QJEEA-2024-09: JEE Advanced 2024 JEEA-2024-09 Solution: Q.9 A disc of mass 𝑀 and radius 𝑅 is free to rotate about its …

Solution:

• Initial Angular Momentum:
• Initially, both discs are at rest, so the initial angular momentum is zero.
• Final Angular Momentum:
• Let the angular speed of the larger disc be ω/n (as given).
• Moment of inertia of the larger disc (I₁): I₁ = (1/2)MR²
• Moment of inertia of the smaller disc (I₂): I₂ = (1/2)M(R/2)² = (1/8)MR²
• Angular momentum of the larger disc (L₁): L₁ = I₁(ω/n) = (1/2)MR²(ω/n)
• Angular momentum of the smaller disc (L₂): L₂ = I₂ω = (1/8)MR²ω
• Conservation of Angular Momentum:
• The total angular momentum of the system must be conserved.
• Initial angular momentum = Final angular momentum
• 0 = L₁ + L₂
• However, the motor is providing torque, so the angular momentum is not conserved.
• However, the motor is fixed to the disc and the small disc is fixed to the motor shaft. The total angular momentum of the two discs must conserve.
• L1 = -L2
• (1/2)MR²(ω/n) = -(1/8)MR²ω.
• Since the discs are rotating in opposite directions, the magnitude of the angular momentum must be equal. Therefore,
• (1/2)MR²(ω/n) = (1/8)MR²ω
• Solve for n:
• (1/2)(1/n) = 1/8
• 1/n = 2/8 = 1/4
• n = 4

Answer:

The value of n is 4.