Question

The angular speed of a flywheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad/s² is:

• A. \(2\pi\)
• B. \(4\pi\)
• C. \(12\pi\)
• D. \(104\pi\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Angular acceleration (\(\alpha\)) is defined as the rate of change of angular velocity (\(\omega\)) with respect to time.
Uniform angular acceleration means the rate of change is constant.
Key Formula or Approach:
The equation for rotational motion with uniform acceleration is:
\[ \omega = \omega_0 + \alpha t \implies \alpha = \frac{\omega - \omega_0}{t} \]
where \(\omega_0\) is initial angular speed and \(\omega\) is final angular speed.
To convert from rpm (revolutions per minute) to rad/s:
\[ \omega = \frac{2\pi N}{60} \]
Step 2: Detailed Explanation:
1. Initial angular speed (\(\omega_0\)):
\[ \omega_0 = \frac{2\pi \times 1200}{60} = 2\pi \times 20 = 40\pi \text{ rad/s} \]
2. Final angular speed (\(\omega\)):
\[ \omega = \frac{2\pi \times 3120}{60} = 2\pi \times 52 = 104\pi \text{ rad/s} \]
3. Time interval (\(t\)) = 16 seconds.
4. Calculating angular acceleration (\(\alpha\)):
\[ \alpha = \frac{104\pi - 40\pi}{16} = \frac{64\pi}{16} \]
\[ \alpha = 4\pi \text{ rad/s}^2 \]
Step 3: Final Answer:
The angular acceleration is \(4\pi \text{ rad/s}^2\).