A pulley of radius $$2 \, \text{m}$$ is rotated about its axis by a force $$F = (20t - 5t^2)$$ Newton (where $$t$$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation made by the pulley before its direction of motion if reversed, is :

Solution & Explanation

1. Understand the Torque-Force Relationship

A tangential force $$F$$ applied to the rim of a pulley creates a torque ($$\tau$$) that drives its angular acceleration. The relationship between the applied force and the resulting torque is defined by:

$$\tau = F \cdot R$$

Where:

• $$R = 2 \,\, \text{m}$$ (Radius of the pulley)
• $$F = 20t - 5t^2 \,\, \text{N}$$ (Time-dependent tangential force)

Substituting the expression for force, we find the torque as a function of time:

$$\tau(t) = (20t - 5t^2) \cdot 2 = 40t - 10t^2 \,\, \text{N}\cdot\text{m}$$

2. Establish the Angular Acceleration ($$\alpha$$)

According to the rotational analog of Newton's Second Law, torque is directly proportional to angular acceleration ($$\alpha$$) through the moment of inertia ($$I$$):

$$\tau = I \cdot \alpha \implies \alpha = \frac{\tau}{I}$$

Given that the moment of inertia of the pulley about its axis of rotation is $$I = 10 \,\, \text{kg}\cdot\text{m}^2$$, we can substitute this value into the equation:

$$\alpha(t) = \frac{40t - 10t^2}{10} = 4t - t^2 \,\, \text{rad/s}^2$$

3. Determine the Time Tracker for Reversal ($$\omega = 0$$)

Angular acceleration is defined as the time rate of change of angular velocity ($$\omega$$):

$$\alpha = \frac{d\omega}{dt} \implies d\omega = \alpha \, dt$$

Assuming the pulley starts from rest ($$\omega = 0$$ at $$t = 0$$), we integrate the angular acceleration to determine the angular velocity at any subsequent time $$t$$:

$$\omega(t) = \int_{0}^{t} \alpha(t) \, dt = \int_{0}^{t} (4t - t^2) \, dt$$

$$\omega(t) = \left[ 2t^2 - \frac{t^3}{3} \right]_{0}^{t} = 2t^2 - \frac{t^3}{3}$$

The direction of motion reverses at the exact moment when the angular velocity momentarily drops to zero ($$\omega = 0$$), after having run in the forward direction:

$$2t^2 - \frac{t^3}{3} = 0 \implies t^2 \left( 2 - \frac{t}{3} \right) = 0$$

Excluding the initial start condition ($t = 0$), the time taken to reach the reversal point is:

$$2 - \frac{t}{3} = 0 \implies t = 6 \,\, \text{seconds}$$

4. Calculate the Total Angular Displacement ($$\theta$$)

Angular velocity is defined as the time rate of change of angular displacement ($$\theta$$):

$$\omega = \frac{d\theta}{dt} \implies d\theta = \omega \, dt$$

To find the total number of radians turned by the pulley before it reverses direction, we integrate the angular velocity from $$t = 0$$ to $$t = 6 \,\, \text{seconds}$$:

$$\theta = \int_{0}^{6} \omega(t) \, dt = \int_{0}^{6} \left( 2t^2 - \frac{t^3}{3} \right) \, dt$$

$$\theta = \left[ \frac{2t^3}{3} - \frac{t^4}{12} \right]_{0}^{6}$$

Evaluating this expression at the upper limit of integration ($$t = 6$$):

$$\theta = \left( \frac{2 \cdot 6^3}{3} \right) - \left( \frac{6^4}{12} \right)$$

$$\theta = \left( \frac{2 \cdot 216}{3} \right) - \left( \frac{1296}{12} \right)$$

$$\theta = (2 \cdot 72) - 108 = 144 - 108 = 36 \,\, \text{rad}$$

5. Convert Radians to Total Revolutions ($$N$$)

A single full revolution around a circle spans exactly $$2\pi$$ radians. Therefore, the total number of complete revolutions ($$N$$) made by the pulley is:

$$N = \frac{\theta}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi}$$

Using the standard numerical approximation for pi ($$\pi \approx 3.1416$$):

$$N \approx \frac{18}{3.1416} \approx 5.73 \,\, \text{revolutions}$$

Concept Check: The value $$5.73$$ falls squarely within the range that is more than 3 but less than 6.

Correct Option Key: Option A (more than 3 but less than 6)