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

To find the angular velocity of the wheel after 10 seconds, we can use the relationship between torque, moment of inertia, and angular acceleration. The torque \( \tau \) produced by the force \( F \) on a wheel of radius \( r \) is given by \( \tau = F \times r \). Given \( F = 40 \, \text{N} \) and \( r = 10 \, \text{cm} = 0.1 \, \text{m} \),

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

The angular acceleration \( \alpha \) is given by the formula \( \alpha = \frac{\tau}{I} \), where \( I \) is the moment of inertia of the wheel. Here, \( I = 0.40 \, \text{kgm}^2 \), so:

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

Knowing that the wheel is initially at rest (initial angular velocity \( \omega_0 = 0 \)), the angular velocity \( \omega \) after time \( t \) is given by the equation:

\[\omega = \omega_0 + \alpha \cdot t.\]

Substituting the known 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 calculated angular velocity \( x = 100 \, \text{rad/s} \) falls within the expected range of 100, 100. Therefore, the solution is verified as correct.

Answer 2

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.