\begin{problem*}{10.72}
A wad of sticky clay with mass $m$ and velocity $\vect{v}_i$ is fired
at a solid cylinder of mass $M$ and radius $R$ (Fig. P10.72).  The
cylinder is initially at rest and is mounted on a fixed horizontal
axle that runs through its center of mass.  The line of motion of the
projectile is perpendixular to the axle and at a distance $d < R$ from
the center.
\Part{a} Find the angular speed of the system just after the clay
strikes and sticks to the surface of the cylinder.
\Part{b} Is the mechanical energy of the clay-cylinder system
conserved in this process?  Explain your answer.
\end{problem*} % problem 10.72

\begin{solution}
\Part{a}
Conserving angular momentum (letting the \ihat\ direction be into the page)
\begin{align}
 \vect{L}_i &= \vect{R} \times \vect{p}
             = R p \sin\theta \ihat
             = R m v_i \frac{d}{R} \ihat = m v_i d \ihat \\
            &= \vect{L}_f = I \omega \ihat = \left(\frac{1}{2} M R^2 + m R^2\right) \omega \ihat = \left(\frac{M}{2} + m\right)R^2 \omega \ihat \\
     \omega &= \ans{v_i \frac{m d}{I_{tot}}}
\end{align}
Where $I_{tot} \equiv \left( M/2 + m \right) R^2$.

\Part{b}
The change in energy is
\begin{align}
 \Delta K &= K_f - K_i = \frac{1}{2} I_{tot} \omega^2 - \frac{1}{2} m v_i^2 \\
          &= \frac{1}{2} \left(\frac{m^2 v_i^2 d^2}{I_tot} - mv_i^2\right)
	  = \frac{1}{2}m v_i^2 \left(\frac{m d^2}{I_tot} - 1\right) \\
          &= K_i \left(\frac{m d^2}{\left(M/2 + m\right)R^2} - 1\right) \\
\end{align}
So for maximum final energy $d=R$ and the 2nd term on the right hand
side reduces to
\begin{equation}
  \frac{m}{M/2 + m} - 1 = \frac{m - (M/2 + m)}{M/2 + m}
          = \frac{-M}{M+2m} < 0
\end{equation}
So the final energy is less than the initial energy unless $m = 0$, in
which case the cylinder just sits still for eternity.  The lost energy
goes to the same types of internal energy that we had in Problem 24 in
Chapter 8: warmer clay, thwacking sound, etc.
\end{solution}
