JEE Advanced 2022 — Paper 2 — PHY-III solution

JEE Advanced2022Paper 2PHY-III

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 $ω$ about the vertical axis passing through the center of $A$ . The angular momentum of $B$ is $n M ω R^{2}$ 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
$\frac{9}{2}$

Solution

Total angular momentum of $B$ about the center of $A$ is the sum of two parts: the orbital part of the center of mass and the spin about the COM of $B$ .

The center of $B$ moves in a circle of radius $2 R$ about the center of $A$ at angular speed $ω$ , so its linear speed is $2 R ω$ . Orbital angular momentum: $L_{orb} = M (2 R)^{2} ω = 4 M R^{2} ω$ .

Rolling without slipping on $A$ 's circumference: the contact point's velocity must vanish. The COM of $B$ moves at $2 R ω$ , and the contact point is on $B$ at distance $R$ from its COM. Therefore the spin angular velocity of $B$ about its own axis is $ω_{B} = 2 R ω / R = 2 ω$ .

Spin angular momentum (thin disk about its central axis): $L_{spin} = \frac{1}{2} M R^{2} \cdot (2 ω) = M R^{2} ω$ .

Both vectors point along the same vertical axis (rolling without slipping fixes the sense). Total: $L = 4 M R^{2} ω + M R^{2} ω = 5 M R^{2} ω$ , so $n = 5$ .

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