A one kg stone attached to the end of a 60 cm chain is revolving at the rate of 3 revolutions/second. It is found that after 30 second, it makes only one revolution per second. What is the mean torque ?

Options

A. 0.35 N-m
B. 0.15 N-m
C. 0.25 N-m
D. 0.45 N-m

Explanation

To solve the problem, we need to calculate the mean torque acting on the stone as it changes its angular velocity over a period of 30 seconds. Here’s a step-by-step solution:

### Step 1: Convert the given values to standard units
- The mass of the stone, \( m = 1 \, \text{kg} \).
- The length of the chain (radius), \( r = 60 \, \text{cm} = 0.6 \, \text{m} \).
- Initial angular velocity, \( \omega_i = 3 \, \text{revolutions/second} = 3 \times 2\pi \, \text{radians/second} = 6\pi \, \text{radians/second} \).
- Final angular velocity after 30 seconds, \( \omega_f = 1 \, \text{revolution/second} = 1 \times 2\pi \, \text{radians/second} = 2\pi \, \text{radians/second} \).

### Step 2: Calculate the moment of inertia
The moment of inertia \( I \) for a point mass rotating about an axis is given by:
\[ I = m r^2 \]
Substituting the values:
\[ I = 1 \, \text{kg} \times (0.6 \, \text{m})^2 = 1 \times 0.36 = 0.36 \, \text{kg m}^2 \]

### Step 3: Calculate the change in angular momentum
The angular momentum \( L \) is given by:
\[ L = I \omega \]
Thus, the initial angular momentum \( L_i \) and final angular momentum \( L_f \) are:
\[ L_i = I \omega_i = 0.36 \, \text{kg m}^2 \times 6\pi \, \text{radians/second} = 2.16\pi \, \text{kg m}^2/\text{s} \]
\[ L_f = I \omega_f = 0.36 \, \text{kg m}^2 \times 2\pi \, \text{radians/second} = 0.72\pi \, \text{kg m}^2/\text{s} \]

### Step 4: Calculate the change in angular momentum
\[ \Delta L = L_f - L_i = 0.72\pi - 2.16\pi = -1.44\pi \, \text{kg m}^2/\text{s} \]

### Step 5: Calculate the mean torque
The mean torque \( \tau \) is given by the change in angular momentum divided by the change in time:
\[ \tau = \frac{\Delta L}{\Delta t} \]
Here, \( \Delta t = 30 \, \text{s} \):
\[ \tau = \frac{-1.44\pi}{30} \, \text{N m} \]
Taking the absolute value:
\[ \tau = \frac{1.44\pi}{30} \approx 0.15 \, \text{N m} \]

### Final Answer
Thus, the mean torque is approximately \( 0.15 \, \text{N m} \).

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