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

To determine the moment of inertia (I) of the wheel, we start by using the relationship between torque (τ), moment of inertia (I), and angular acceleration (α). The equation is:

\( \tau = I \cdot \alpha \)

Given the force applied (F) and the radius (r) of the wheel, the torque can be calculated as:

\( \tau = F \cdot r \)

Substituting the given values, we have:

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

Next, use this torque in the relation \( \tau = I \cdot \alpha \) to solve for the moment of inertia (I):

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

Thus, the moment of inertia of the wheel is \( 1 \, \text{kg m}^2 \). This result fits exactly within the given range (1, 1).