JEE Advanced 2022 Paper 2 · Q15 · Angular Momentum

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 centre of mass of $B$ has fixed angular speed $\omega$ about the vertical axis passing through the centre of $A$. The angular momentum of $B$ is $n M \omega R^2$ with respect to the centre of $A$. Which of the following is the value of $n$?

[Figure: Top view of disk A glued to table; disk B of same radius R rolls without slipping around A's circumference. Centre of B traces a circle of radius 2R about centre of A with angular speed omega.]

SolveKar AI · Accepted Answer

Centre of $B$ moves in a circle of radius $2R$ about $A$'s centre, with speed $v_{CM} = 2R\omega$. Rolling without slipping at point of contact: $v_{CM} = R\omega_0$, where $\omega_0$ is the spin angular speed of $B$ about its own centre, so $\omega_0 = 2\omega$.
Angular momentum of $B$ about centre of $A$ = (spin about its own CM) + (orbital about $A$):
$L_{spin} = I_{CM}\omega_0 = (MR^2/2)(2\omega) = MR^2\omega$.
$L_{orbital} = M v_{CM} \cdot (2R) = M(2R\omega)(2R) = 4MR^2\omega$.
Total $L = MR^2\omega + 4MR^2\omega = 5 MR^2\omega$, so $n = 5$.

JEE Advanced 2022 Paper 2 · Q15 · Angular Momentum

Solution