\begin{problem*}{10.51}
An object with a mass $m=5.10\U{kg}$ is attached to the free end of a
light string wrapped around a reel of radius $R=0.250\U{m}$ and mass
$M=3.00\U{kg}$.  The reel is a solid disk, free to rotate in a
vertical plane about the horiizontal axis passing through its center
as shown in Figure~P10.51.  The suspended object is released from rest
$6.00\U{m}$ above the floor.  Determine \Part{a} the tension in the
string, \Part{b} the acceleration of the object, and \Part{c} the
speed with which the object hits the floor.  \Part{d} Verify your
answer to \Part{c} by using the isolated system (energy) model.
\end{problem*}

\begin{solution}
\Part{a}
Summing the forces and tourques,
\begin{align}
  \sum F &= mg - T = ma \\
  \sum \tau &= RT = I\alpha
    = \p({\frac{1}{2}MR^2})\cdot\frac{a}{R}
    = \frac{MRa}{2} \;.
\end{align}
Solving for tension
\begin{align}
  a &= g - \frac{T}{m} \\
  RT &= \frac{MRa}{2} \\
  2T &= Ma \\
  2T &= Mg - \frac{M}{m}T \\
  T \p({2 + \frac{M}{m}}) &= Mg \\
  T &= \frac{Mg}{2 + \frac{M}{m}} = \ans{11.4\U{N}} \;.
\end{align}

\Part{b}
Plugging back in for $a$,
\begin{equation}
  a = g - \frac{T}{m} = 9.80\U{m/s$^2$} - \frac{11.4\U{N}}{5.10\U{kg}}
    = \ans{7.57\U{m/s$^2$}}
\end{equation}

\Part{c}
Finding the floor-hitting speed is a constant acceleration problem
\begin{align}
  v^2 &= v_0^2 + 2a\Delta x \\
  v &= \sqrt{2a\Delta x} = \ans{9.53\U{m/s}}
\end{align}

\Part{d}
Conserving energy, the initial gravitational energy is converted into
linear and angular kinetic energies.
\begin{align}
  E_i = mgh &= E_f = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2
    = \frac{1}{2}v^2 + \frac{1}{2} \p({\frac{1}{2}MR^2}) \p({\frac{v}{R}})^2
    = \frac{m}{2}mv^2 + \frac{M}{4} v^2
    = \frac{2m + M}{4}v^2 \\
  v^2 &= \frac{4mgh}{2m+M} \\
  v &= \sqrt{\frac{4mgh}{2m+M}} = \ans{9.53\U{m/s$^2$}} \;,
\end{align}
which is the same speed we got in \Part{c}.
\end{solution}
