1 \begin{problem*}{10.44}
2 Consider the system shown in Figure~P10.44 with $m_1=20.0\U{kg}$,
3 $m_2=12.5\U{kg}$, $R=0.200\U{m}$, and the mass of the pulley
4 $M=5.00\U{kg}$. Object $m_2$ is resting on the floor, and object
5 $m_1$ is $4.00\U{m}$ above the floor when it is released from rest.
6 The pulley axis is frictionless. The cord is light, does not stretch,
7 and does not slip on the pulley. \Part{a} Calculate the time interval
8 required for $m_1$ to hit the floot. \Part{b} How would your answer
9 change if the pulley were massless?
14 Let the positive direction to be the direction of motion (down for
15 $m_1$, up for $m_2$, and counter clockwise for the pulley). Balancing
16 forces and torques on each object we have
18 \sum F_1 &= m_1 g - T_1 = m_1 a \\
19 \sum F_2 &= T_2 - m_2 g = m_2 a \\
20 \sum \tau &= T_1 R - T_2 R = I\alpha
21 = \p({\frac{1}{2}MR^2})\cdot\frac{a}{R}
24 where we used the moment of interia for a solid cylinder
25 $I=\frac{1}{2}MR^2$ and the relationship between linear and angular
26 acceleration $a = r\alpha$. Now we have three equations for our three
27 unknowns ($a$, $T_1$, and $T_2$) and we can solve for acceleration.
31 T_1 - T_2 &= \frac{Ma}{2} \\
32 m_1 g - m_1 a - m_2 g - m_2 a &= \frac{Ma}{2} \\
33 (m_1 - m_2)g - (m_1 + m_2)a &= \frac{M}{2}a \\
34 (m_1 - m_2)g &= a \p({\frac{M}{2} + m_1 + m_2}) \\
35 a &= g\frac{m_1 - m_2}{\frac{M}{2} + m_1 + m_2} \;.
38 The time-to-floor is the a constant acceleration problem.
40 h &= \frac{1}{2} a t^2 \\
41 t^2 &= \frac{2h}{a} \\
42 t &= \sqrt{\frac{2h}{a}}
43 = \sqrt{\frac{2h}{g}\frac{\frac{M}{2} + m_1 + m_2}{m_1 - m_2}}
48 If the mass of the pulley was zero,
50 t = \sqrt{\frac{2h}{g}\frac{m_1 + m_2}{m_1 - m_2}}
53 which is slightly faster than the original time.