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. 100 Nm
B. 150 Nm
C. 200 Nm
D. 250 Nm

Answer

To solve the problem of how much torque will be applied to stop a rotating body in 10 seconds, we can follow these steps:

### Step 1: Identify the given data
- Moment of inertia (I) = 50 kg·m²
- Initial angular speed (ω_i) = 20 rad/s
- Final angular speed (ω_f) = 0 rad/s (since the body is brought to a stop)
- Time (t) = 10 s

### Step 2: Use the angular kinematics equation
We can use the equation of motion for angular motion, which is similar to the linear motion equation:
\[ \omega_f = \omega_i + \alpha t \]
where:
- \(\alpha\) is the angular acceleration.

### Step 3: Rearrange the equation to find angular acceleration (α)
Substituting the known values into the equation:
\[ 0 = 20 + \alpha \cdot 10 \]
Now, solve for \(\alpha\):
\[ \alpha \cdot 10 = -20 \]
\[ \alpha = -\frac{20}{10} = -2 \text{ rad/s}^2 \]

### Step 4: Calculate the torque (τ)
The torque required to produce this angular acceleration can be calculated using the formula:
\[ \tau = I \cdot \alpha \]
Substituting the values of I and α:
\[ \tau = 50 \cdot (-2) = -100 \text{ Nm} \]

### Step 5: Interpret the result
The negative sign indicates that the torque is applied in the opposite direction of the angular velocity to bring the body to rest. Therefore, the magnitude of the torque required is:
\[ \tau = 100 \text{ Nm} \]

### Final Answer
The torque applied to stop the body in 10 seconds is **100 Nm**.

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