Physics Problems & Solutions: Classical Mechanics

The cylinder shown above, with mass M and radius R, has a radially dependent density. The cylinder starts from from rest and rolls without slipping down an inclined plane of height H. At the bottom of the plane its translational speed is (8gH/7)^{1/2. Which of the following is the rotational inertia of the cylinder?}

$\\$A.$ \hspace{5 mm} \frac{1}{2}MR^2 \\\\$B.$ \hspace{5 mm} \frac{3}{4}MR^2 \\\\$C.$ \hspace{5 mm} \frac{7}{8}MR^2 \\\\$D.$ \hspace{5 mm} MR^2 \\\\$E.$ \hspace{5 mm} 4MR^2$

(GR0177 #91)

Solution:

Initally, it starts from rest, from height H,

$E_i=T_i+V_i=0+MgH$

At the bottom of the plane,

$\begin{aligned} E_f&=T_f+V_f=T_{\text{trans}}+T_{\text{rot}}+0 \\&=\frac{1}{2}Mv^2+\frac{1}{2}I \omega^2 \end{aligned}$

Conservation of Energy,

$\begin{aligned} E_i&=E_f \\MgH&=\frac{1}{2}Mv^2+\frac{1}{2}I \omega^2 \\I&=\frac{2MgH-Mv^2}{\omega^2} \end{aligned}$

with $v=R\omega$ and $(8gH/7)^{1/2}$

$\begin{aligned} I&=\frac{R^2}{v^2}\left( 2MgH-Mv^2 \right) \\&=\frac{2MgHR^2}{v^2}-MR^2 \\&=\frac{14MgHR^2}{8gH}-MR^2 \\&=\frac{7}{4}MR^2-MR^2 \\&=\frac{3}{4}MR^2 \end{aligned}$

Answer: B

Physics Problems & Solutions

Classical Mechanics - Rotational Motion

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