Question

A body is moving along a circular path of radius $r$ with a frequency of revolution numerically equal to the radius of the circular path. What is the acceleration of the body if radius of the path is $\frac{5}{\pi} \, \text{m}?$

• A. \( 100\pi \, \text{ms}^{-2} \)
• B. \( 500\pi \, \text{ms}^{-2} \)
• C. \( 25\pi \, \text{ms}^{-2} \)
• D. \( \left( \frac{500}{\pi} \right) \, \text{ms}^{-2} \)

Hint:

Remember that the centripetal acceleration depends on both the radius of the path and the frequency of revolution. When the frequency is equal to the radius, you can use this simplified formula for angular velocity and acceleration.

Answer 1

The Correct Option is D

The centripetal acceleration \( a \) for a body moving in a circular path is given by the formula: \[ a = \omega^2 r \] Where: - \( \omega \) is the angular velocity, and - \( r \) is the radius of the circular path. The angular velocity \( \omega \) is related to the frequency of revolution \( f \) by the relation: \[ \omega = 2\pi f \] Given that the frequency of revolution \( f \) is numerically equal to the radius of the circular path \( r \), we can write: \[ f = r \] So, the angular velocity becomes: \[ \omega = 2\pi r \] Substituting this into the formula for acceleration: \[ a = (2\pi r)^2 r = 4\pi^2 r^3 \] Now, substitute the given radius \( r = \frac{5}{\pi} \, \text{m} \): \[ a = 4\pi^2 \left( \frac{5}{\pi} \right)^3 = 4\pi^2 \times \frac{125}{\pi^3} = \frac{500}{\pi} \, \text{ms}^{-2} \] Thus, the acceleration of the body is \( \left( \frac{500}{\pi} \right) \, \text{ms}^{-2} \).