A disc of mass M kg and radius R metre is rotating at an angular speed of `omega` rad/s when the motor is switched off. Neglecting the friction at the axie, the force that must be applied tangentially to the wheel to bring it to rest time t is

Options

A. `(MR omega)/(2t)`
B. `(MR omega)/(t)`
C. `(2MR omega)/(t)`
D. `(MR^(2)omega)/(2t)`

Answer

To solve the problem of determining the force that must be applied tangentially to a rotating disc to bring it to rest in a given time \( t \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The disc has a mass \( M \) and radius \( R \).
- It is rotating with an initial angular speed \( \omega \) (in radians per second).

2. **Calculate the Moment of Inertia:**
- The moment of inertia \( I \) of a solid disc about its central axis is given by the formula:
\[ I = \frac{1}{2} M R^2 \]

3. **Determine the Initial Angular Momentum:**
- The initial angular momentum \( L_i \) of the disc can be calculated using:
\[ L_i = I \cdot \omega = \frac{1}{2} M R^2 \cdot \omega \]

4. **Final Angular Momentum:**
- The final angular momentum \( L_f \) when the disc comes to rest is:
\[ L_f = 0 \]

5. **Calculate the Change in Angular Momentum:**
- The change in angular momentum \( \Delta L \) is:
\[ \Delta L = L_f - L_i = 0 - \frac{1}{2} M R^2 \omega = -\frac{1}{2} M R^2 \omega \]

6. **Relate Torque to Change in Angular Momentum:**
- The torque \( \tau \) applied to the disc is related to the change in angular momentum over time \( t \):
\[ \tau = \frac{\Delta L}{\Delta t} = \frac{-\frac{1}{2} M R^2 \omega}{t} \]

7. **Express Torque in Terms of Force:**
- The torque can also be expressed as:
\[ \tau = R \cdot F \]
- Where \( F \) is the tangential force applied.

8. **Set the Two Expressions for Torque Equal:**
- Equating the two expressions for torque gives:
\[ R \cdot F = -\frac{1}{2} \frac{M R^2 \omega}{t} \]

9. **Solve for the Force \( F \):**
- Rearranging the equation to solve for \( F \):
\[ F = -\frac{1}{2} \frac{M R \omega}{t} \]
- The negative sign indicates that the force is applied in the opposite direction of the rotation.

10. **Final Expression for the Force:**
- The magnitude of the force required to bring the disc to rest in time \( t \) is:
\[ F = \frac{M R \omega}{2 t} \]

### Summary:
The force that must be applied tangentially to the wheel to bring it to rest in time \( t \) is given by:
\[ F = \frac{M R \omega}{2 t} \]