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Suppose that a very simple ballistic pendulum is made by attaching a 2.5 kg solid wooden cube 0.17 m on each side to a 0.25 m long metal rod with a mass of 0.75 kg.  The rod is attached to the block in the center of one of the faces. The assembled pendulum (rod plus block) is hung straight down from a smooth axle so that the block is at the bottom.  The pendulum can rotate freely about the axle.  A 4.0 g bullet moving perfectly horizontally (and perpendicular to the axle) at 920 m/s strikes the block in the center of one of the faces.  The bullet embeds itself in the exact center of the block.  What is the angular speed of the entire system about the axle immediately after the collision?

System:  Bullet plus pendulum.  External influences are neglected under the assumption that the collision forces dominate.

Model:  [Constant Angular Momentum].

Approach:  Angular momentum conservation implies:

{latex}\begin{large}\[ m_{b}v_{b,{\rm i}}(L+d/2) + I_{p}\omega_{\rm i} = (m_{b}(L+d/2)^{2} + I_{p}) \omega_{\rm f} \]\end{large}{latex}

To use this equation to find the final angular speed, we must first determine the [moment of inertia] of the pendulum.  The pendulum is composed of two parts: a rod (assumed thin) rotated about one end and a block rotated about an axis displaced by a distance _L_+_d_/2 from its center of mass.  The contribution of the rod is a standard one:

{latex}\begin{large}\[ I_{r} = \frac{1}{3} m_{r}L^{2} \]\end{large}{latex}

To get the contribution of the cubic block we must use the [parallel axis theorem]:

{latex}\begin{large}\[ I_{c} = I_{c,cm} + m_{c}(L+d/2)^{2} = \frac{1}{12}m_{c}(d^{2}+d^{2}) + m_{c}(L+d/2)^{2} = \frac{1}{6}m_{c}d^{2}+m_{c}(L+d/2)^{2}\]\end{large}{latex}

We therefore know that:

{latex}\begin{large}\[ I_{p} = I_{r} + I_{c} = \frac{1}{3}m_{r}L^{2} +\frac{1}{6}m_{c}d^{2} + m_{c}(L+d/2)^{2})\]\end{large}{latex}

Inputting into the conservation of momentum equation and solving for ω, we find:

{latex}\begin{large}\[ \omega_{\rm f} = \frac{m_{b}v_{b,{\rm i}}(L+d/2)}{m_{b}(L+d/2)^{2} + \frac{1}{3}m_{r}L^{2}+ \frac{1}{6}m_{c}d^{2}+m_{c}(L+d/2)^{2}} =\left(\frac{1}{L+d/2}\right)\frac{m_{b}v_{b,{\rm i}}}{m_{b}+ m_{c} + \frac{1}{3}m_{r}\frac{\displaystyle L^{2}}{\displaystyle (L+d/2)^{2}} + \frac{1}{6}m_{c}\frac{\displaystyle d^{2}}{\displaystyle (L+d/2)^{2}}}\]\end{large}{latex}

{tip}You should check that by taking the simultaneous limits: \\

{latex}\[ d << L \qquad \mbox{and} \qquad m_{r} << m_{c}\]{latex} \\

the expression turns into the result obtained using linear momentum alone.  (In these simultaneous limits, the rod has no effectinertia onat the motionall and the block becomes a point mass that has no resistance to rotation.){tip}