CHAPTER 6: ROTATION OF RIGID BODY

6.2 Equilibrium of a uniform rigid body

Torque ( ) or (moment of a force)

Meaning :

• Torque is the turning effect produced by a force .
• Torque is measure of how much a force acting on

an object causes that object to rotate.

Definition :
• Torque is defined as the vector product between

lever arm, r and the force, F.

 rF

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Physical meaning of cross (vector) product for torque :

• is magnitude of torque :   r F sin θ

F : force
r : lever arm (distance from axis of rotation to F)

θ : angle between r and F

• Torque is a vector quantity.
• Unit of torque is Nm

Notes :
i. r and F sin θ must perpendicular to each other
ii. θ ≤ 90 o

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Magnitude of torque depends on :

i. the magnitude of force applied,
ii. the length of the lever arm,
iii. the angle between the force and the lever arm.

Sign convention of torque :
• Depends on direction of rotation it would cause

~ in clockwise rotation (Ʈ = -ve)
~ in anticlockwise rotation ( Ʈ = +ve)

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Direction of torque :
Use right hand grip rule

i) thumb ; direction of torque
ii) fingers ; direction of rotation

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Torque is (+) when the Torque is (–) when the
force tends to produce force tends to produce a
a anticlockwise clockwise rotation about
rotation about the axis. the axis.

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55

Difficult to open the door.

Easier to open the door.

56

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Example 6.9 :

(a) A mechanic applies a force of 200 N as shown in figure
above. Calculate the torque produced?

(b) Would it be easier to undo/loosen the nut with a longer or
shorter spanner? Explain your answer.

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Solution : This is the point where we
take the axis of rotation

r

(a) Torque, τ  r F sin θ

 200(0.28) sin 50  43Nm

(b) From : τ  r F sin θ F  cons tan t,  r 

This makes it easier to undo/loosen the nut with a longer
spanner.

Example 6.10 : 2.

Calculate torque on each case
1.

3. 4.

Solution : 2.

1.

= θ = θ
= 1 60 90 = 3 100 90
= 60 = 300
= + = +

Solution : 4.

3.

= θ = θ
= 5 90 0 = 1 50 30
= = 25
= −

Example 6.11 :

A +
90º
80 cm –

100 cm

Calculate the net torque at point A.

Solution :

  rF sin

1(100102 )(4)sin 90

 4Nm
 2 (80102 )(6)sin 60

 4.157Nm

  1  2  0.157 Nm

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Exercise :

1. For the each figure below, calculate the torque at point O. Given
the force applied F is 30 N and the length of the rod is 4 m.

(a) O (b) O 30°
F F
(c)
F (d)
60° F

O 60° 2m
O

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(e) (f) 2.0 m

135° FO

O 140°

2.5 m F

(g) (h) F
F O 60°

O
2.5 m

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Equilibrium of a uniform rigid body
What is meant by rigid body ?
Rigid body [Jasad Tegar] is any real body which has a
definite size and shape and can move.

In ideal cases, rigid bodies do not change shapes & sizes.
They do not bend, dent or stretch when ever any force or
forces are applied on them.

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A rigid body need not to be in static equilibrium
although the resultant force on it is zero.
Consider a body under the action of 2 equal & opposite
forces, F ( as shown in figure (a) & (b) )

The forces cancel out to give zero net force ( ∑F = 0 )
But these two forces acting at different points on the body. The
two forces will create a turning effect. ( known as TORQUE)

Conditions for equilibrium of a rigid body
(1) Vector sum of all forces acting on a rigid the body is zero.


F  0


Fx  0, Fy  0

(2) Vector sum of all torques about any axis of rotation is zero.

 any point 0

τx  0,τy  0

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Problem – Solving Strategy

1. Sketch a free body diagram showing & labeling all
external forces acting on the body.

2. Resolving the forces into x & y components.

Then apply the 1st condition : 
Fx  0 ; Fy  0

3. Take torque about any convenient point in the system.
Then apply the 2nd condition.

  0

4. Solve for the unknown quantities.

Some useful tips for you !

If there are 2 or more unknown
forces, chooses a point so that the
lines of action of one or more of
the unknown forces pass through
that point.
Such choice simplifies the torque
equation because the torques
produced by these forces will
equal to zero.

The words ‘rough’ & ‘smooth’ Weight of the object is
have a particular meaning in
physics. acting at its center of
‘rough’ need to take account of
friction. gravity. For uniform
‘smooth’ treat it as friction free
( no friction ) object, weight always

acting at the center of the

object. 71

PROBLEM SOLVING
EXAMPLE

Example 6.12 :
The see saw in the diagram is balanced. Determine the
value of weight, W

N

+ ––

Solution :
Taking torque about the pivot.
W tends to rotate the system ACW 300N & 550N tend to rotate
the system CW

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N

+ ––

  rF sin

Since system in equilibrium, thus we have:  0pivot

    0W 30N0 55N0 N

W(1.5)s9i3n0(1.0)0s9i50n0(1.5)5s9i0n0
1.5W300825

W  750 N

Example 8.13 : 75 cm B
A 35 cm O

W2 W1

A hanging flower basket having weight, W2 =23 N is hung out over the
edge of a balcony railing on a uniform horizontal beam AB of length
110 cm that rests on the balcony railing. The basket is
counterbalanced by a body of weight, W1 as shown in Figure 8.13. If
the mass of the beam is 3.0 kg, calculate

a. the weight, W1 needed,
b. the force exerted on the beam at point O.
(Given g =9.81 m s2)



Solution :

m  3 kg;W2  23 N

Free body diagram of the beam A 0.35 m 0.20 m B

0.75 m 1–

  rF sin CG

O
+ 2 0.55 m m– റ 0.55 m

Let point O as the rotation axis.

a. Since the beam remains at rest thus the system in equilibrium

τO  0

−0.75 1 sin 90 + (0.35)(23)sin90 − (0.20)(3)(9.81)sin90 = 0

W1  2.89 N



Solution : 0.20 m 0.75 m
b.
A 0.35 m B

CG 1–

O

+ 2 0.55 m m– റ 0.55 m

Fy  0

N  23  (3)(9.81) W1  0

N  23  29.43 - 2.89  0

= 55.32 N

Example 8.13 : A

A ladder of length 12 m and mass 150 N rest smooth B
against a smooth wall so that it makes an wall
angle of 60o with the rough floor.
a) Find the reaction force exerted on ladder rough floor

from the floor.
b) Calculate the force exerted by the wall.
c) Find the frictional force between the

ladder and the floor.
d)Find the coefficient of static friction

between the rod and the floor.

Solution :   rF sin A NA

Since the ladder in equilibrium thus

a) F  0 N B  150N
y
NB 150  0

b) τ  0 N A  43.30N 12 m
B
(6)(150)sin 30  (12)N A sin 60  0 C NB
G
c) F  0 B
x 150N
NA  fs  0 fs  N A  43.30N
6.0 m 60o


fs

d) fs  N B   0.29
43.30  (150)

Example 8.14 :

A floodlight of mass 20.0 kg in a park is
supported at the end of a 10.0 kg uniform
horizontal beam that is hinged to a pole as
shown in Figure. A cable at an angle 30 with
the beam helps to support the light.
a. Sketch a free body diagram of the beam.
b. Determine

i. the tension in the cable,
ii. the force exerted on the beam by the pole.

Solution : m f  20.0 kg; mb  10.0 kg റ
a. The free body diagram of the beam :
0.5l 30

O CG
റ

l

b. Let point O as the rotation axis.   rF sin റ

i. τ  0
O

 (0.5l)(10)(9.81) sin 90  (l)(20)(9.81) sin 90  (l)T sin 30  0

T  490.5N

Solution : റ

b. ii. F 0 O 0.5l 30
x

 T cos30  Sx  0 CG
റ
Sx  424 N
l

റ

F 0
y

S y  (490.5) sin 30  (10)(9.81)  (20)(9.81)  0

= 49.05N

Solution :
b. ii. Therefore the magnitude of the force is

S  Sx2  Sy2

= 424 2 + 49.05 2
= 426.83 N

and its direction is given by y

θ  tan 1 S xy49.05 റ
S
θ = tan−1 424 6.600
x

θ = 6.60∘

Exercise : റ 1 B

1. a

A റ 2

b

D Cγ

റ 3

Figure above shows the forces, F1 =10 N, F2= 50 N and F3= 60 N are
applied to a rectangle with side lengths, a = 4.0 cm and b = 5.0 cm.
The angle  is 30. Calculate the resultant torque about point D.
ANS. : -3.7 N m

2.

A see-saw consists of a uniform board of mass 10 kg and length 3.50 m
supports a father and daughter with masses 60 kg and 45 kg,
respectively as shown in Figure. The fulcrum is under the centre of
gravity of the board. Determine
a. the magnitude of the force exerted by the fulcrum on the board,
b. where the father should sit from the fulcrum to balance the system.
ANS. : 1128 N; 1.31 m

3.

A traffic light hangs from a structure as show in Figure 8.18. The uniform
aluminum pole AB is 7.5 m long has a mass of 8.0 kg. The mass of the
traffic light is 12.0 kg. Determine
a. the tension in the horizontal massless cable CD,
b. the vertical and horizontal components of the force exerted by the
pivot A on the aluminum pole.
ANS. : 248 N; 197 N, 248 N

4. 30.0 cm 50.0

15.0 cm

A uniform 10.0 N picture frame is supported by two light റ string as shown
in Figure 8.19. The horizontal force, F is applied for holding the frame in
the position shown.
a. Sketch the free body diagram of the picture frame.
b. Calculate

i. the tension in the ropes,
ii. the magnitude of the horizontal force, F .
ANS. : 1.42 N, 11.2 N; 7.20 N

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5. A

A uniform ladder AB of length 10 m and mass 5.0 kg smooth B
leans against a smooth wall as shown in Figure. wall
The height of the end A of the ladder is 8.0 m from
the rough floor. rough floor
a. Determine the horizontal and vertical forces

the floor exerts on the end B of the ladder
when a firefighter of mass 60 kg is 3.0 m from B.
b. If the ladder is just on the verge of slipping when
the firefighter is 7.0 m up the ladder, Calculate
the coefficient of static friction between ladder
and floor.

[Ans; 151 N, 638 N, 0.514]

Learning Outcome:

6.3 Rotational dynamics

At the end of this chapter, students should be able to:

a) Define and use moment of inertia, I  mr 2

b) Use the moment of inertia of a uniform rigid body

(sphere, cylinder, ring, disc, and rod).

c) Determine the moment of inertia of a flywheel.

(Experiment 4 : Rotational motion of rigid body)

d) State and use net torque,   I

6.3 Rotational dynamics

Moment of inertia, I
• Figure shows a rigid body about a fixed axis O with angular velocity

. 

m1
mn r1
rn r2 m2

Or3 m3

• is defined as the sum of the products of the mass of each
particle and the square of its respective distance from the
rotation axis.

90

n

I  m1r12  m2r22  m3r32  ...mn rn2  miri2
OR
i1

where I : moment of inertia of a rigid body about rotation axis

m : mass of particle
r : distance from the particle to the rotation axis
• It is a scalar quantity.

• Moment of inertia, I in the rotational kinematics is analogous to
the mass, m in linear kinematics.

• The S.I. unit of moment of inertia is kg m2.

• The factors which affect the moment of inertia, I of a rigid body:

a. the mass of the body,

b. the shape of the body,

c. the position of the rotation axis.

Moments of inertia of various bodies

• Table shows the moments of inertia for a number of objects about
axes through the centre of mass.

Shape Diagram Equation

Hoop or ring or CM ICM  MR 2
thin cylindrical

shell

Solid cylinder or CM I CM  1 MR 2
disk 2

Moments of inertia of various bodies

• Table shows the moments of inertia for a number of objects about
axes through the centre of mass.

Shape Diagram Equation

Uniform rod or CM I CM  1 ML2
long thin rod 12
with rotation axis
through the
centre of mass.

Solid Sphere CM I CM  2 MR 2
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Moments of inertia of various bodies

• Table shows the moments of inertia for a number of objects about
axes through the centre of mass.

Shape Diagram Equation

Hollow Sphere CM I CM  2 MR 2
or thin spherical 3

shell

Relationship between torque, and angular acceleration, 

• Consider a force, F acts on a rigid body freely pivoted on an axis
through point O as shown in Figure.

a1 m1 റ

an mnrn r1
aO r2 m2 2

• The body rotates in the anticlockwise direction and a nett torque is
produced.

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• A particle of mass, m1 of distance r1 from the rotation axis O will
experience a nett force F1 . The nett force on this particle is

F1  m1a1 and a1  r1α

• The torque on the mass m1 is F1  m1r1α

1  r1F1 sin 90

• The total (nett) torque on the rigid 1  m1r12

body is given by

  m1r12  m2r22  ...  mnrn2

    n  n
  i 1 mi ri 2 and
 miri2  I

i 1

  I

• A particle of mass, m1 of distance r1 from the rotation axis O will
experience a nett force F1 . The nett force on this particle is

F1  m1a1 and a1  r1α

• The torque on the mass m1 is

F1  m1r1α

1  r1F1 sin 90

1  m1r12

• The total (nett) torque on the rigid body is given by

  m1r12  m2r22  ...  mnrn2

  n  and n
  i 1 mi ri 2
 miri2  I

i 1

  I

• From the equation, the nett torque acting on the rigid body is
proportional to the body’s angular acceleration.

• Note : Nett torque ,   I

is analogous to the

Nett force,  F  ma

PROBLEM SOLVING
EXAMPLE