# Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises: 10.76

(a) $\omega = 1.28~rad/s$ (b) $\omega = 0.631~rad/s$ (c) $h = 1.26~m$

#### Work Step by Step

We can find the moment of inertia of the vine when Jane is swinging on it. $I = \frac{1}{3}M_vR^2+M_jR^2$ $I = \frac{1}{3}(30.0~kg)(8.00~m)^2+(60.0~kg)(8.00~m)^2$ $I = 4480~kg~m^2$ We can use conservation of energy to find the angular speed at the bottom. Note that the center of mass of the vine drops a height of 2.50 meters when Jane falls 5.00 meters. $\frac{1}{2}I\omega^2 = M_j~gh+M_v~g(\frac{h}{2})$ $\omega^2 = \frac{2M_j~gh+M_v~gh}{I}$ $\omega = \sqrt{\frac{2M_j~gh+M_v~gh}{I}}$ $\omega = \sqrt{\frac{(2)(60.0~kg)(9.80~m/s^2)(5.00~m)+(30.0~kg)(9.80~m/s^2)(5.00~m)}{4480~kg~m^2}}$ $\omega = 1.28~rad/s$ (b) We can find the moment of inertia when Tarzan is included. $I = \frac{1}{3}M_vR^2+(M_j+M_t)R^2$ $I = \frac{1}{3}(30.0~kg)(8.00~m)^2+(60.0~kg+72.0~kg)(8.00~m)^2$ $I = 9088~kg~m^2$ We can use conservation of angular momentum to find the angular speed after Jane grabs Tarzan. $L_2=L_1$ $I_2\omega_2=I_1\omega_1$ $\omega_2=\frac{I_1\omega_1}{I_2}$ $\omega_2=\frac{(4480~kg~m^2)(1.28~rad/s)}{9088~kg~m^2}$ $\omega_2 = 0.631~rad/s$ (c) We can use conservation of energy to find the maximum height that Jane and Tarzan swing up. Note that the center of mass of the vine rises a height of $\frac{h}{2}$ meters when Jane rises $h$ meters. $(M_j+M_t)~gh+M_v~g(\frac{h}{2}) = \frac{1}{2}I_2\omega_2^2$ $(M_j+M_t)~2gh+M_v~gh = I_2\omega_2^2$ $h = \frac{I_2\omega_2^2}{(M_j+M_t)~2g+M_v~g}$ $h = \frac{(9088~kg~m^2)(0.631~rad/s)^2}{(60.0~kg+72.0~kg)(2)(9.80~m/s^2)+(30.0~kg)(9.80~m/s^2)}$ $h = 1.26~m$

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