A wheel has moment of inertia `5 xx 10^(-3) kg m^(2)` and is making `20 "rev" s^(-1)`. The torque needed to stop it in `10 s` is….. `xx 10^(-2) N-m`

Options

A. `2 pi`
B. `2.5 pi `
C. `4 pi`
D. `4.5 pi`

Explanation

To solve the problem, we need to find the torque required to stop a wheel with a given moment of inertia and initial angular velocity in a specified time. Let's break it down step by step.

### Step 1: Identify the given values
- Moment of inertia (I) = \(5 \times 10^{-3} \, \text{kg m}^2\)
- Initial angular velocity (\(\omega_i\)) = \(20 \, \text{rev/s}\)
- Time (\(t\)) = \(10 \, \text{s}\)

### Step 2: Convert angular velocity from revolutions per second to radians per second
1 revolution = \(2\pi\) radians, so:
\[ \omega_i = 20 \, \text{rev/s} \times 2\pi \, \text{rad/rev} = 40\pi \, \text{rad/s} \]

### Step 3: Determine the final angular velocity
Since the wheel is coming to a stop, the final angular velocity (\(\omega_f\)) is:
\[ \omega_f = 0 \, \text{rad/s} \]

### Step 4: Use the angular kinematics equation
We can use the equation for angular motion:
\[ \omega_f = \omega_i + \alpha t \]
Where \(\alpha\) is the angular acceleration. Rearranging gives:
\[ \alpha = \frac{\omega_f - \omega_i}{t} \]
Substituting the known values:
\[ \alpha = \frac{0 - 40\pi}{10} = -4\pi \, \text{rad/s}^2 \]

### Step 5: Calculate the torque
The torque (\(\tau\)) is related to the moment of inertia and angular acceleration by the equation:
\[ \tau = I \alpha \]
Substituting the values:
\[ \tau = (5 \times 10^{-3}) \times (-4\pi) = -20\pi \times 10^{-3} \, \text{N m} \]
Since we are interested in the magnitude of torque:
\[ |\tau| = 20\pi \times 10^{-3} \, \text{N m} \]

### Step 6: Express the torque in the required format
To express the torque in terms of \(10^{-2} \, \text{N m}\):
\[ |\tau| = 2\pi \times 10^{-2} \, \text{N m} \]

### Final Answer
The torque needed to stop the wheel in \(10\) seconds is:
\[ 2\pi \, \text{(where the answer is expressed as } xx \times 10^{-2} \text{ N m)} \]

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