A turntable of radius R=10 m is rotation making 98 rev in 10 s with a boy of mass m=60 kg standing at its centre. He starts running along a radius. Find the frequency of the turntable when the boy is 4m from the centre. The moment of inertia of the turntable about its axis `1000 kg - m^(2) ` .

Options

A. 10 Hz
B. 2.5 Hz
C. 5 Hz
D. 4Hz

Explanation

To solve the problem, we need to find the frequency of the turntable when the boy is 4 meters from the center. We will use the conservation of angular momentum to do this.

### Step-by-Step Solution:

1. **Calculate the Initial Angular Velocity (ω₁)**:
The turntable makes 98 revolutions in 10 seconds. We can convert this to angular velocity in radians per second.
\[ \text{Revolutions per second} = \frac{98 \text{ rev}}{10 \text{ s}} = 9.8 \text{ rev/s} \]
Since \(1 \text{ rev} = 2\pi \text{ rad}\),
\[ \omega_1 = 9.8 \times 2\pi \text{ rad/s} = 61.5 \text{ rad/s} \]

2. **Calculate the Initial Moment of Inertia (I₁)**:
The moment of inertia of the turntable is given as \(I_{\text{turntable}} = 1000 \text{ kg m}^2\). Since the boy is standing at the center, his contribution to the moment of inertia is zero.
\[ I_1 = I_{\text{turntable}} + 0 = 1000 \text{ kg m}^2 \]

3. **Calculate the Moment of Inertia when the Boy is 4m from the Center (I₂)**:
When the boy runs to 4 meters from the center, his moment of inertia can be calculated as:
\[ I_{\text{boy}} = m r^2 = 60 \text{ kg} \times (4 \text{ m})^2 = 60 \times 16 = 960 \text{ kg m}^2 \]
Therefore, the total moment of inertia when the boy is at 4m is:
\[ I_2 = I_{\text{turntable}} + I_{\text{boy}} = 1000 + 960 = 1960 \text{ kg m}^2 \]

4. **Apply Conservation of Angular Momentum**:
According to the conservation of angular momentum:
\[ I_1 \omega_1 = I_2 \omega_2 \]
Substituting the known values:
\[ 1000 \times 61.5 = 1960 \times \omega_2 \]
Solving for \(\omega_2\):
\[ \omega_2 = \frac{1000 \times 61.5}{1960} \approx 31.4 \text{ rad/s} \]

5. **Convert Angular Velocity to Frequency (f)**:
Frequency \(f\) in hertz can be calculated from angular velocity:
\[ f = \frac{\omega_2}{2\pi} = \frac{31.4}{2\pi} \approx 5 \text{ Hz} \]

### Final Answer:
The frequency of the turntable when the boy is 4m from the center is approximately **5 Hz**.