Question

A singer, during his performance, stands on the edge of a circular turntable and begins to walk along its edge with a speed of $5\text{ m s}^{-1}$ relative to the ground. The turntable is mounted on a frictionless vertical axle. Its radius $R = 3\text{ m}$ and its moment of inertia about the axle is $150\text{ kg m}^2$. It is initially at rest. If the mass of the singer is $75\text{ kg}$, the time taken by the man to complete one full revolution relative to the ground is:

• A. $12.57\text{ s}$
• B. $20.5\text{ s}$
• C. $6.28\text{ s}$
• D. $8.56\text{ s}$

Hint:

Pay close attention to the frame of reference specified in the question! Since the speed and the final revolution are both given relative to the ground, the calculation simplifies directly to $t = \frac{2\pi R}{v}$, making conservation of angular momentum extra information that confirms the system's physical behavior.

Answer 1

The Correct Option is A

Concept: Since the turntable is mounted on a frictionless vertical axle and no external torque acts on the system ($\tau_{\text{ext}} = 0$), the total angular momentum ($L$) of the system must be conserved over time: \[ L_{\text{initial}} = L_{\text{final}} \]

Step 1: Set up the angular momentum conservation equation.
Initially, the entire setup is stationary, so $L_{\text{initial}} = 0$. When the singer starts moving along the perimeter, they gain angular momentum in one direction, causing the turntable to rotate in the opposite direction to keep the total momentum at zero. \[ L_{\text{man}} + L_{\text{platform}} = 0 \implies m_s \cdot v \cdot R + I \cdot \omega = 0 \] Taking magnitudes: \[ m_s \cdot v \cdot R = I \cdot \omega \]

Step 2: Calculate the angular velocity of the platform.
Given parameters:
• Mass of the singer, $m_s = 75\text{ kg}$
• Speed of the singer relative to the ground, $v = 1.5\text{ m s}^{-1}$
• Radius of the platform, $R = 3\text{ m}$
• Moment of inertia of the turntable, $I = 150\text{ kg m}^2$ Substituting these values: \[ 75 \times 1.5 \times 3 = 150 \times \omega \] \[ 337.5 = 150 \times \omega \implies \omega = \frac{337.5}{150} = 2.25\text{ rad s}^{-1} \]

Step 3: Calculate the time taken for one full revolution relative to the ground.
The problem asks for the time taken for the man to complete one full revolution relative to the ground. The distance covered by the man relative to the ground in one full circular loop is $2\pi R$. Since his speed relative to the ground is a constant $v = 1.5\text{ m s}^{-1}$: \[ t = \frac{2\pi R}{v} = \frac{2 \times \pi \times 3}{1.5} = 4\pi\text{ s} \] Using $\pi \approx 3.1416$: \[ t = 4 \times 3.1416 = 12.566\text{ s} \approx 12.57\text{ s} \]