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

Step 1: Torque and angular acceleration The torque (τ) acting on the wheel is:

$\tau = F \times R.$

Substitute $F = 40 \, \text{N}$ and $R = 0.10 \, \text{m}$:

$\tau = 40 \times 0.1 = 4 \, \text{Nm}.$

From the rotational dynamics equation:

$\tau = I \times \alpha,$

where $I = 0.40 \, \text{kgm}^2$ is the moment of inertia and $\alpha$ is the angular acceleration. Solving for $\alpha$:

$\alpha = \frac{\tau}{I} = \frac{4}{0.4} = 10 \, \text{rad/s}^2.$

Step 2: Angular velocity after 10 seconds Using the kinematic relation for angular motion:

$\omega_f = \omega_i + \alpha \times t.$

Here, the initial angular velocity $\omega_i = 0$, $\alpha = 10 \, \text{rad/s}^2$, and $t = 10 \, \text{s}$. Substituting these values:

$\omega_f = 0 + 10 \times10 = 100 \, \text{rad/s}.$

Final Answer: 100 rad/s.