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

Explanation

To solve the problem step-by-step, we will follow the reasoning provided in the video transcript.

### Step 1: Determine the Torque
Given the force \( F(t) = 20t - 5t^2 \) N and the radius \( R = 2 \) m, we can calculate the torque \( \tau \) applied to the pulley.

\[ \tau = F \cdot R = (20t - 5t^2) \cdot 2 = 40t - 10t^2 \text{ N m} \]

### Step 2: Calculate the Angular Acceleration
The moment of inertia \( I \) of the pulley is given as \( 10 \, \text{kg m}^2 \). Using the relation \( \tau = I \alpha \), we can find the angular acceleration \( \alpha \).

\[ \alpha = \frac{\tau}{I} = \frac{40t - 10t^2}{10} = 4t - t^2 \text{ rad/s}^2 \]

### Step 3: Relate Angular Acceleration to Angular Velocity
We know that angular acceleration \( \alpha \) is the derivative of angular velocity \( \omega \) with respect to time \( t \):

\[ \alpha = \frac{d\omega}{dt} \]

Thus, we can express this as:

\[ d\omega = (4t - t^2) dt \]

### Step 4: Integrate to Find Angular Velocity
Now, we will integrate both sides from \( t = 0 \) to \( t = T \) (where \( T \) is the time when the direction of motion reverses) and from \( \omega = 0 \) to \( \omega \):

\[ \int_0^\omega d\omega = \int_0^T (4t - t^2) dt \]

Calculating the right-hand side:

\[ \int (4t - t^2) dt = 2t^2 - \frac{t^3}{3} \]

Evaluating from \( 0 \) to \( T \):

\[ \omega = 2T^2 - \frac{T^3}{3} \]

### Step 5: Determine When the Direction Reverses
The direction of motion reverses when \( \omega = 0 \). Setting the equation for \( \omega \) to zero:

\[ 2T^2 - \frac{T^3}{3} = 0 \]

Factoring out \( T^2 \):

\[ T^2 \left( 2 - \frac{T}{3} \right) = 0 \]

This gives us \( T^2 = 0 \) or \( T = 6 \) seconds (since \( 2 - \frac{T}{3} = 0 \) leads to \( T = 6 \)).

### Step 6: Calculate the Angular Displacement
Now, we need to find the angular displacement \( \theta \) during this time. We know:

\[ d\theta = \omega dt \]

Thus, we will integrate \( \omega \) from \( 0 \) to \( 6 \):

\[ \theta = \int_0^6 \left( 2t^2 - \frac{t^3}{3} \right) dt \]

Calculating this integral:

\[ \theta = \left[ \frac{2t^3}{3} - \frac{t^4}{12} \right]_0^6 = \left( \frac{2 \cdot 6^3}{3} - \frac{6^4}{12} \right) \]

Calculating the values:

\[ \theta = \left( \frac{2 \cdot 216}{3} - \frac{1296}{12} \right) = 144 - 108 = 36 \text{ radians} \]

### Step 7: Convert Radians to Rotations
To find the number of rotations, we convert radians to rotations using the relation \( 1 \text{ rotation} = 2\pi \text{ radians} \):

\[ \text{Number of rotations} = \frac{36}{2\pi} = \frac{36}{6.2832} \approx 5.73 \text{ rotations} \]

### Final Answer
The number of rotations made by the pulley before its direction of motion is reversed is approximately **5.73 rotations**.