Question

A flat surface of a thin uniform disk $A$ of radius $R$ is glued to a horizontal table. Another thin uniform disk $B$ of mass $M$ and with the same radius $R$ rolls without slipping on the circumference of $A$, as shown in the figure. A flat surface of $B$ also lies on the plane of the table. The center of mass of $B$ has fixed angular speed $\omega$ about the vertical axis passing through the center of $A$. The angular momentum of $B$ is $n M \omega R^2$ with respect to the center of $A$. Which of the following is the value of $n$ ?

• A. 2
• B. 5
• C. $\frac{7}{2}$
• D. $\frac{9}{2}$

Answer 1

The Correct Option is B

Given that the velocity of the center of mass (VCM) is:

\( V_{\text{CM}} = 2R \omega \)

At the point of contact, we have:

\( V_{\text{CM}} = \omega_0 R \Rightarrow \omega_0 = 2 \omega \)

Now, the angular momentum of disk B with respect to the center of disk A is calculated as:

\( L = \left( \frac{MR^2}{2} \right) (\omega_0) + M (2R \omega) (2R) \)

Substitute the values of \( \omega_0 \) and simplify:

\( L = \left( \frac{MR^2}{2} \right) (2\omega) + 4MR^2 \omega \)

\( L = 5MR^2 \omega \) and hence, \( n = 5 \).