JEE Advanced 2024 Paper 1 · Q09 · Angular Momentum

Question

A disc of mass $M$ and radius $R$ is free to rotate about its vertical axis as shown in the figure. A battery operated motor of negligible mass is fixed to this disc at a point on its circumference. Another disc of the same mass $M$ and radius $R/2$ is fixed to the motor's thin shaft. Initially, both the discs are at rest. The motor is switched on so that the smaller disc rotates at a uniform angular speed $\omega$. If the angular speed at which the large disc rotates is $\omega/n$, then the value of $n$ is _____.

SolveKar AI · Accepted Answer

Angular momentum about the fixed vertical axis is conserved (zero net external torque about it; motor exerts only internal torques). Initially total angular momentum is zero. Let the large disc spin at $-\Omega$ (opposite to the small disc) and the small disc spin at $\omega$ relative to the lab frame. The small disc's centre is at distance $R$ from the axis (rim of the large disc). Angular momentum of large disc about axis: $I_L\cdot(-\Omega) = -\dfrac{1}{2}MR^2\,\Omega$. Angular momentum of small disc about axis = spin part + orbital part. Spin: $I_S\,\omega = 	frac{1}{2}M(R/2)^2\,\omega = 	frac{1}{8}MR^2\,\omega$ (about its own centre, but the small disc's symmetry axis coincides with the lab vertical axis up to a translation; using parallel-axis the orbital contribution is $MR^2\cdot(-\Omega)$ since its centre orbits with the large disc). Setting total $= 0$: $-	frac{1}{2}MR^2\Omega + 	frac{1}{8}MR^2\omega - MR^2\Omega = 0$, hence $	frac{3}{2}\Omega = 	frac{1}{8}\omega$ ⇒ $\Omega = \omega/12$. Therefore $n = 12$.

JEE Advanced 2024 Paper 1 · Q09 · Angular Momentum

Solution