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

Question

A solid disk of mass 2.47 kg and diameter of 28.6 cm is attached to a rod through a diameter of the disk. The disk is free to rotate about the rod, but it is initially at rest. A small lead projectile of mass 67.1 g 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. The projectile sticks in the disk. Determine the rotational rate of the disk in radians per second after the lead projectile strikes the disk.

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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.