Question

A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius $9 \, \text{m}$ and completes $120$ revolutions in $3$ minutes. The magnitude of centripetal acceleration of the monkey is (in $\text{m/s}^2$):

• A. zero
• B. $16 \pi^2 \, \text{m/s}^2$
• C. $4 \pi^2 \, \text{m/s}^2$
• D. $57600 \pi^2 \, \text{m/s}^2$

Answer 1

Answer: (B)

$16 \pi^2 \, \text{m/s}^2$

Answer 2

Given:
- Radius of the circular track: $R = 9 \, \text{m}$
- Number of revolutions completed: 120 revolutions
- Time taken: 3 minutes

Step 1: Calculate the Angular Velocity
The angular velocity $\omega$ is given by:

$\omega = \frac{\text{Total revolutions}}{\text{Time taken}} \times 2\pi \, \text{rad/s}.$

Substituting the given values:

$\omega = \frac{120 \, \text{revolutions}}{3 \, \text{minutes}} \times \frac{2\pi \, \text{rad}}{1 \, \text{revolution}}.$

Converting time to seconds:

$\omega = \frac{120 \times 2\pi}{3 \times 60} \, \text{rad/s} = \frac{4\pi}{3} \, \text{rad/s}.$

Step 2: Calculate the Centripetal Acceleration
The centripetal acceleration $a_{\text{centripetal}}$ is given by:

$a_{\text{centripetal}} = \omega^2 R.$

Substituting the values of $\omega$ and $R$:

$a_{\text{centripetal}} = \left(\frac{4\pi}{3}\right)^2 \times 9.$

Simplifying:

$a_{\text{centripetal}} = \frac{16\pi^2}{9} \times 9 = 16\pi^2 \, \text{m/s}^2.$

Therefore, the magnitude of the centripetal acceleration of the monkey is $16\pi^2 \, \text{m/s}^2$.