A solid wheel of mass 2.0 kg and radius 18 cm is free to rotate without friction

Question

A solid wheel of mass 2.0 kg and radius 18 cm is free to rotate without friction about a horizontal axis through its center. It begins at rest. A bullet of mass 5.0 g and velocity 330 m/s enters the wheel at its rim and passes through it on a line 15 cm from the center and parallel to the line representing its diameter. (I.e., the path is in the plane of the wheel.) The bullet emerges with a nal speed of 220 m/s. What are the angular velocity, angular momemtum, and kinetic energy of the wheel afterwards? Is the energy of motion conserved?

Explanation / Answer

Momentum conservation
5*10-3 * 15*10-2 *330 = (1/2)2*182*10-4 + 5*10-3 * 15*10-2 *220

182*10-4 = 0.0825

= 2.5 rad/s

Angular momentum = (1/2)2*182*10-4* 2.5

                             = 0.081

kE of wheel =(1/2) I2

                 =(1/2)(1/2)2*182*10-4 *2.52

                 = 0.10125

Initial KE =(1/2)mv2 = (1/2)5*10-3 * 3302 = 272.25

final KE = 0.10125 + (1/2)5*10-3 * 2202 = 121.10125

Not conserved energy

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