A constant torque of `1000 N-m` turns a wheel of moment of inertia `200 kg-m^2` about an axis through its centre. Its angular velocity after `3` seconds is.

Options

A. `1 rad//s`
B. `5 rad//s`
C. `10 rad//s`
D. `15 rad//s`

Explanation

To find the angular velocity of the wheel after 3 seconds under the influence of a constant torque, we can use the following steps:

### Step 1: Understand the relationship between torque, moment of inertia, and angular acceleration.
The relationship is given by Newton's second law for rotation:
\[ \tau = I \alpha \]
where:
- \(\tau\) is the torque,
- \(I\) is the moment of inertia,
- \(\alpha\) is the angular acceleration.

### Step 2: Calculate the angular acceleration (\(\alpha\)).
Rearranging the equation gives us:
\[ \alpha = \frac{\tau}{I} \]
Substituting the given values:
\[ \alpha = \frac{1000 \, \text{N-m}}{200 \, \text{kg-m}^2} = 5 \, \text{rad/s}^2 \]

### Step 3: Use the angular acceleration to find the final angular velocity (\(\omega\)).
The angular velocity after a time \(t\) can be calculated using the formula:
\[ \omega = \omega_0 + \alpha t \]
where:
- \(\omega_0\) is the initial angular velocity (which is 0 since the wheel starts from rest),
- \(t\) is the time in seconds.

Since the wheel starts from rest, \(\omega_0 = 0\):
\[ \omega = 0 + (5 \, \text{rad/s}^2)(3 \, \text{s}) = 15 \, \text{rad/s} \]

### Final Answer:
The angular velocity of the wheel after 3 seconds is \(15 \, \text{rad/s}\).

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