the blades, why?
[HSEB 2067 S 2051]
A fan with blades has more moment of inertia than a fan without blades. Thus, the inertia of motion is
more for a fan with blades than a fan without blades. Hence, the fan with blades takes longer time to
come to rest than without the blades.
Explain why spokes are fitted in the cycle wheel. [HSEB 2056]
When spokes are fitted in the cycle wheel, then the most of the mass of the cycle is concentrated on its
rim. This increases the moment of inertia of the cycle. Such a large value of moment of inertia results
the uniform motion on the cycle reducing jerks.
What is the counterpart of the mass and force in rotational motion? [HSEB 2055]
We know, in linear motion, Force = mass × linear acceleration
i.e. F = ma . . . (1)
In rotational motion, Torque = (moment of inertia) × angular acceleration
τ = Iα . . . (2)
From equations (1) and (2), we see that in rotational motion, moment of inertia and torque are the
counter parts of mass and force respectively. They play the same role in rotational motion as mass and
force play in translational (linear motion).
Suppose that only two external forces act on a rigid body and the two forces are equal in
magnitude but opposite in direction. Under what conditions will the body rotate? [HSEB 2054]
Suppose, two equal and opposite forces are acting on a body, then these two forces will rotate the body
if their lines of action are different. In such case, the two forces form a couple.
If the earth is struck by meteorites, the earth will slow down slightly, why? [HSEB 2053]
Our earth is rotating about its polar axis by following the principle of conservation of angular
momentum, i.e. Iω = constant
If the earth is struck by meteorites, then its effective mass increases and hence its moment of inertia also
increases. Consequently, the angular velocity (ω) is decreased. So, the earth will slow down slightly.
Why is it easier to hold down a 10 kg body in your hand at your side than to hold it with your arm
extended horizontally? [HSEB 2050]
If a 10 kg body is hold in our hand with our arm extended horizontally, the force i.e. weight of 10 kg
mass and the distance of force from the axis of rotation are perpendicular . Hence, a maximum torque is
produced. But, if the body is hold at our side, then the lines of action of force (i.e. weight) passes
Rotational Dynamics | 237
through the rotation axis. So, a very small turning effect about the shoulder joint (rotation axis) is
produced. Hence, it is easier to hold down a 10 kg mass in our hand at our side than to hold it with our
arm extended horizontally.
12. Can a single force applied to a body change both its translational and rotational motion? Explain.
Ans:
[HSEB 2068, S]
13.
Ans: Yes, when a body is rolling without slipping on a horizontal plane surface under the application of a
14. force, then it rotates about the horizontal axis through its C.M. and it also undergoes displacement in the
Ans:
15. forward direction. Here, the body rotates about an horizontal axis and at the same time the C.M. of the
Ans:
body covers linear displacement. Hence, the body can have the change in both translational (linear) and
16.
Ans: rotational motion when a single force is applied.
17. If the ice on the polar caps of the earth melts, how will it affect the duration of the day? Explain.
Ans:
[HSEB 2073 C]
18.
Ans: Since, no external torque is acting on the earth, its total angular momentum remains conserved, i.e.
Iω = constant. Also, I = mr2. When ice on the polar caps of the earth melts, then there is redistribution
of its mass i.e. mass shifts towards the equatorial region, i.e. r increases from the polar axis of rotation
of the earth. So, I increases and hence angular velocity (ω) of earth should decrease. When ω decreases,
then the period (T) of rotation of earth increases. Hence, duration of the day increases.
Moment of inertia is also called the rotational inertia. Why?
We know, the moment of inertia plays the same role in rotational motion as mass plays in translational
(linear) motion. Greater the value of moment of inertia of a body, higher is the tendency to continue in
the state of rotational motion. Hence, the moment of inertia is also called the rotational inertia.
Why is the most of the mass of a flywheel is concentrated at its rim?
The main purpose of a flywheel is to maintain uniform motion in spite of the changes in the accelerating
or decelerating torques. For this, the moment of inertia (I) of the flywheel should be large. When the
mass is concentrated at its rim, then the moment of inertia of the flywheel increases as given by the
relation I = Σmr2. When such a wheel gains or loses some K.E. of rotation (K.E. = 1 Iω2) then the large
2
value of 'I' causes the small change in angular velocity of rotation 'ω'. This helps to maintain uniform
rotational motion of the flywheel.
If earth shrinks suddenly, how will be the duration of a day affected?
The radius 'r' of earth will decrease if earth shrinks suddenly. Since, I = Σmr2, I will also be less. From
principle of conservation of angular momentum Iω = constant. So, 'ω' should increase. Since, T = 2π
ω,
the time period will decrease. The decrease in time period 'T' makes the duration of a day shorter than
now.
How is a swimmer jumping from a height in a swimming pool able to increase the no. of loops
made in the air?
From angular momentum conservation principle, Iω = constant. To increase the no. of loops in air, 'ω'
should increase. To increase ω, moment of inertia (I) has to be decreased. This can be done by
decreasing 'r' of the body. Since, I = Σmr2, the swimmer has to curl his/her body in order to achieve
more no. of loops in air.
A circular platform is rotating with a uniform angular velocity. What will be the change in
motion of the platform if a person sits near the edge and then starts moving towards the center of
the platform?
From angular momentum conservation principle, Iw = constant. Here, total moment of inertia of the
system is, Itotal = Iplatform + mr2. If a person sits near the edge of the platform, then 'r' and hence mr2 will
be more. This makes total moment of inertia Itotal more. This makes angular velocity 'ω' less and hence
the circular platform rotates at a slower rate. When the person starts moving towards the center of the
platform, then r and hence mr2 decreases. It reduces Itotal and increases 'ω'. So, the platform starts
rotating at the faster rate.
238 | Essential Physics
19. A helicopter has two propellers, why?
Ans:
In order that the helicopter rise and fly, it has to have just one propeller which is able to throw air
20.
Ans: downwards. In doing so, the helicopter would turn in a certain direction. From angular momentum
21.
Ans: conservation principle, I1ω1 = I2ω2., the helicopter would then rotate in opposite direction, which is not
practical. To stop this reverse rotation, a second propeller is used at the tail which pushes the air in
22.
Ans: opposite direction and hence stops the rotation. This second propeller at the tail makes the helicopter
23. stable and safe.
Ans:
24. Explain how a cat is able to land on its feet when thrown in to air?
Ans:
25. The cat uses the law of conservation of angular momentum, i.e. Iω = constant. When the cat is thrown
in air, it stretches its body together with the tail and feet so that the moment of inertia (I) increases.
When I increase, then the angular velocity ω will be small and hence the cat can land on safely on its
feet.
How will you distinguish between a hardboiled egg and a raw egg by spinning each on a table
top?
Here, the torques applied on both the eggs are equal, i.e. τ = I1α1 = I2α2. When moment of inertia (I) is
high, than angular acceleration α is small and vice–versa. During the rotation of raw egg, the liquid
matter moves away from the axis of rotation and hence moment of inertia (I) increases. So, α decreases
and hence raw egg comes to rest first. The boiled egg behaves as a rigid body and spins more than a raw
egg.
If the earth contracts to half its radius, what would be the length of one day?
Earth is assumed to be a perfect sphere and its moment of inertia I is, I = 25MR2
If the earth contracts to half of its radius, than its moment of inertia becomes, I' = 2 M R2 2 = 2 MR2
5 5 4
From angular momentum conservations principle, I'w' = Iw
or, 2 MR2 .ω' = 2 MR2. ω
5 4 5
or, ω ' = 4 ω
or, 2π = 42Tπ
T'
T 24
or, T' = 4 = 4 = 6 hrs.
Thus, the length of one day would be 6 hrs if earth contracts to half radius.
When calculating the moment of inertia of an object, can we treat all its mass as if it were
concentrated at the center of mass of the object? Justify your answer.
No, we can't consider the total mass to be concentrated at the centre of mass in rotational motion. This
is because the moment of inertia of a body is equal to the sum of moments of inertia of all the particles
of the body and its value depends on the locations of the axis of rotations, total mass and the distribution
of mass w.r.t. axis of rotation.
A small girl standing on a turn table with her hands fully extended horizontally brings her hands
close together to say 'namaste'; what happens?
From angular momentum conservation principle, we have, I1ω1 = I2ω2 . When the small girl brings her
hands close together to say 'namaste', the moment of inertia of the system decreases. To keep Iω =
constant in the absence of external torque, angular velocity 'ω' of the system increases. So, turn table
rotates faster.
The work done by a force is the product of force and distance. The torque due to the force is the
product of force and distance. Does this mean that torque and work are equivalent? Explain .
[HSEB 2071 S]
Rotational Dynamics | 239
Ans: No, they are not equivalent. We have, work done W is, W = → . → = FScosθ, where S is displacement
F S
26.
Ans: →→→
and θ is angle between force and displacement. Work is a scalar quantity. But, torque τ is, τ = F × d =
→
Fdsinθ n^ ,where d is the distance between the line of action of force and axis of rotation. Torque is a
τ
vector quantity. Unit of work is Nm or Joule but unit of torque is Nm but not Joule.
A flywheel rotates with constant angular velocity. Does a point on its rim have a tangential
acceleration ? A radial acceleration ? Are these acceleration constant in magnitude? In direction?
In each case, give the reasoning behind your answer. [HSEB 2070]
Here, angular velocity, ω = constant. So, tangential acceleration, aT = r α= r dω radial
dt = 0 and
acceleration, a = rω2 = constant. Thus, the tangential component of acceleration is zero and radial
component of acceleration has a constant magnitude. The direction of radial acceleration changes at
each point and it is directed towards the radius through each point of the circular path. The radial
acceleration changes the direction of velocity and tangential acceleration changes the magnitude of
velocity in rotational motion.
Numerical Examples
1. A ballet dancer spins with 2.4 rev/sec with her arms outstretched when the moment of inertia about
the axis of rotation is I. With her arms folded, the moment of inertia about the same axis becomes
0.6I. Calculate, the new rate of spin. [Model Question; HSEB 2069, A]
Solution:
Given, Initial frequency (f1) = 2.4 rev/sec Initial moment of inertia (I1) = I
Final moment of inertia (I2) = 0.6 I Final frequency (f2) = ?
From angular momentum conservations principle,
I2ω2 = I1ω1
or, I2 × 2πf2 = I1 × 2πf1
∴ f2 = I1 × f1 = I × 2.4 = 4 rev/sec.
I2 0.6 I
2. A constant torque 200 Nm turns a wheel about its centre. The moment of inertia about the axis is 100
kg m2. Find the kinetic energy gained after 20 revolutions when it starts from rest.
[HSEB 2069 A, 2055]
Solution: Moment of inertia (I) = 100 kg m2
Given, Constant torque (τ) = 200 Nm
Number of revolution (n) = 20 Kinetic energy gained (K.E) = ?
Since, τ = Iα
or, α = τ = 200 = 2 rad s–2
I 100
If θ be the angular displacement in 20 revolutions,
then, θ = 2πn = 2π × 20 = 40π rad
Here, ω0 = 0 (initial angular velocity)
ω = angular velocity after 20 revolutions.
Therefore, ω2 = ω02 + 2αθ
or, ω2 = 02 + 2 × 2 × 40π = 160π
∴ ω2 = 160π.
So, kinetic energy gained, K.E. = 1 Iω2 = 1 × 100 × 160π = 8000π Joules = 25132.82 Joules.
2 2
240 | Essential Physics
3. An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200
rev/min in 4.00s. (a) Find the angular acceleration and the number of revolutions made by the motor
in 4.00 s. interval. (b) How many more seconds are required for the fan to come to rest if the angular
acceleration remains constant? [HSEB 2067]
Solution:
500 25
Given, Initial frequency (f0) = 500 rev/min = 60 rev/sec = 3 rev/sec.
200 10
Final frequency (f) = 200 rev/min = 60 rev/sec = 3 rev/sec.
Time (t) = 4 sec
Initial angular velocity (ω0) = 2πf0 = 2π × 25 = 52.3 rad/sec.
3
Final angular velocity (ω) = 2πf = 2π × 10 = 20.9 rad/sec.
3
(a) Angular acceleration (α) = ?
Number of revolutions made in 4 sec, (n) = ?
We have, α = ω – ω0 = 20.9 – 52.3 = –7.85 rad/s2
t 4
If θ be the angular displacement, then , θ = ω0t + 12αt2 = 52.3 × 4 – 1 × 7.85 × 42 = 146.4 rad
2
Again, θ = 2πn
or, n = θ = 146.4 = 23.3 revolutions
2π 2 × 3.14
(b) When the fan comes at rest, then, ω' = 0, ω0' = 20.9 rad/sec.
t' = ? (time taken to come to rest)
Since, ω' = ω0' + αt'
ω' – ω0' 0 – 20.9
or, t' = α = –7.85 = 2.66 sec
So, the fan takes 2.66 sec more to come to rest.
4. A ballet dancer spins about a vertical axis at 1 revolution per second with her arms stretched. With
her arms folded, her moment of inertia about this axis decreases by 40%, calculate the new rate of
revolution. [HSEB 2067 S]
Solution:
Given, Initial frequency (f1) = 1 rev s–1 Initial moment of inertial (I1) = I (suppose)
Final moment of inertia (I2) = I – 40% of I = 0.6 I ; Final frequency (f2) = ?
From angular momentum conservation principle,
I2ω2 = I1ω1
or, I2 × 2πf2 = I1 × 2πf1
or, f2 = I1 × f1 = I×1 = 1.67 rev s–1
I2 0.6I
Hence, the final frequency (rate of revolutions) is 1.67 rev s–1
5. A disc of moment of inertia 5 × 10–4 kgm2 is rotating freely about the axis through its centre at 40
rpm. Calculate the new revolutions per minute if some wax of mass 0.02 kg is dropped gently on to
the disc 0.08 m from the axis. [HSEB 2066]
Solution: Initial frequency of the disc (f1) = 40 rpm
Given, Initial moment of inertia (I1) = 5 × 10–4 kgm2 Mass of the wax (m) = 0.02 kg
Final frequency of the disc (f2) = ? Final moment of inertia (I2) = I1 + mr2
Distance of wax from axis (r) = 0.08 m
From angular momentum conservation principle,
I2ω2 = I1ω1
Rotational Dynamics | 241
or, (I1 + mr2) 2π f2 = I1 (2πf1)
or, f2 = I1 I1 f1 = 5 × 5× 10–4 × 40 = 32 rpm.
+ mr2 10–4 + 0.02 × 0.082
Hence, new frequency (revolution per minute) of the disc is f2 = 32 rpm
6. A constant torque of 500 Nm turns a wheel about its centre. The moment of inertia about this axis is
20 kgm2. Find the angular velocity and kinetic energy gained in 2 seconds.
[HSEB 2061, 2054, 2072 C, 2070]
Solution: Moment of inertia (I) = 20 kgm2.
Constant torque (τ) = 500 Nm
Angular velocity gained (ω) = ? K.E. gained (K.E.) = ?
Time taken (t) = 2 sec Initial angular velocity (ω0) = 0
We have, τ = Iα
or, τ = I (ω – ω0)
t
or, 500 = 20 (ω – 0) = 10ω
2
∴ ω = 50 rad/sec
Again, K.E. = 1 Iω2 = 1 × 20 × 502 = 25000 Joules.
2 2
Hence, angular velocity and K.E. gained in 2 sec are 50 rad/sec and 25,000 Joules respectively.
7. Speed of a body spinning about an axis increases from rest to 100 rev.min–1 in 5 sec, if a constant
torque of 20 Nm is applied. The external torque is then removed and the body comes to rest in 100
sec due to friction. Calculate the frictional torque. [HSEB 2052]
Solution:
First case: When a constant torque is applied,
Torque (τ) = 20 Nm Initial frequency (f0) = 0
Final frequency (f) = 100 rev. min–1 = 100 rev/s = 5 rev/s–1
60 3
Time (t) = 5 sec;
Here, ω0 = 2πf0 = 0 ω = 2πf = 2π × 5 = 10π rad. s–1
3 3
We know, τ = Iα
or, τ = I (ω – ω0)
t
or, 20 = I 103π – 0 = I × 10π × 1 = I × 2π
3 5 3
5
∴ I = 30 kg m2
π
Second case: When the torque is removed,
t' = 100 sec ω' = 0 ω0' = 10π rad. s–1
3
Frictional torque (τ') = ?
Therefoe, τ' = Iα' = I (ω' – ω0' )
t'
= 30 (0 – 10π/3)
π 100
or, τ' = 30 × 10π × 1 = –1 Nm.
–π 3 100
Hence, the magnitude of the frictional torque is 1 Nm.
242 | Essential Physics
8. What is the power output in horsepower of an electric motor turning at 4800 rev/min and developing
a torque of 4.30 Nm?
Solution:
Power output (P) = ?
4800
Angular frequency (f) = 4800 rev/min = 60 rev/s= 80 rev/s
Angular velocity (ω) = 2πf = 2π × 80 = 503 rad s–1
Torque (τ) = 4.30 Nm
Since, P = τ.ω = 4.30 × 503 = 2163 watts = 2163 HP = 2.9 HP
746
9. The flywheel of a gasoline engine is required to give up 500 J of kinetic energy while its angular
velocity decreases from 650 rev/min to 520 rev/min. What moment of inertia is required?
Solution:
Decrease in K.E. = 500 J
650
Initial frequency (f0) = 650 rev/min = 60 rev/s = 10.83 rev/s
520
Final frequency (f) = 520 rev/min = 60 rev/s = 8.67 rev/s
Moment of inertia (I) = ?
Initial angular velocity (ω0) = 2πf0 = 2 × 3.14 × 10.83 = 68.01 rad s–1
Final angular velocity (ω) = 2πf = 2 × 3.14 × 8.67= 54.45 rad s–1
Since, change in K.E. = 1 Iω02 – 1 Iω2
2 2
or, 500 = 1 I (ω20 – ω2)
2
∴ I = 500 × 2 = 1000 = 0.602 kgm2.
ω20 – ω2 (68.01)2 – (54.45)2
10. The main rotor of a helicopter is turning in a horizontal plane at 90.0 rev/min, the distance from the
centre of the rotor shaft to each blade tip is 5.00 m. Calculate the speed through the air of the blade
tip if (a) the helicopter is sitting on the ground; (b) the helicopter is rising vertically at 4.00 m/s.
Solution:
90 3
Angular frequency (f) = 90 rev/min = 60 rev/s = 2 rev/s.
Angular velocity (ω) = 2πf = 2 × 3.14 × 3 = 9.42 rad/s
2
Radius of the path of blade (r) = 5 m
(a) Speed of the blade at ground (v) = ? (b) Speed of blade rising (vT) = ?
Rising velocity of the helicopter (v1) = 4 ms–1
Since, v = rω = 5 × 9.42 = 47.10 ms–1
Again, vT = v2 + v21 = (47.1)2 + 42 = 47.3 ms–1
11. When drilling a 12.7 mm diameter hole in wood the required speed is 1250 rev/min. Determine its
linear speed and radial acceleration.
Solution:
Diameter of hole (d) = 12.7 mm = 12.7 × 10–3m
Radius of hole (r) = d = 6.35 × 10–3m
2
1250 125
Angular frequency (f) = 1250 rev/min = 60 rev/s = 6 rev/s
Angular speed (ω) = 2πf = 2 × 3.14 × 125 = 131 rads–1
6
Rotational Dynamics | 243
Linear speed (v) = ? Radial acceleration (arad) = ?
Since, v = rω = 6.35 × 10–3 × 131 = 0.832 ms–1
Again, arad = rω2 = (6.35 × 10–3) × (131)2 = 109 ms–2
12. The earth, which is not a uniform sphere, has a moment of inertia of 0.3308 MR2 about an axis
through its north and south poles. It takes the earth 86,164 s to spin once about this axis. Calculate
(a) the earth's kinetic energy due to its rotation about this axis and (b) the earth's kinetic energy due
to its orbital motion around the sun. (mass of earth = 5.97 × 1024 kg, Radius of the earth = 6.38 × 106
m, Radius of orbit = 1.50 × 1011 m, orbital period = 365.3 days).
Solution:
Moment of inertia of earth (I) = 0.3308 MR2 Time period of rotation (Tr) = 86164 sec
Orbital period (T0) = 365.3 × 24 × 60 × 60 secs = 31.56 × 106 secs.
Rotational Kinetic energy (K.E.r) = ? Orbital kinetic energy (K.E.0) = ?
Mass of the earth (M) = 5.97 × 1024 kg; Radius of the earth (R) = 6.38 × 106m
Radius of orbit of earth around the sun (R0) = 1.50 × 1011m
Since, K.Er = 1 Iω2 = 1 0.3308MR2 2Tπr 2 = 1 × 0.3308 × 5.97 × 1024 × (6.38 × 106)2 × 28×6134.1442
2 2 2
∴ K.Er = 2.14 × 1029 J
And, K.E0 = 1 MV2 = 1 (R0ω0)2 = 1 MR022Tπ02
2 2M 2
or, K.E0 = 1 × 5.97 × 1024 × (1.50 × 1011)2 × 2× 3.14 2
2 31.56 × 106
∴ K.E0 = 2.66 × 1033 J.
13. Forces F1 = 7.50 N and F2 = 5.30 N are applied tangentially to a wheel with radius 0.330 m, as
shown in figure. What is the net torque on the wheel due to these two F1
forces for an axis perpendicular to the wheel and passing through its
center? R
Solution: O
F1 = 7.50 N, F2 = 5.30 N R
Radius of the wheel (R) = 0.330 m 0.330m
Net torque on the wheel (τ) = ? F2
Since, τ = τ1 + τ2 = – F1R + F2R = (–F1 + F2) R = (–7.50 + 5.30) × 0.330
∴ τ = –0.726 Nm
14. Emilie's potter's wheel rotates with a constant 2.25 rad/s2 angular acceleration. After 4.00s the wheel
has rotated through an angle of 60.0 rad. What was the angular velocity of the wheel at the
beginning of the 4.00s interval?
Solution:
Constant angular acceleration (α) = 2.25 rad s–2
Time (t) = 4 sec Angular displacement (θ) = 60 rad
Initial angular velocity (ω0) = ?
Using, θ = ω0t + 1 αt2
2
or, 60 = ω0 × 4 + 1 × 2.25 × 42
2
∴ ω0 = 10.5 rad s–1
15. A computer disk drive is turned on starting from the rest and has angular acceleration, (a) how long
did it take to make the first complete rotation, and (b) what is the angular acceleration? Given that
the disk took 0.750 sec for the drive to make its second complete revolution. [HSEB 2070 S]
Solution: Time for 1st complete rotation (t1) = ?
Angular acceleration (α) = ?
Given, Initial angular velocity (ω0) = 0
Time for 2nd complete rotation = 0.75 sec.
Time for two complete rotations (t2) = (t1 + 0.750) secs.
244 | Essential Physics
For 1st rotation, we have For two rotations, we have
θ1 = ω0t1 + 12α t12 θ2 = ω0t2 + 1 α t22
2
or, 2π = 0 + 1 α t12 or, 4π = 0 + 1 α (t1 + 0.75)2
2 2
∴ 4π = αt12 . . . (1) ∴ 4π = 1 α [t12 + 2 × 0.75 × t1 + (0.75)2] . . . (2)
2
From equations (1) and (2), we get
1 α [t12 + 1.50t1 + 0.5625] = α t12
2
or, t12 + 1.50 t1 + 0.5625 = 2t12 . . . (3)
∴ t12 – 1.50t1 – 0.5625 = 0
Solving equation (3), we get, t1 = 1.81 sec and t1 = – 0.31sec (Impossible)
∴ Time for 1st rotation (t1) = 1.81 sec
Again, using value of t1 in equation (1), we get,
4π = α (1.81)2
∴ α = 4π = 3.84 rad/sec2
3.2761
Important Numerical Problems
1. A child is pushing a merry–go–round. The angle through which the merry–go–round has turned
varies with time according to θ(t) = γt + βt3, where γ = 0.400 rad/s and β = 0.0120 rad/s3. (a) Calculate
the angular velocity of the merry–go–round as a function of time. (b) What is the initial value of the
angular velocity? (c) Calculate the instantaneous value of the angular velocity ω at t = 5.00 s and the
average angular velocity ωav–z for the time interval t = 0 to t = 5.00 s.
Solution:
Given, θ(t) = γt + βt3 . . . (1)
where, γ = 0.400 rad s–1 and β = 0.0120 rad s–3
(a) Angular velocity (ω) = ?
We have, ω = dθ (t) = ddt(γt + βt3) = γ + 3βt2 = 0.400 + 3 × 0.0120 × t2
dt
∴ ω = 0.400 + 0.360 t2
(b) ω0 = ? at t = 0 ∴ ω0 = 0.400 + 0.360 × 02 = 0.400 rad s–1
(c) ω5 = ? at t = 5.00 sec
We know that, ω = 0.400 + 0.0360 t2
∴ ω5 = 0.400 + 0.360 × 52 = 1.300 rad s–1
Also, average angular velocity , ωav–z = ?
ωav–z = total angular displacement
total time
θ(5) – θ(0) [γt + βt3]5– [γt + βt3] = 0 [0.400 × 5 + 0.0120 × 53] – 0
t5 – t0 =5
or, ωav–z = = 5–0 t
∴ ωav–z = 0.700 rad s–1
2. A flywheel is spinning at 500 rpm when a power fails. The power is off for 30.0s and during this time
the flywheel makes 200 complete revolutions. (a) What is the angular speed after 30s? (b) What is the
time taken to stop the wheel and how many revolutions are made during this time?
Solution:
500 25
Initial frequency (f0) = 500 rpm = 60 rev/s = 3 rev/s
Time (t) = 30.0 sec. Number of complete revolutions made (n) = 200
Rotational Dynamics | 245
(a) ω = ? after 30 secs (b) Time for stop (t1) = ?
No. of revolutions (n1) = ?
ω0 = 2πf0 = 2 × 3.14 × 25 Using, ω1 = ω0 + αt1
3
or, 0 = 52.4 – 0.701 × t1
= 52.4 rad s–1 ∴ t1 = 74.8 sec
Again, we have,
Angular displacement,
θ = 2πn = 2 × 3.14 × 200 = 1256.6 rad
Now, θ = ω0t + 1 αt2 ω12 = ω20 + 2αθ1
2
or, 0 = (52.4)2 – 2 × 0.701 × θ1
1 302
or, 1256.6 = 52.4 × 30 + 2 × α × ∴ θ1 = 1958.4 rad
∴ α = –0.701 rad s–2 Since, θ1 = 2πn1
Again, we have or, n1 = θ1 = 1958.4 = 312 revoltions.
2π 2 × 3.14
ω = ω0 + αt = 52.4 – 0.701 × 30
∴ ω = 31.4 rad s–1 ∴ n1 = 312 revoltions.
And, f = ω = 31.4 rev s–1
2π 2π
∴ f = 31.4 × 60 rpm = 300 rpm.
2π
3. The rotating blade of a blender turns with constant angular acceleration 1.50 rad/s2. (a) How much
time does it take to reach an angular velocity of 36.0 rad/s, starting from rest? (b) Through how
many revolutions does the blade turn in this time interval?
Solution:
Constant angular acceleration (α) = 1.50 rad/s2
(a) Time required (t) = ? (b) No. of revolutions made (n) = ?
Final angular velocity (ω) = 36 rad/s Therefore, ω2 = ω02 + 2αθ
Initial angular velocity (ω0) = 0 or, (36)2 = 02 + 2 × 1.50 × θ
Therefore, ω = ω0 + αt ∴ θ = 432 rad
or, t = ω – ω0 = 36 – 0 Now, θ = 2πn
α 1.5
θ 432
∴ t = 24.0 sec. or, n = 2π = 2 × 3.14
∴ n = 68.8 revolutions
4. A safety device brings the blade of a power mower from an initial angular speed of ω1 to rest in 1.00
revolution. At the same constant acceleration, how many revolutions would it take the blade to come
to rest from an initial angular speed ω3 that was three times as great, ω3 = 3ω1?
Solution:
Case I Case II
Initial angular speed = ω1, number of revolution n Now, initial angular speed (ω3) = 3ω1
= 1rev, Final angular speed = ω2 = 0 Final angular speed (ω4) = 0
Angular displacement(θ)= 2πn = 2π × 1 = 2π rad Number of revolution (n1) = ?
Let, α be the constant angular acceleration, then Using,
ω22 = ω12 + 2αθ ω42 = ω23 + 2 αθ1
or, 02 = ω21 + 2α × 2π
∴ α = –ω4π12 rev/s2 or, 02 = (3ω1)2 + 2–ω4π21 .2πn1
∴ n1= 9 revs.
246 | Essential Physics
5. A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular
acceleration of 0.600 rad/s2. Compute the magnitude of the tangential acceleration, the radial
acceleration and the resultant acceleration of a point on its rim (a) at the start; (b) after it has turned
through 60.0°
Solution:
Radius of the flywheel (r) = 0.3 m (b) Angular displacement (θ) = 60°
Initial angular velocity (ω0) = 0 = 60 × π = π rad
Constant angular acceleration (α) = 0.6 rad s–2 180 3
(a) At the start, t = 0, Let, w be the angular velocity when the
Tangential acceleration (atan) = ? angular displacement is θ. Then,
Radial acceleration (arad) = ? ω2 = ω20 + 2αθ
Resultant acceleration (a) = ? ∴ ω2 = 2αθ
Now, arad = rω2 = r × 2αθ
Since,atan = rα = 0.3 × (2 × 0.6 × π3) = 0.377 ms–2
or, atan = 0.3 × 0.6 = 0.18 ms–2 Also, atan = rα = 0.3 × 0.6 = 0.18 ms–2
Also, arad = rω02 = 0
And, a = ata2n + ara2d
So, a = ata2n + ara2d = (0.18)2 + 02 = 0.18 ms–2
= (0.18)2 + (0.377)2 = 0.418 ms–2
6. A centrifuge can produce a radial acceleration of 3000g at 5000 rev/min. Calculate the required
radius of the centrifuge.
Solution:
Radial acceleration (arad) = 3000g ms–2
5000 250
Angular frequency (f) = 5000 rev/min = 60 rev/s = 3 rev/s.
Angular velocity (ω) = 2πf = 2 × 3.14 × 250 = 524 rads–1.
3
Radius (r) = ?
Since, arad = rω2
∴ 3000g = r × (524)2
3000 × 9.8
or, r = (524)2 = 0.107 m = 10.7 cm.
7. Energy is to be stored in a 70.0 kg flywheel in the shape of a uniform solid disk with radius
R = 1.20m To prevent structural failure of the flywheel, the maximum allowed radial acceleration of
a point on its rim is 3500 m/s2. What is the maximum kinetic energy that can be stored in the
flywheel?
Solution:
Mass of the flywheel (M) = 70 kg Radius (r) = 1.20 m
Maximum radial acceleration (arad) = 3500 ms–2
Maximum kinetic energy stored (K.E.) = ?
Now, arad = rω2
or, 3500 = 1.20 × ω2
∴ ω2 = 3500 = 2917.rad2.s–2
1.20
And, I = 1 mr2 = 1 × 70 × (1.2)2 = 50.4 kgm2
2 2
So, K.E. = 1 Iω2 = 1 × 50.4 × 2917= 7.35 × 104 J
2 2
Rotational Dynamics | 247
8. Your car's speedometer converts the angular speed of the wheels to the linear speed of the car,
assuming standard size tyres and no slipping on the pavement. If your car's standard tyres are 24
inches in diameter, at what rate (in rpm) are your wheels rotating when you are driving at a freeway
speed of 60 mph?
Solution:
Tyres diameter (d) = 24 inches; 1 mile = 1609 m.
Tyres radius (r) = d = 12 inches = 12 × 2.54 cm = 30.5 cm = 0.305 m
2
Linear speed (v) = 60 mph (miles per hour) = 60 × 1609 = 26.81 ms–1
60 × 60
Angular frequency (f) = ?
We have v = rω
or, ω = v = 26.81 = 87.92 rad s–1
r 0.305
Also, ω = 2πf
∴ f = ω
2π
87.92
= 2 × 3.14
= 13.99 rev s–1
= 13.99 × 60 rpm = 840 rpm
9. A flywheel having a moment of inertia of 16.0 kg.m2 and runs at 300 rev/min. (a) Find the speed in
rev/min to which the flywheel will be reduced when 4000J of work is required. (b) What must the
constant power supply to the flywheel (in watts) be to bring it back to its initial speed in a time of
5.00s?
Solution:
M.I. of flywheel (I) = 16.0 kg m2
300
Initial frequency (f0) = 300 rev/min = 60 rev/s = 5 rev/s
Initial angular velocity (ω0) = 2πf0
= 2 × 3.14 × 5= 31.4 rad s–1
(a) Work required (ω) = 4000 J ;
The required value of angular velocity (ω) = ? Time taken (t) = 5 sec
Since, Change in K.E. = work required
or, 1 Iω20 – 1 Iω2 = W
2 2
or, 1 Iω2 = 1 Iω02 – W
2 2
∴ω= ω20 – 2W
I
= (31.4)2 – 2 × 4000 = 22.04 rad s–1
16
and, f = ω
2π
= 22.04 = 3.50 rev/sec = 3.50 × 60 rev/min = 210.5 rev/min
2 × 3.14
W 4000
(b) Again, Power, P = t = 5 = 800 watts.
248 | Essential Physics
→
10. Calculate the torque (magnitude and direction) about point 0 due to the force F in each of the
→
situations sketched in figures. In each case, the force F and the rod lie in the plane of the page, the
rod has length 4.00 m, and the force has magnitude F = 10.0 N.
0 0 F
F
F 30.0°
90.0°
120.0°
(a) (b) (c)
F
60.0° F 60.0° F
0 0
0
(d) 2.00m (e) (f)
Solution:
Length of the rod (l) = r = 4 m ; Applied force (F) = 10 N
Since, →τ = →r × →
F
∴ τ = r F sin θ (in magnitude)
(a) τ = r F sin θ = 4 × 10 × sin 90° = 40 Nm (out of page)
(b) τ = r F sin θ = 4 × 10 × sin 120° = 34.64 Nm (out of page)
(c) τ = r F sin θ = 4 × 10 × sin 30° = 20 Nm (out of page)
(d) τ = r F sin θ = 2 × (–10) × sin 60° = –17.32 Nm (into the page)
(e) τ = r F sin θ = 0 × 10 × sin 60° = 0
(f) τ = r F sin θ = 4 × 10 × sin 0° = 0
11. Calculate the net torque about point 0 for the two forces applied as in figure. The rod and both forces
are in the plane of the page. F2=12.0N F1= 8.00N
Solution:
F1 = 8N θ1 = 90° r1 = (2 + 3) m = 5m
F2= 12N θ2 = 30° r2 = 2m 30.0°
Net torque (τ) = ? 0
Here, τ = τ1 + τ2 2.00m 3.00m
Since, τ1 = r1F1 sin θ1 = 5 × (–8) × Sin 90° = – 40 Nm (Clockwise).
And, τ2 = r2F2 sin θ2 = 2 × 12 × sin 30° = 12 Nm (anti–clockwise)
∴ τ = τ1 + τ2 = – 40 + 12 = – 28 Nm (Clockwise)
12. The flywheel of an engine has a moment of inertia 2.50 kgm2 about its rotation axis. (a) What
constant torque is required to bring up to an angular speed of 400 rev/min in 8.00s, starting from
rest? (b) What is its final kinetic energy?
Solution:
M.I. of flywheel (I) = 2.50 kgm2
(a) Constant torque (τ) = ? Initial angular velocity (ω0) = 0
400 20
Final frequency (f) = 400 rev/min = 60 rev/sec = 3 rev/sec
Final angular velocity (ω) = 2πf = 2 × 3.14 × 20 = 41.89 rads–1
3
Time (t) = 8.00 sec. Since, τ = Iα
or, τ = I (ω – ω0) = 2.50 × (41.89 – 0) = 13.09 Nm.
t 8
(b) Final Kinetic energy (K.E.) = ?
Since, K.E. = 1 Iω2 = 1 × 2.50 × (41.89)2 = 2193 J = 2.193 × 103 J
2 2
Rotational Dynamics | 249
13. A cord is wrapped around the rim of a wheel 0.250 m in radius, and a steady pull of 40.0N is exerted
on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its centre.
The moment of inertia of the wheel about this shaft is 5.00 kgm2. T0
2.00 kg
Compute the angular acceleration of the wheel.
Solution: m1
Radius of the wheel (R) = 0.250 m; Force applied (F) = 40 N
M.I. of the wheel (I) = 5 kgm2; Angular acceleration of the wheel (α) = ? TB
Since, τ = Iα
Also, τ = FR
∴ Iα = FR m2 3.00kg
or, α = FR = 40 × 0.250 = 2 rad s–2
I 5
m2g
14. A 2.00 kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over
a pulley whose diameter is 0.150 m, to a hanging book with mass 3.00 kg. The system is released
from rest, and the books are observed to move 1.20 m in 0.800s. (a) What is the tension in each part
of the cord? (b) What is the moment of inertia of the pulley about its rotation axis?
Solution:
Mass of textbook (m1) = 2kg
Diameter of pulley (d) = 0.150 m
∴ d
Radius of pulley (r) = 2 = 0.075 m
Mass of hanging book (m2) = 3kg
Distance moved by the system (s) = 1.20 m
Time (t) = 8.00 secs
(a) Tension on horizontal part (T0) = ? (b) M.I. of the pulley (I) = ?
Tension on vertical part (TB) = ? We have, τ = Iα
Here, T0 = m1a → (i) Also, τ = rTB – rT0
And, m2g – TB = m2a → (ii) ∴ Iα = r(TB – T0)
Also, using s = ut + 1 a t2 or, I ar = r(TB – T0) [ a = rα]
2
or, s = 1 a t2, [since, u = 0] or, I = r2(TB – T0) = (0.075)2 × (18.2 – 7.5)
2 a 3.75
or, 1.20 = 1 × a × (0.8)2 ∴ I = 0.0161 kgm2
2
∴ a = 3.75 ms–2
∴ From (i), T0 = 2 × 3.75 = 7.5 N
And, from (ii), TB = m2 (g – a) = 3(9.8 – 3.75)
∴ TB = 18.2 N
15. A playground merry–go–round has radius 2.40 m and a moment of inertia 2100 kgm2 about a
vertical axis through its centre, and turns with negligible friction. (a) A child applies an 18.0 N force
tangentially to the edge of the merry–go–round for 15.0 s. If the merry–go–round is initially at rest,
what it its angular speed after this 15.0 s interval? (b) How much work did the child do on the merry–
go–round? (c) What is the average power supplied by the child?
Solution: Moment of inertia (I) = 2100 kgm2
Radius (r) = 2.40 m
(a) Force applied (F) = 18 N Time (t) = 15 sec
Initial angular speed (ω0) = 0 Final angular speed (ω) = ?
Since, τ = Iα Also, τ = rF
∴ Iα = rF
250 | Essential Physics
or, α = rF = 2.40 × 18 = 0.0206 rads–2
I 2100
Again, θ = ω0t + 12αt2 = 0 × 15 + 1 × 0.0206 × (15)2 = 2.32 rad
2
Now, ω2 = ω02 + 2αθ
or, ω2 = 0 + 2 × 0.0206 × 2.32
∴ ω = 0.309 rad s–1
(b) Work done (W) = ?
(c) Average Power (Pav) = ?
Since, W = τθ = rFθ = 2.40 × 18 × 2.32 = 100.2 J
W 100.2
And, Pav = t = 15 = 6.68 watts
16. A diver comes off a board with arms straight up and legs straight down, giving her a moment of
inertia about her rotation axis of 18 kgm2. She then tucks into a small ball, decreasing this moment
of inertia to 3.6 kgm2. While tucked, she makes two complete revolutions in 1.0s. If she hadn't tucked
at all, how many revolutions would she have made in the 1.5s from board to water?
Solution: I2 = 3.6 kgm2
I1 = 18 kgm2 f2 = 2 revolutions/sec
ω2 = 2πf2 = 2π × 2 = 4π rad s–1
n1 = ? t2 = 1.0 sec
ω1 = ?
t1 = 1.5 sec
Using angular momentum conservation principle,
I1ω1 = I2ω2
or, 18 × ω1 = 3.6 × 4π
∴ ω1 = 2.51 rad s–1
Now, θ1 = 2πn1
or, ω1t1 = 2πn1
∴ n1 = ω1t1 = 2.51 × 1.5 = 0.60 rev.
2π 2× 3.14
17. A solid disk is rolling without slipping on a level surface at a constant speed of 2.50 m/s. If the disk
rolls up a 30.0° ramp, how far along the ramp will it move before it stops?
Solution:
Speed of solid disk (v) = 2.50 ms–1, Angle of inclination (θ) = 30°
Distance travelled along the incline (s) = ?
Here, Total K.E. is converted in to P.E. That
is,
Loss in K.E. = Gain in P.E.
or, 1 mv2 + 1 Iω2 = mgh S h
2 2
ωV
1 mv2 1 1 mr2) ω2
or, 2 + 2 (2 = mgh
or, v2 + 1 r2ω2 = 2gh 30°
2
or, v2 + 1 v2 = 2gh
2
3v2
or, 2 = 2gh
∴ 3v2 3 × (2.50)2
h = 4g = 4 × 9.8 = 0.478 m
Now, sin θ = h ∴ s = h = 0.478 = 0.956 m
s sinθ sin 30°
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