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Conservation of angular momentum (application)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Solving problems is an essential part of the understanding process.

Note: You are viewing an old version of this document. The latest version is available here.

Questions and their answers are presented here in the module text format as if it were an extension of the theoretical treatment of the topic. The idea is to provide a verbose explanation to the answer, detailing the application of theory. Solution presented here, therefore, is treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Hints on solving problems

  • The first thing in attempting question, based on conservation of angular momentum, is to know the torques operating on the system. If there is no torque on the system, then we can apply law of conservation of angular momentum in any convenient direction as we choose in accordance with the inputs given in the problem.
  • In most of the situations, we find that there is external torque in certain direction. However, if the torque is perpendicular to the axis of rotation, then it does not have a component in the direction of the axis of rotation. Therefore, we can employ component form of law of conservation of angular momentum about the axis of rotation.
  • Application of law of conservation of angular momentum in component form can be used in scalar form as there are only two directions, which can be represented by appropriate sign convention.
  • We need to specify direction (assign sign) of angular quantities like angular velocity, momentum and torque etc applying right hand rule.
  • When the system involves both rotation and translation of individual particles or particle like objects, then we should assign angular momentum with respect to an axis for rotation and a point for non-rotational motion of particles.
  • The force at the axis of rotation is external force on the system. However, this force does not have moment arm and, therefore, does not constitute a torque on the system.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the conservation of angular momentum. For this reason, questions are categorized in terms of the characterizing features pertaining to the questions :

  • Change in the distribution of mass about the axis of rotation
  • Conservation of angular momentum about two parallel axes
  • System consisting of both rotational and translational motion

Change in the distribution of mass about the axis of rotation

Example 1

Problem : A person sitting on a turn-table 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 "turn-table, 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).

Figure 1: A person sitting on a turn-table rotates about a vertical axis.
Rotation of the system
 Rotation of the system  (cs3.gif)

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 in parts in this fashion, the torque acts in horizontal direction (apply right hand rule of vector product). The direction of torque is perpendicular to the axis of rotation. 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.

Figure 2: A person folds his hands drawing weights towards the axis.
Rotation of the system
 Rotation of the system  (cs4.gif)

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.

Figure 3: The rod is free to move within the tube.
Rotation of coaxial tub and rod
 Rotation of coaxial tub and rod  (cs1.gif)

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.

Figure 4: The rod slips away from the axis of rotation.
Rotation of coaxial tub and rod
 Rotation of coaxial tub and rod  (cs2.gif)

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 inter-molecular force on the particle that acts towards the center of rotation. In this case, the inter-molecular 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 mid-point + mass of the system x perpendicular distance between the axes I i = MI of the system about perpendicular at mid-point + 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 :

Figure 5: The rod slips away from the axis of rotation.
Rotation of coaxial tub and rod
 Rotation of coaxial tub and rod  (cs2.gif)

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”.

Conservation of angular momentum about two parallel axes

Example 3

Problem : Two uniform disks of masses “4M” and “M” and radius “2R” and “R” respectively are connected with a mass-less rod as shown in the figure. Initially, the smaller disk is rotating clockwise with angular velocity “ω” with the help of an electric motor mounted on it, whereas the larger disk is stationary. At an instant, the smaller disk reverses the direction of rotation, keeping its angular speed same. Find the angular velocity of the larger disk.

Figure 6: The question involves two axes of rotation : one for larger plus smaller disks (z-axis) and other for smaller disk (z’-axis).
Rotations about two parallel axes
 Rotations about two parallel axes  (cs5.gif)

Solution : This question involves two axes of rotation : one for larger plus smaller disks (z-axis) and other for smaller disk (z’-axis). The treatment of angular momentum, in this case, differs from other cases in one important aspect. The constituents of the system are defined with respect to axis of rotation – not with respect to objects.

The body about the z - axis is a composite body comprising of larger and smaller disks. The smaller disk is part of this composite body. It (part) rotates about its own axis (z’) parallel to z-axis. This smaller disk is another constituent of the system. Thus, system comprises of two rotating bodies :

  1. Composite body comprising of larger and smaller disk rotating about z-axis
  2. Smaller disk rotating about z’-axis

The MI of the composite body is constant and is independent of the motion of smaller disk, because mass distribution about z-axis does not change by the rotation of smaller disk.

Now, let us check external torques on the system. The weight of smaller disk constitutes a torque, but it is perpendicular to the axis of rotation. As such, we can employ law of conservation of momentum about z and z’ axes. Since, two axes point in the same direction, the net angular momentums before and after are simply the arithmetic sum of angular momentums about the two axes. Let “C” and “S” subscripts denote the composite body and smaller disk respectively, then :

L i = L f L Ci + L Si = L Cf + L Sf L i = L f L Ci + L Si = L Cf + L Sf

The moment of inertia of the smaller disk about z’-axis is :

I S = M R 2 2 I S = M R 2 2

Now, the moment of inertia, “ I C I C ”, of the composite body is given by the theorem of parallel axes as :

I C = 4 M x ( 2 R ) 2 2 + M R 2 2 + M ( 2 R ) 2 = 12.5 M R 2 I C = 4 M x ( 2 R ) 2 2 + M R 2 2 + M ( 2 R ) 2 = 12.5 M R 2

Let “ ω f ω f ” be the final angular velocity of the composite body. Substituting in the equation of conservation of angular momentum,

I C x 0 - I S x ω = I C x ω f + I S x ω I C x ω f = 2 I S ω I C x 0 - I S x ω = I C x ω f + I S x ω I C x ω f = 2 I S ω

ω f = 2 I S x ω I C = 2 ( M R 2 2 ) ω 12.5 M R 2 = ω 12.5 ω f = 2 I S x ω I C = 2 ( M R 2 2 ) ω 12.5 M R 2 = ω 12.5

System consisting of rotational and translational motion

Example 4

Problem : A rod of mass "M" and length, "a" is pivoted at middle point "O" such that it can rotate in horizontal plane. A mud ball of mass "M/12", moving in horizontal plane strikes the rod at the point "P" with a speed "v" as shown in the figure. If the mud ball sticks to the rod after collision, find the magnitude of the angular velocity of the rod immediately after the ball strikes the rod.

Figure 7: A mud ball, moving in horizontal plane strikes the rod at the point "P" with a speed "v".
Collision with rotating rod
 Collision with rotating rod  (cs6.gif)

Solution : We first need to determine whether there is any external torque on the system of rod and mud ball. The forces during collision constitute internal torques. On the other hand, the force at the pivot passes through the axis of rotation. It has no moment arm about it and as such, it does not constitute a torque. The gravitation pull on the mud ball is in vertical direction. This force constitute a torque in x-direction (apply right hand rule to assess the direction). We can, therefore, conclude that there is no external torque in z -direction on the system before, during and after the collision.

We observe here that the rod is capable to rotate in the horizontal plane. Also, the motion of the mud ball is in the horizontal plane. The directions of angular momentums for both rod and the mud ball are in vertical direction (z-direction). We can, therefore, apply conservation of angular momentums along the direction of axis of rotation i.e. perpendicular to the planes of motion. This enables us to use the component form of conservation law in z-direction as scalar expression with only two directions. Let “R” and “B” subscripts denote rod and the mud ball respectively, then :

L i = L f L Ri + L Bi = L Rf + L Bf L i = L f L Ri + L Bi = L Rf + L Bf

Here, the initial angular momentum of the rod is zero as it is stationary. Initial angular momentum of the mud ball just before it strikes the rod is given by,

L Bi = - m r v = - M v a 12 x 4 L Bi = - m r v = - M v a 12 x 4

Note in the above equation that mass of the mud ball is M/12 and its moment arm is "a/4". Negative sign to show that initial angular momentum of the mud ball is clockwise.

Since the mud ball sticks to the rod, two entities become one and move with a common angular velocity. Let us now consider their common angular velocity is "ω". The final angular momentum of the rod is given by :

L Rf = I R ω = M a 2 ω 12 L Rf = I R ω = M a 2 ω 12

The angular momentum of the mud ball as part of the rotating rod is given by :

L Bf = M 12 x ( a 4 ) 2 L Bf = M 12 x ( a 4 ) 2

Putting these values in equation of conservation of angular momentum, we have :

0 - M v a 12 x 4 = ( M a 2 12 + M 12 x a 2 16 } ω = 17 M a 2 ω 12 x 16 0 - M v a 12 x 4 = ( M a 2 12 + M 12 x a 2 16 } ω = 17 M a 2 ω 12 x 16

ω = - 12 x 16 M v a 48 x 17 M a 2 = - 4 v 17 a ω = - 12 x 16 M v a 48 x 17 M a 2 = - 4 v 17 a

Thus, magnitude of angular velocity of the mud ball just after the collision is :

ω = 4 v 17 a ω = 4 v 17 a

Note:

This example illustrates application of conservation of angular momentum of a system that consists of a rotating body and a particle like mud ball in translation. The angular momentum of rotating rod is about vertical axis and that of mud ball is about the point on this axis. The angular momentums are each in z-direction. As such, it is possible here to apply conservation of angular momentum in component form along z-axis, in which component of external torque is zero. It is important to realize here that the moving ball has torque about “O”, but along z-direction.

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