# Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 352: 79

The final angular velocity of the door just after the impact is 1.2 rad/s

#### Work Step by Step

We can find the moment of inertia of the door (and bullet) about the axis of rotation. Let $M_d$ be the door's mass and let $M_b$ be the bullet's mass. $I = \frac{1}{3}M_dR^2+M_bR^2$ $I = \frac{1}{3}(10~kg)(1.0~m)^2+(0.010~kg)(1.0~m)^2$ $I = 3.34~kg~m^2$ We can use conservation of angular momentum to find the final angular velocity of the door just after the impact. The final angular momentum of the door will be equal to the initial angular momentum of the bullet about the axis of rotation. $L_f = L_0$ $I~\omega_f = M_b~v~R$ $\omega_f = \frac{M_b~v~R}{I}$ $\omega_f = \frac{(0.010~kg)(400~m/s)(1.0~m)}{3.34~kg~m^2}$ $\omega_f = 1.2~rad/s$ The final angular velocity of the door just after the impact is 1.2 rad/s.

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