Question

Wheels A and B are connected by a belt that does not slip $(R_A=3R_B)$. What would be the ratio of the rotational inertia, if the two wheels had same angular momentum about their central axis?

• A. $3$
• B. $\frac{1}{3}$
• C. $1$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $\frac{1}{3}$

$\begin{array}{c}\text{Given}{R}_B=3RA\Rightarrow \frac{R_A}{RB}=\frac{1}{3}\\ \text{angular Momentum}L=I\omega \\ \text{Kinetic Energy}K=\frac{1}{2}I{\omega }^2\\ \text{Velocities on both wheels}A\&B\text{must}be\end{array}$

$\begin{array}{c}\text{Same}\\ \begin{array}{cc}{V}_A=& V_B\Rightarrow {\omega }_A{\gamma }_A={\omega }_B{r}_B\\ & \frac{W_B}{{\omega }_A}=\frac{V_A}{r_B}\\ & \text{If}{L}_A=L_B=L\text{then}\end{array}\end{array}$

$\begin{array}{c}\frac{I_A}{I_B}=\frac{L/{\omega }_A}{L/{\omega }_B}\Rightarrow \frac{{\omega }_B}{{\omega }_A}=\frac{V_A}{V_B}=\frac{1}{3}\\ \text{Hence option}B\text{is correct}\\ I=4\times M{\left(\frac{\sqrt{2}L}{2}\right)}^2\text{since distance of the}\end{array}$

$\begin{array}{c}\text{Point mass from center is half of the}\\ \text{diagonal}\\ \Rightarrow I=2\pi L^2\\ \text{Hence option}B\text{is corret}\end{array}$