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11.3 Conservation of angular momentum  (Page 3/7)

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Conservation of angular momentum of a collision

A bullet of mass m = 2.0 g is moving horizontally with a speed of 500.0 m / s . The bullet strikes and becomes embedded in the edge of a solid disk of mass M = 3.2 kg and radius R = 0.5 m . The cylinder is free to rotate around its axis and is initially at rest ( [link] ). What is the angular velocity of the disk immediately after the bullet is embedded?

Illustrations of a bullet before and after striking a disk. On the left is the before illustration. The bullet is travelling to the left at 500 meters per second, toward the front edge of a horizontal disk of radius R. The vertical axis through the center of the disc is shown as a vertical line connecting points A above and A prime below the center. On the right is the after illustration. The bullet is embedded in the edge of the disk, which is rotating about the vertical axis through the center. The rotation is counterclockwise as viewed from above.
A bullet is fired horizontally and becomes embedded in the edge of a disk that is free to rotate about its vertical axis.

Strategy

For the system of the bullet and the cylinder, no external torque acts along the vertical axis through the center of the disk. Thus, the angular momentum along this axis is conserved. The initial angular momentum of the bullet is m v R , which is taken about the rotational axis of the disk the moment before the collision. The initial angular momentum of the cylinder is zero. Thus, the net angular momentum of the system is m v R . Since angular momentum is conserved, the initial angular momentum of the system is equal to the angular momentum of the bullet embedded in the disk immediately after impact.

Solution

The initial angular momentum of the system is

L i = m v R .

The moment of inertia of the system with the bullet embedded in the disk is

I = m R 2 + 1 2 M R 2 = ( m + M 2 ) R 2 .

The final angular momentum of the system is

L f = I ω f .

Thus, by conservation of angular momentum, L i = L f and

m v R = ( m + M 2 ) R 2 ω f .

Solving for ω f ,

ω f = m v R ( m + M / 2 ) R 2 = ( 2.0 × 10 −3 kg ) ( 500.0 m / s ) ( 2.0 × 10 −3 kg + 1.6 kg ) ( 0.50 m ) = 1.2 rad / s .

Significance

The system is composed of both a point particle and a rigid body. Care must be taken when formulating the angular momentum before and after the collision. Just before impact the angular momentum of the bullet is taken about the rotational axis of the disk.

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Summary

  • In the absence of external torques, a system’s total angular momentum is conserved. This is the rotational counterpart to linear momentum being conserved when the external force on a system is zero.
  • For a rigid body that changes its angular momentum in the absence of a net external torque, conservation of angular momentum gives I f ω f = I i ω i . This equation says that the angular velocity is inversely proportional to the moment of inertia. Thus, if the moment of inertia decreases, the angular velocity must increase to conserve angular momentum.
  • Systems containing both point particles and rigid bodies can be analyzed using conservation of angular momentum. The angular momentum of all bodies in the system must be taken about a common axis.

Conceptual questions

What is the purpose of the small propeller at the back of a helicopter that rotates in the plane perpendicular to the large propeller?

Without the small propeller, the body of the helicopter would rotate in the opposite sense to the large propeller in order to conserve angular momentum. The small propeller exerts a thrust at a distance R from the center of mass of the aircraft to prevent this from happening.

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Suppose a child walks from the outer edge of a rotating merry-go-round to the inside. Does the angular velocity of the merry-go-round increase, decrease, or remain the same? Explain your answer. Assume the merry-go-round is spinning without friction.

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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