A solid disk of mass 2.47 kg and diameter 28.6 cm is attached to a rod through a

Question

A solid disk of mass 2.47 kg and diameter 28.6 cm is attached to a rod through a
diameter of the disk as shown in the accompanying diagram. The disk is free to
rotate about the rod, but it is initially at rest. A small lead projectile of mass 67.1
grams is shot into the disk at a speed of 57.1 m/s. The lead projectile strikes the
disk at a point 8.25 cm to the right of the centerline of the disk (as shown by the X
in the diagram). The projectile sticks in the disk. Determine the rotational rate of
the disk (in radians per second) after the lead projectile strikes the

Explanation / Answer

   Mass of solid disc is m1 = 2.47 kg diameter of disc d is = 28.6 cm = 0.286 m radius of disc is   r1 = d/2 = 0.286 / 2 =0.143 m from given system the moment of inertia I1 = m r12/2                                                               = 1/2 (2.47)(0.143)2                                                               =0.02525 kg m2 when  small lead projectile of mass m2= 67.1 g = 0.067 kg is shot into the disk at a speed is v = 57.1 m/s. the lead projectile strikes the disk at a point from the center of disc is   r2 = 8.25 cm =0.0825 m at this point moment of inertia of disc is I2 =1/2 m2 r22                                                               =1/2 ( 2.47 kg )(0.0825 m )2                                                               = 0.008405 kg m2 at this striking point velocity v = r2 2                                           2 = v / r2 =57.1 / 0.0825                                                      = 692.12 rad/s according to law of conservation of angular momentum             I1 1 = I2 2             1 = I2 2 /   I1                  = (0.008405 kg m2 )( 692.12 rad/s ) / (0.02525 kg m2)                  =230.38 rad/s Mass of solid disc is m1 = 2.47 kg diameter of disc d is = 28.6 cm = 0.286 m radius of disc is   r1 = d/2 = 0.286 / 2 =0.143 m from given system the moment of inertia I1 = m r12/2                                                               = 1/2 (2.47)(0.143)2                                                               =0.02525 kg m2 when  small lead projectile of mass m2= 67.1 g = 0.067 kg is shot into the disk at a speed is v = 57.1 m/s. the lead projectile strikes the disk at a point from the center of disc is   r2 = 8.25 cm =0.0825 m at this point moment of inertia of disc is I2 =1/2 m2 r22                                                               =1/2 ( 2.47 kg )(0.0825 m )2                                                               = 0.008405 kg m2 at this striking point velocity v = r2 2                                           2 = v / r2 =57.1 / 0.0825                                                      = 692.12 rad/s according to law of conservation of angular momentum             I1 1 = I2 2             1 = I2 2 /   I1                  = (0.008405 kg m2 )( 692.12 rad/s ) / (0.02525 kg m2)                  =230.38 rad/s Mass of solid disc is m1 = 2.47 kg diameter of disc d is = 28.6 cm = 0.286 m radius of disc is   r1 = d/2 = 0.286 / 2 =0.143 m from given system the moment of inertia I1 = m r12/2                                                               = 1/2 (2.47)(0.143)2                                                               =0.02525 kg m2 when  small lead projectile of mass m2= 67.1 g = 0.067 kg is shot into the disk at a speed is v = 57.1 m/s. the lead projectile strikes the disk at a point from the center of disc is   r2 = 8.25 cm =0.0825 m at this point moment of inertia of disc is I2 =1/2 m2 r22                                                               =1/2 ( 2.47 kg )(0.0825 m )2                                                               = 0.008405 kg m2 at this striking point velocity v = r2 2                                           2 = v / r2 =57.1 / 0.0825                                                      = 692.12 rad/s according to law of conservation of angular momentum             I1 1 = I2 2             1 = I2 2 /   I1                  = (0.008405 kg m2 )( 692.12 rad/s ) / (0.02525 kg m2)                  =230.38 rad/s             1 = I2 2 /   I1                  = (0.008405 kg m2 )( 692.12 rad/s ) / (0.02525 kg m2)                  =230.38 rad/s

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