A pulley os radius 2m is rotated about its axis by a force `F= (20 t- 5t^2)` newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is `10kg m^2` the number of rotaitons made by the pulley before its direction of motion is reversed, is:

Options

A. less than 3
B. more than 3 but less than 6
C. more than 6 but less than 9
D. more than 9

Explanation

To solve the problem, we need to find the number of rotations made by the pulley before its direction of motion is reversed. Here’s the step-by-step solution:

### Step 1: Determine the Torque
The force applied tangentially to the pulley is given by:
\[ F = 20t - 5t^2 \, \text{N} \]

The torque (\( \tau \)) produced by this force is given by:
\[ \tau = R \cdot F \]
where \( R \) is the radius of the pulley.

Given that the radius \( R = 2 \, \text{m} \):
\[ \tau = 2 \cdot (20t - 5t^2) = 40t - 10t^2 \, \text{N m} \]

### Step 2: Calculate Angular Acceleration
Using the relation between torque and angular acceleration (\( \alpha \)):
\[ \tau = I \alpha \]
where \( I \) is the moment of inertia of the pulley.

Given \( I = 10 \, \text{kg m}^2 \):
\[ \alpha = \frac{\tau}{I} = \frac{40t - 10t^2}{10} = 4t - t^2 \, \text{rad/s}^2 \]

### Step 3: Relate Angular Acceleration to Angular Velocity
We know that:
\[ \alpha = \frac{d\omega}{dt} \]
Thus,
\[ d\omega = (4t - t^2) dt \]

### Step 4: Integrate to Find Angular Velocity
Integrating both sides from \( t = 0 \) to \( t \) and \( \omega = 0 \) to \( \omega \):
\[ \int_0^{\omega} d\omega = \int_0^{t} (4t - t^2) dt \]
Calculating the integral on the right:
\[ \omega = \left[ 2t^2 - \frac{t^3}{3} \right]_0^{t} = 2t^2 - \frac{t^3}{3} \]

### Step 5: Determine When Angular Velocity is Zero
The direction of motion will reverse when \( \omega = 0 \):
\[ 2t^2 - \frac{t^3}{3} = 0 \]
Factoring out \( t^2 \):
\[ t^2 \left( 2 - \frac{t}{3} \right) = 0 \]
This gives us:
1. \( t = 0 \)
2. \( 2 - \frac{t}{3} = 0 \) → \( t = 6 \, \text{s} \)

### Step 6: Calculate the Total Angular Displacement
Now, we need to find the total angular displacement (\( \theta \)) from \( t = 0 \) to \( t = 6 \):
\[ d\theta = \omega dt = \left( 2t^2 - \frac{t^3}{3} \right) dt \]
Integrating from \( 0 \) to \( 6 \):
\[ \theta = \int_0^{6} \left( 2t^2 - \frac{t^3}{3} \right) dt \]
Calculating the integral:
\[ \theta = \left[ \frac{2t^3}{3} - \frac{t^4}{12} \right]_0^{6} \]
Substituting \( t = 6 \):
\[ \theta = \left( \frac{2 \cdot 6^3}{3} - \frac{6^4}{12} \right) = \left( \frac{2 \cdot 216}{3} - \frac{1296}{12} \right) \]
Calculating:
\[ = 144 - 108 = 36 \, \text{radians} \]

### Step 7: Convert Radians to Rotations
To find the number of rotations:
\[ \text{Number of rotations} = \frac{\theta}{2\pi} = \frac{36}{2\pi} \approx 5.73 \]

### Conclusion
Thus, the number of rotations made by the pulley before its direction of motion is reversed is approximately **5.73 rotations**.

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