A disc of radius 1m and mass 1 kg is rotating with 40 radians/sec. the torque required to stop it in 10sec will be

Options

A. `IN-m`
B. `2N-m`
C. `0.5 N-m`
D. `4N - m`

Answer

To solve the problem of finding the torque required to stop a disc rotating at 40 radians/sec in 10 seconds, we can follow these steps:

### Step 1: Identify the given values
- Radius of the disc (r) = 1 m
- Mass of the disc (m) = 1 kg
- Initial angular velocity (ω_initial) = 40 rad/s
- Final angular velocity (ω_final) = 0 rad/s (since we want to stop the disc)
- Time (t) = 10 s

### Step 2: Calculate the angular deceleration (α)
We can use the formula for angular acceleration (deceleration in this case) given by:
\[ \alpha = \frac{\Delta \omega}{t} \]
where \(\Delta \omega = \omega_{final} - \omega_{initial}\).

Substituting the values:
\[ \Delta \omega = 0 - 40 = -40 \text{ rad/s} \]
\[ \alpha = \frac{-40 \text{ rad/s}}{10 \text{ s}} = -4 \text{ rad/s}^2 \]

### Step 3: Calculate the moment of inertia (I) of the disc
The moment of inertia for a solid disc about its central axis is given by:
\[ I = \frac{1}{2} m r^2 \]
Substituting the values:
\[ I = \frac{1}{2} \times 1 \text{ kg} \times (1 \text{ m})^2 = \frac{1}{2} \text{ kg m}^2 = 0.5 \text{ kg m}^2 \]

### Step 4: Calculate the torque (τ) required to achieve this angular deceleration
The torque is related to the moment of inertia and angular acceleration by the equation:
\[ \tau = I \alpha \]
Substituting the values we calculated:
\[ \tau = 0.5 \text{ kg m}^2 \times (-4 \text{ rad/s}^2) = -2 \text{ N m} \]

### Step 5: Interpret the result
The negative sign indicates that the torque is applied in the opposite direction of the rotation to bring the disc to a stop. Therefore, the magnitude of the torque required to stop the disc in 10 seconds is:
\[ \tau = 2 \text{ N m} \]

### Final Answer
The torque required to stop the disc in 10 seconds is **2 N m**.

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