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

To find the angular acceleration, we use the formula for uniform angular acceleration: \(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega\) is the change in angular velocity and \(\Delta t\) is the change in time. 

Given:

• Initial angular speed, \(\omega_i = 1200 \text{ rpm}\)
• Final angular speed, \(\omega_f = 3120 \text{ rpm}\)
• Time interval, \(\Delta t = 16 \text{ s}\)

First, convert angular speeds from rpm to rad/s using the conversion factor \(1 \text{ rpm} = \frac{2\pi}{60} \text{ rad/s}\):

\(\omega_i = 1200 \times \frac{2\pi}{60} = 40\pi \text{ rad/s}\)

\(\omega_f = 3120 \times \frac{2\pi}{60} = 104\pi \text{ rad/s}\)

The change in angular velocity, \(\Delta \omega = \omega_f - \omega_i = 104\pi - 40\pi = 64\pi \text{ rad/s}\)

Now, using the formula for angular acceleration:

\(\alpha = \frac{\Delta \omega}{\Delta t}\)

\(\alpha = \frac{64\pi}{16} = 4\pi \text{ rad/s}^2\)

Therefore, the angular acceleration is \(4\pi \text{ rad/s}^2\).

Answer 2

We have two rotational velocities:
\(ω_1\), which is 1200 revolutions per minute, and \(ω_2\), which is 3120 revolutions per minute.
We also know that the time it takes for the rotation to go from \(ω_1\) to \(ω_2\) is 16 seconds.
Using this information, we can calculate the angular acceleration \((\alpha)\) of the rotation as follows:

\(\alpha = [\frac{(ω_2 - ω_1)}{t}] \times [\frac{2\pi}{60}]\)

\(\alpha = [\frac{(3120 - 1200)}{16}] \times [\frac{2\pi}{60}]\)

\(\alpha = (\frac{1920}{16}) \times (\frac{2\pi}{60})\)

\(\alpha = 4\pi\)

Therefore, the angular acceleration is 4π radians per second squared.