\begin{problem*}{10.44}
Consider the system shown in Figure~P10.44 with $m_1=20.0\U{kg}$,
$m_2=12.5\U{kg}$, $R=0.200\U{m}$, and the mass of the pulley
$M=5.00\U{kg}$.  Object $m_2$ is resting on the floor, and object
$m_1$ is $4.00\U{m}$ above the floor when it is released from rest.
The pulley axis is frictionless.  The cord is light, does not stretch,
and does not slip on the pulley.  \Part{a} Calculate the time interval
required for $m_1$ to hit the floot.  \Part{b} How would your answer
change if the pulley were massless?
\end{problem*}

\begin{solution}
\Part{a}
Let the positive direction to be the direction of motion (down for
$m_1$, up for $m_2$, and counter clockwise for the pulley).  Balancing
forces and torques on each object we have
\begin{align}
  \sum F_1 &= m_1 g - T_1 = m_1 a \\
  \sum F_2 &= T_2 - m_2 g = m_2 a \\
  \sum \tau &= T_1 R - T_2 R = I\alpha
    = \p({\frac{1}{2}MR^2})\cdot\frac{a}{R}
    = \frac{MRa}{2} \;,
\end{align}
where we used the moment of interia for a solid cylinder
$I=\frac{1}{2}MR^2$ and the relationship between linear and angular
acceleration $a = r\alpha$.  Now we have three equations for our three
unknowns ($a$, $T_1$, and $T_2$) and we can solve for acceleration.
\begin{align}
  T_1 &= m_1 (g-a) \\
  T_2 &= m_2 (g+a) \\
  T_1 - T_2 &= \frac{Ma}{2} \\
  m_1 g - m_1 a - m_2 g - m_2 a &= \frac{Ma}{2} \\
  (m_1 - m_2)g - (m_1 + m_2)a &= \frac{M}{2}a \\
  (m_1 - m_2)g &= a \p({\frac{M}{2} + m_1 + m_2}) \\
  a &= g\frac{m_1 - m_2}{\frac{M}{2} + m_1 + m_2} \;.
\end{align}

The time-to-floor is the a constant acceleration problem.
\begin{align}
  h &= \frac{1}{2} a t^2 \\
  t^2 &= \frac{2h}{a} \\
  t &= \sqrt{\frac{2h}{a}}
    = \sqrt{\frac{2h}{g}\frac{\frac{M}{2} + m_1 + m_2}{m_1 - m_2}}
    = \ans{1.95\U{s}} \;.
\end{align}

\Part{b}
If the mass of the pulley was zero,
\begin{equation}
  t = \sqrt{\frac{2h}{g}\frac{m_1 + m_2}{m_1 - m_2}}
    = \ans{1.88\U{s}} \;,
\end{equation}
which is slightly faster than the original time.
\end{solution}
