1 \begin{problem*}{10.51}
2 An object with a mass $m=5.10\U{kg}$ is attached to the free end of a
3 light string wrapped around a reel of radius $R=0.250\U{m}$ and mass
4 $M=3.00\U{kg}$. The reel is a solid disk, free to rotate in a
5 vertical plane about the horiizontal axis passing through its center
6 as shown in Figure P10.51. The suspended object is released from rest
7 $6.00\U{m}$ above the floor. Determine \Part{a} the tension in the
8 string, \Part{b} the acceleration of the object, and \Part{c} the
9 speed with which the object hits the floor. \Part{d} Verify your
10 answer to \Part{c} by using the isolated system (energy) model.
15 Summing the forces and tourques,
17 \sum F &= mg - T = ma \\
18 \sum \tau &= RT = I\alpha
19 = \p({\frac{1}{2}MR^2})\cdot\frac{a}{R}
24 a &= g - \frac{T}{m} \\
25 RT &= \frac{MRa}{2} \\
27 2T &= Mg - \frac{M}{m}T \\
28 T \p({2 + \frac{M}{m}}) &= Mg \\
29 T &= \frac{Mg}{2 + \frac{M}{m}} = \ans{11.4\U{N}} \;.
33 Plugging back in for $a$,
35 a = g - \frac{T}{m} = 9.80\U{m/s$^2$} - \frac{11.4\U{N}}{5.10\U{kg}}
36 = \ans{7.57\U{m/s$^2$}}
40 Finding the floor-hitting speed is a constant acceleration problem
42 v^2 &= v_0^2 + 2a\Delta x \\
43 v &= \sqrt{2a\Delta x} = \ans{9.53\U{m/s}}
47 Conserving energy, the initial gravitational energy is converted into
48 linear and angular kinetic energies.
50 E_i = mgh &= E_f = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2
51 = \frac{1}{2}v^2 + \frac{1}{2} \p({\frac{1}{2}MR^2}) \p({\frac{v}{R}})^2
52 = \frac{m}{2}mv^2 + \frac{M}{4} v^2
53 = \frac{2m + M}{4}v^2 \\
54 v^2 &= \frac{4mgh}{2m+M} \\
55 v &= \sqrt{\frac{4mgh}{2m+M}} = \ans{9.53\U{m/s$^2$}} \;,
57 which is the same speed we got in \Part{c}.