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ROTATIONAL DYNAMICS

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Illustration: 1

A one – miece cylinder is shaped as in figure with a core section producing from the larger drum. The cylinder

is free to rotate around the central axis shown in the drawing. A rope wrapped around the drum, of radius R
1
,

enters a force T
1
to the right on the cylinder. A rope warped around the core, of radius R

2
, enters a force T

2

downward on the cylinder. (a) what is the net torque acting on the cylinder about the rotation axis (which is the

z – axis in figure)? (b) suppose T
1
= 5N, R

1
= 1.0 m, T

2
= 6n, and R

2
= 0.50 m. what is the net torque about the

rotation axis and which way does the cylinder rotate if it starts from rest

Figures

Solution:

(a) The torque due to T
i
is – R

1
T

1
. It is negative because it tends to produce a clockwise rotation from the point of

view in figure. The torque due to T
2
is T

2
= R

2
T

2
and is positive because it tends to produce a counter

clockwise 1 2 2 2 1 1net R T R T� � �� � � � �

(b) 2 2 1 1net R T R T� � �

= (0.5) (6) – (1) (5) = - 2 N.m

Because the net torque is negative, the cylinder rotates clockwise from rest.

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Illustration: 2

A triangular plate of uniform thickness and density is made to rotate about an axis perpendicular to the plane of

the paper and (a) Passing through A (b) passing through D by the application of the same force F at c (mid –

point of AB) as shown in fig. In which case the torque is more?

Solution:

2 2
A D

l l
F and F� �
� � � �

� � �� �




� Both are equal in magnitude

Illustration: 3

A rectangular plate in the x – y plane is shown in the figure. A force F = 30 N is applied at point B. Find the

moment of F about (a) the origin O (b) the point C (c) x – axis, y – axis, and z – axis.
0

30� �

Solution :

(a) 0 cos 30 2 sin 30 1F F� � �


3 1
30 2 30

2 2

� �
� �
� �




1
30 3 .

2
N m

� �
� � ��




(b) cos 2 30 3c F Nm� � �
� �

0 0x yand� �� �

1
30 3 .

2
z N m�

� �
� � ��




Illustration: 4

A force 2 3 4F i j k N� � � is applied to a point having position vector 3 2 .r i j k m� � � Find the torque due to

the force about the axis passing through origin. Figure

Solution:

r F� �

3 2 2 9 4i j k i j k� � �
� �

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2 2 21 1

2 2
R i i i i i i

i i

K K m V m r �� � �� � �

2 2 21 1

2 2
i im r I� �� �� .

Therefore, we can express the kinetic energy of the rotating rigid body as

21

2
RK I ��

Although we shall commonly refer to the quantity
21

2
I� as the rotational kinetic energy, it is not a new form

of energy. It is ordinary kinetic energy because it was derived from a sum over individual kinetic energies of the

particles contained in the rigid body. It is a new role for kinetic energy for us, however, because we have only

considered kinetic energy associated with translation through space so far. On the storage side of the continuity

equation of energy we should now consider that the kinetic energy term should be the sum of the changes in

both translational and rotational kinetic energy. Thus, in energy versions of the system models, we should keep

in mind the possibility of rotational kinetic energy.

Check your Concept :

1. Two spheres, one hollow and one solid, are rotating with the same angular speed about their centers.

Both spheres have the same mass and radius. Which one, if either, has the higher rotational kinetic

energy?

Answer: The hollow sphere has the higher rotational kinetic energy. Its mass is located near the spherical surface,

and the mass of the solid sphere is located throughout the volume. This results in a higher moment of inertia for

the hollow sphere.

Illustration 1: Four small spheres are fastened to the corners of a frame of ineligible in the xy plane. (a) If the

rotation of the system occurs about the y axis with an angular speed � , find the rotational kinetic energy about
this axis. (b) Suppose the system rotate in xy plane about an axis through O (the z axis). Calculate the rotational

energy about this axis.

Figure.

Solution:

a. The two spheres of mass m that lie on the y axis do not contribute to I
y
. Because they are modeled as particles,

r
i
= 0 for these spheres about this axis. We have for the tow spheres on the x axis,

2 2 2 2
2y i i

i

I m r Ma Ma Ma� � � ��

Therefore, the rotational kinetic energy about the y axis is

2 2 2 2 21 1
2

2 2
R yK I Ma Ma� � �� � �

The fact that the spheres of mass do not enter into this result makes sense, because they have no

motion about the chose axis of rotation; hence, they have no kinetic energy.

b.
2 2 2 2 2 2

2 2y i i
i

I m r Ma Ma Mb Ma Mb� � � � � ��

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2 2 2 2 2 21 1
2 2

2 2
R zK I Ma Mb Ma Mb� �� � � � �

Comparing the result for (a) and (b), we see that the moment of inertia and therefore the rotational

energy associated with a given angular speed depend on the axis of rotation.

Illustration 2: A uniform rod of length L and mass M is free to rotate on a frictionless pin through

one end. The rod is released from rest in horizontal position. (a) What is the angular speed of the

rod at its lowest position? (b) Determine the tangential speed of the center of mass and the

tangential speed of the lowest point on the rod in the vertical position.

Solution: Reasoning. We consider the rod and the Earth as an isolated system. Consider the

mechanical energy of the system. When the rod is horizontal, it has no rotational kinetic energy. Let

us also define this position of the rod as representing the zero gravitational potential energy of the

system. When the center of the mass of the rod is at the lowest position, the potential energy of the

system is – Mg L/2 and the rod has rotational kinetic energy (1/2) I 2� , where I is the moment of

inertia about the pivot. Using conservation f mechanical energy, we have

i i f fK U K U� � �

1 1
0 0

2 2
MgL MgL� � � �

2 21 1 1

2 3 2
M L Mgl�

� �
��




3 g

L
� � .

1
3

2 2
CM

L
v r g L� �� � �

The lowest point on the rod, because it is twice as far from the pivot as the center of mass, has a

tangential speed equal to

2 3CMv g L�

A rod of length L is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the

rod when it is in vertical position is:

a. 2 /g L b. 3 /g L c. / 2g L d. /g L

Problem 40. A thin rod of length L and mass M is held vertically with one end on the floor and is allowed to fall. Find

the velocity of the other end when it hits the floor, assuming that the end of the floor does not slip.

Solution: Let M be the mass L the length of the stick. When it is held vertically, its centre of mass is at a higher (L/

2) from the floor, so that the potential energy of the stick is Mg (L/2). On releasing, the stick falls, i.e., it rotates

about the end on the floor and the potential energy is converted into rotational kinetic energy
21

2
I� , where I is

the moment of inertia of the rod about the lower end and � the angular velocity when it hits the floor. Thus by
conservation of mechanical energy.

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a. v
1
= v

2
b. v

1
> v

2
c. v

1
< v

2
d. data is insufficient .

solution : (C)

From conservation of angular momentum about point of contact:

0I I m Rv� �� �

Or, 0
v

I I m R v
R

� � � �

v

v = R

Or,

0I
v

I
m r

R






Or,

0
v

I m R

R I






Now, solid sphere hollowI I�

solid hollowv v� �

1 2v v� �

21. Let t
1
and t

2
be the times when pure rolling of solid sphere and of hollow sphere is started. Then

a. 1 2t t� b. 1 2t t� c. 1 2t t� d. None of these.

Solution: (B) Friction force ' '� mg acts in forward direction till pure rolling is started. Hence linear acceleration .

m g
a g

m


��

� v = at

Or,

0

1

1

v
t

m Ra
g

R




� �

� �
��




Again solid hollowI I�

solid hollowt t� �

Or, t
1
< t

2

22. Which of the following quantity has same magnitude for the solid sphere at start and when it start

pure rolling.

a. kinetic energy

b. momentum

c. angular momentum about CM

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d. angular momentum about bottom most point

Solution: (D) Clearly, angular momentum about the bottom most point is conserved, since the net torque about that

point is zero.

23. Now solid sphere and hollow sphere’s centre are joined through a light rigid rod and placed on a highly

rough inclined plane, what type of force will be present in rod?

a. Tensile b. Compressive

c. Impulsive d. Bidirectional.

Solution: (B)

Motion of both the object is rolling.

sin sin
,

2 2
1 1

5 3

solid hollow

g g
a a

� �
� �

� �

solid hollowa a� �

Hence force will be compressive.

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