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

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.