Question

A string is wrapped around the rim of a wheel of moment of inertia $0.40 \, \text{kgm}^2$ and radius $10 \, \text{cm}$. The wheel is free to rotate about its axis. Initially, the wheel is at rest. The string is now pulled by a force of $40 \, \text{N}$. The angular velocity of the wheel after $10 \, \text{s}$ is $x \, \text{rad/s}$, where $x$ is ______.

Answer 1

Correct Answer: 100

The angular velocity of the wheel after 10 seconds is determined by the relationship between torque, moment of inertia, and angular acceleration. The torque \( \tau \) generated by a force \( F \) on a wheel with radius \( r \) is calculated as \( \tau = F \times r \). With \( F = 40 \, \text{N} \) and \( r = 10 \, \text{cm} = 0.1 \, \text{m} \),

\[\tau = 40 \, \text{N} \times 0.1 \, \text{m} = 4 \, \text{Nm}.\]

Angular acceleration \( \alpha \) is derived from \( \alpha = \frac{\tau}{I} \), where \( I \) is the wheel's moment of inertia. Given \( I = 0.40 \, \text{kgm}^2 \), the acceleration is:

\[\alpha = \frac{4 \, \text{Nm}}{0.40 \, \text{kgm}^2} = 10 \, \text{rad/s}^2.\]

Assuming the wheel starts from rest (\( \omega_0 = 0 \)), its angular velocity \( \omega \) after time \( t \) is given by \( \omega = \omega_0 + \alpha \cdot t \).

Substituting the values \( \omega_0 = 0 \), \( \alpha = 10 \, \text{rad/s}^2 \), and \( t = 10 \, \text{s} \):

\[\omega = 0 + 10 \, \text{rad/s}^2 \times 10 \, \text{s} = 100 \, \text{rad/s}.\]

The computed angular velocity, \( x = 100 \, \text{rad/s} \), is consistent with the expected value of 100. The solution is thus confirmed.