If `I=50kg m^(2)`, then how much torque will be applied to stop it in 10 sec. Its initial angular speed is 20 rad/sec?

Options

A. (a)100 Nm
B. (b)150 Nm
C. (c)200 Nm
D. (d)250 Nm

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the given values
- Moment of inertia (I) = 50 kg m²
- Initial angular speed (ω) = 20 rad/s
- Time to stop (t) = 10 s

### Step 2: Calculate the angular acceleration (α)
To find the angular acceleration required to stop the object, we can use the formula:
\[ \alpha = \frac{\Delta \omega}{t} \]
Since we want to stop the object, the final angular speed (ω_f) will be 0 rad/s. Therefore, the change in angular speed (Δω) is:
\[ \Delta \omega = \omega_f - \omega = 0 - 20 = -20 \text{ rad/s} \]
Now, substituting the values into the formula:
\[ \alpha = \frac{-20 \text{ rad/s}}{10 \text{ s}} = -2 \text{ rad/s}^2 \]

### Step 3: Calculate the torque (τ)
Using the relationship between torque, moment of inertia, and angular acceleration:
\[ \tau = I \cdot \alpha \]
Substituting the known values:
\[ \tau = 50 \text{ kg m}^2 \cdot (-2 \text{ rad/s}^2) = -100 \text{ N m} \]
The negative sign indicates that the torque is applied in the opposite direction of the angular motion to stop it.

### Step 4: Conclusion
The magnitude of the torque required to stop the object in 10 seconds is:
\[ \tau = 100 \text{ N m} \]

### Final Answer
The torque that must be applied to stop it in 10 seconds is **100 N m**.

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