Example 1
Problem : A person sitting on a turntable is free to rotate about a vertical axis. Initially, he rotates with angular velocity, “ω”, holding two weights in his outstretched hands. The person, then, folds his hand horizontally to move weights towards axis of rotation. In the process, MI of the system of "turntable, person and weights" is reduced from I to I/2. Find (i) the final angular velocity and (ii) work done by the hands in moving the weights (neglect friction, air resistance).
Solution : The system in question is composed of rigid bodies. However, internally the system can change its MI by virtue of changing positions of hands and weights held in them. The hand maneuvering, however, is internal to the system and does not affect rotation of the system.
Now, let us determine if there is external force and thereby external torque on the system. Force, resulting from rotation at the axis, does not constitute torque. On the other hand, we can consider the whole system as one, the weight of which acts at the center of mass, which falls on the central axis i.e. axis of rotation in this case. Thus, we can say that the weights of the system do not constitute torque.
We may argue that the weights of outstretched hands and weights kept in them constitute torque about the center of mass. Even if we look at the system parts in this fashion, two torques acts in opposite directions due to each of the weights in horizontal direction (apply right hand rule of vector product) and balance each other. We, therefore, can safely conclude that there is no external torque on the system in the direction of rotation and apply law of conservation of momentum in component form in the vertical direction.
L
i
=
L
f
I
i
ω
i
=
I
f
ω
f
L
i
=
L
f
I
i
ω
i
=
I
f
ω
f
⇒
ω
f
=
I
i
ω
i
I
f
⇒
ω
f
=
I
i
ω
i
I
f
From the question,
I
i
=
I
I
i
= I
and
I
f
=
I
2
I
f
=
I
2
,
⇒
ω
f
=
2
ω
i
=

2
ω
⇒
ω
f
=
2
ω
i
=

2
ω
Negative sign indicates that angular velocity is clockwise.
In order to answer the second part of the question, we need to understand the process of folding hands. We see that folding of hands has resulted in the increase of angular kinetic energy of the system as angular speed has doubled. The change in angular kinetic energy of the system is given by :
Δ
K
=
1
2
I
f
ω
f
2

1
2
I
i
ω
i
2
Δ
K
=
1
2
I
f
ω
f
2

1
2
I
i
ω
i
2
Putting values,
⇒
Δ
K
=
1
2
x
I
2
x
(
2
ω
)
2

1
2
I
ω
2
=
1
2
I
ω
2
⇒
Δ
K
=
1
2
x
I
2
x
(
2
ω
)
2

1
2
I
ω
2
=
1
2
I
ω
2
Note that expression of kinetic energy involves magnitude of angular velocity. Hence, sign is not considered.
Now, the basic question is “where from this kinetic energy has come?” We see here that the person is applying force on the weights on his hands. This results in doing work on a part of the system i.e. weights. Work, as we know, transfers energy from one system to another, draws muscular energy of the hands (internal energy of the person) and transfers the same to the system as angular kinetic energy. As no friction is involved in the process – there is no dissipation of energy either. Further, there is no vertical elevation of the weights involved as they are moved horizontally. As such, there is no change in gravitational potential energy of the system. Thus, the kinetic energy change is equal to the work done in accordance with Work – angular kinetic energy theorem (in order to conserve energy of the system) :
⇒
W
=
Δ
K
=
1
2
I
ω
2
⇒
W
=
Δ
K
=
1
2
I
ω
2
Example 2
Problem : A thin rod is placed coaxially within a thin hollow tube, which lies on a smooth horizontal table. The rod and tube are of same mass “M” and length “L”. The rod is free to move within the tube. The system of tube and rod is given an initial angular velocity, “ω”, about a vertical axis at one of its end. Considering negligible friction between surfaces, find the angular velocity of the rod, when it just slips out of the tube.
Solution : The first question that we need to answer is “why does rod slip out of the tube?”. The rod along with the tube is forced to rotate about vertical axis. To understand the process, let us concentrate on the motion of the rod only. Each particle of the rod, say at the far end, tends to move straight (natural tendency). However, there is no radial force available that could bend its path toward the axis of rotation. Hence, the particle tends to keep its path in tangential direction. As such, the particle and all successive particles closer to the axis have the tendency to keep themselves away from the axis of rotation. Eventually, the rod slips away from the axis of rotation.
We may question why the same does not happen with a particle of rigid body in rotation. We know that rigid body does not allow relative displacement of particles. The particles are in place due to intermolecular forces operating on them. The tendency of the particle to move away from the axis of rotation is counteracted by the net intermolecular force on the particle that acts towards the center of rotation. In this case, the intermolecular force meets the requirement of centripetal force for circular motion of the particle about the axis of rotation.
Now, we must check about external torque on the system. The system lies on the smooth horizontal plane. It means that there is no net vertical force (weights of the bodies are balanced by normal forces). The force at the axle does not constitute torque as its moment arm is zero. We, therefore, conclude that there is no external torque on the system and that the problem can be analyzed in terms of conservation of angular momentum.
In the beginning, when the system is given initial angular velocity, the system has certain angular momentum, “
L
i
L
i
”,. During the motion, however, rod begins to slip away from the axis. As such, mass distribution of the system changes. In other words, MI of the system changes. But, the angular momentum of the system ("
L
f
L
f
") remains unchanged as there is no external torque on the system.
Now, applying conservation of angular momentum in vertical direction :
L
i
=
L
f
I
i
ω
i
=
I
f
ω
f
L
i
=
L
f
I
i
ω
i
=
I
f
ω
f
Here,
ω
i
=
ω
ω
i
= ω
. According to theorem of parallel axes, the initial MI of the system, “
I
i
I
i
”, is :
I
i
=
MI of the system about perpendicular at midpoint
+
mass of the system x perpendicular distance between the axes
I
i
=
MI of the system about perpendicular at midpoint
+
mass of the system x perpendicular distance between the axes
I
i
=
{
(
2
M
L
2
12
+
2
M
(
L
2
)
2
}
I
i
=
{
(
2
M
L
2
12
+
2
M
(
L
2
)
2
}
⇒
I
i
=
M
L
2
6
+
M
L
2
2
=
2
M
L
2
3
⇒
I
i
=
M
L
2
6
+
M
L
2
2
=
2
M
L
2
3
Note that we have used "2M" to account for both tube and rod. Now, we again employ theorem of parallel axes in order to obtain MI of the system for the final position, when the rod just slips out of the tube. Final MI of the system, “
I
f
I
f
”, is :
I
f
=
{
(
M
L
2
12
+
M
(
L
2
)
2
+
M
L
2
12
+
M
(
3
L
2
)
2
}
I
f
=
{
(
M
L
2
12
+
M
(
L
2
)
2
+
M
L
2
12
+
M
(
3
L
2
)
2
}
⇒
I
f
=
M
L
2
6
+
M
L
2
4
+
9
M
L
2
4
⇒
I
f
=
M
L
2
6
+
M
L
2
4
+
9
M
L
2
4
⇒
I
f
=
2
M
L
2
+
3
M
L
2
+
27
M
L
2
12
=
8
M
L
2
3
⇒
I
f
=
2
M
L
2
+
3
M
L
2
+
27
M
L
2
12
=
8
M
L
2
3
Putting values in the equation of angular momentum, we have :
⇒
2
M
L
2
3
x
ω
=
8
M
L
2
3
x
ω
f
⇒
2
M
L
2
3
x
ω
=
8
M
L
2
3
x
ω
f
⇒
ω
f
=
ω
4
⇒
ω
f
=
ω
4
QBA (Question based on above) : A thin rod is placed coaxially within a thin hollow tube, which lies on a smooth horizontal table. The rod and tube are of same mass “M” and length “L”. The rod is free to move within the tube. The system of tube and rod is given an initial angular velocity, “ω”, about a vertical axis passing through the center of the system. Considering negligible friction between surfaces, find the angular velocity of the rod, when it just slips out of the tube.
Hint : The question differs in one respect to earlier question. The axis of rotation here is vertical axis through the center of the system instead of the axis at the end of the system. This changes the calculation of MIs before and after. Answer is “ω/7”.