Question

A disk of 10 cm radius has a moment of inertia of 0.02 kg.m\(^2\). A force of 15 N is applied tangentially to the periphery of the disk to give it an angular acceleration \( \alpha \) of magnitude:

• A. 25 rad/s\(^2\)
• B. 35 rad/s\(^2\)
• C. 45 rad/s\(^2\)
• D. 75 rad/s\(^2\)

Hint:

The torque is the product of the force applied tangentially to the radius of the disk. The resulting angular acceleration depends on the moment of inertia.

Answer 1

The Correct Option is D

To find the angular acceleration \( \alpha \), we use the relation between torque \( \tau \), moment of inertia \( I \), and angular acceleration:

\[\tau = I \cdot \alpha\]

The torque \( \tau \) is given by the product of the applied force \( F \) and the radius \( r \) of the disk: 

\[\tau = F \cdot r\]

Substitute the given values: \( F = 15 \, \text{N} \), \( r = 0.1 \, \text{m} \), and \( I = 0.02 \, \text{kg} \cdot \text{m}^2 \).

Calculate the torque:

\[\tau = 15 \, \text{N} \times 0.1 \, \text{m} = 1.5 \, \text{Nm}\]

Now, substitute \( \tau \) into the first equation to solve for \( \alpha \):

\[1.5 = 0.02 \cdot \alpha\]

Solve for \( \alpha \):

\[\alpha = \frac{1.5}{0.02} = 75 \, \text{rad/s}^2\]

The angular acceleration \( \alpha \) is 75 rad/s\(^2\).

Answer 2

The torque \( \tau \) produced by the force is given by: \[ \tau = F \times r \] where \( r = 0.1 \, \text{m} \) (radius of the disk) and \( F = 15 \, \text{N} \). So, \[ \tau = 15 \times 0.1 = 1.5 \, \text{N m} \] Now, the angular acceleration \( \alpha \) can be found using the relation: \[ \tau = I \alpha \] where \( I = 0.02 \, \text{kg m}^2 \). Solving for \( \alpha \): \[ \alpha = \frac{\tau}{I} = \frac{1.5}{0.02} = 75 \, \text{rad/s}^2 \]