Question

A wheel of radius $ 0.2 \, \text{m} $ rotates freely about its center when a string that is wrapped over its rim is pulled by a force of $ 10 \, \text{N} $. The established torque produces an angular acceleration of $ 2 \, \text{rad/s}^2 $. Moment of inertia of the wheel is............. kg m².

Hint:

In problems involving rotational motion, torque and moment of inertia are related by the equation \( \text{Torque} = I \alpha \), where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration.

Answer 1

Correct Answer: 1

The wheel's moment of inertia (I) is determined using the equation \( \tau = I \cdot \alpha \), which relates torque (τ), moment of inertia (I), and angular acceleration (α).

Torque (τ) is calculated from the applied force (F) and wheel radius (r) using the formula \( \tau = F \cdot r \).

With the given values, the torque is:

\( \tau = 10 \, \text{N} \times 0.2 \, \text{m} = 2 \, \text{Nm} \)

This torque value is then used in \( \tau = I \cdot \alpha \) to find the moment of inertia (I):

\( I = \frac{\tau}{\alpha} = \frac{2 \, \text{Nm}}{2 \, \text{rad/s}^2} = 1 \, \text{kg m}^2 \)

The moment of inertia of the wheel is therefore \( 1 \, \text{kg m}^2 \), which falls within the specified range of (1, 1).