Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 352: 75

Answer

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Work Step by Step

a) We can assume that the axle of rotation is frictionless, so energy is conserved. $$E_i=E_f$$ where $E_i$ is for the hoop when its center of mass is just above the axle of rotation, while $E_f$ is when its center of mass is below the axle exactly. $$K_i+U_{ig}=K_f+U_{gf}$$ $$\frac{1}{2}I\omega^2_i+mgy_i=\frac{1}{2}I\omega^2_f+mgy_f$$ The hoop was initially at rest; $$\frac{1}{2}I(0^2)+mgy_i=\frac{1}{2}I\omega^2_f+mgy_f$$ the hoop's center of mass moved down a distance $2R$, thus $y_i=2R$ while $y_f=0$. $$ 2mgR=\frac{1}{2}I\omega^2_f \tag 1$$ Now we need to find the moment of inertia of the hoop around its edge. According to the parallel-axis theorem; $$I=I_{cm}+mR^2= mR^2+mR^2=2mR^2$$ Plugging into (1); $$ 2\color{red}{\bf\not} mg\color{red}{\bf\not} R=\frac{1}{\color{red}{\bf\not} 2}(\color{red}{\bf\not} 2\color{red}{\bf\not} mR^{\color{red}{\bf\not} 2})\omega^2_f $$ Thus, $$\boxed{\omega_f=\sqrt{\dfrac{2g}{R}}}$$ --- b) Hence, the speed of the lowest point is given by $$v_T=\omega_f R_T$$ where the lowest point is moving in a vertical circle of diameter of $4R$. Hence, $R_T=2R$ $$v_T=\omega_f (2R)$$ Plugging from the boxed formula above; $$v_T=\sqrt{\dfrac{2g}{R}}(2R)=\sqrt{\dfrac{2g(4R^{\color{red}{\bf\not} 2})}{\color{red}{\bf\not} R}}$$ $$\boxed{v_T=\sqrt{8gR}}$$
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