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JEE Advanced 2024 · System of Particles and Rotational Motion

A disc of mass MM and radius RR 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 MM and radius R2\frac{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 ω\omega. If the angular speed at which the large disc rotates is ωn\frac{\omega}{n}, then the value of nn is ______.

Answer: 1212

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Written solution

Step 1: Identify the parameters of the system.

The system consists of a large disc of mass MM and radius RR, and a small disc of mass MM and radius R2\frac{R}{2} attached to a motor on the circumference of the large disc.

Step 2: Apply conservation of angular momentum.

Since there are no external torques acting on the system about the vertical axis of rotation, the total angular momentum is conserved.

Linitial=Lfinal=0L_{\text{initial}} = L_{\text{final}} = 0

Step 3: Determine the moments of inertia.

Calculate the moment of inertia of the large disc about its central axis, and the small disc about its own axis.

Ibig=12MR2I_{\text{big}} = \frac{1}{2}MR^2 Ismall=12M(R2)2=MR28I_{\text{small}} = \frac{1}{2}M\left(\frac{R}{2}\right)^2 = \frac{MR^2}{8}

Step 4: Express the components of the final angular momentum.

The final angular momentum consists of the spin of the large disc, the spin of the small disc, and the orbital motion of the small disc's center of mass at a distance RR from the main axis.

Lbig=IbigΩL_{\text{big}} = I_{\text{big}}\Omega Lspin=IsmallωL_{\text{spin}} = I_{\text{small}}\omega Lorb=MR2ΩL_{\text{orb}} = MR^2\Omega

Step 5: Set up the conservation equation.

Sum the individual angular momentum components to zero.

Lbig+Lspin+Lorb=0L_{\text{big}} + L_{\text{spin}} + L_{\text{orb}} = 0 12MR2Ω+MR28ω+MR2Ω=0\frac{1}{2}MR^2\Omega + \frac{MR^2}{8}\omega + MR^2\Omega = 0

Step 6: Solve for the angular speed of the large disc.

Group the terms containing Ω\Omega and simplify.

32MR2Ω=−18MR2ω\frac{3}{2}MR^2\Omega = -\frac{1}{8}MR^2\omega Ω=−ω12\Omega = -\frac{\omega}{12}

Step 7: Determine the value of nn.

Compare the magnitude of the angular speed to the given expression.

∣Ω∣=ω12=ωn|\Omega| = \frac{\omega}{12} = \frac{\omega}{n} n=12n = 12

Answer: 12

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