Question 237: Work, Energy and Power & Circular Motion

Question

A particle is performing uniform circular motion along the circumference of the circle of diameter 1 m with frequency 4 Hz . The acceleration of the particle in $\mathrm{m} / \mathrm{s}^2$ is

Accepted Answer

In uniform circular motion, the acceleration of a particle, also called the centripetal acceleration, is given by the formula:
$$ a = \omega^2 r $$
where $ \omega $ is the angular velocity and $ r $ is the radius of the circle.
Find the radius, $ r $:
Since the diameter is given as 1 m, the radius $ r $ is:
$$ r = \frac{	ext{diameter}}{2} = \frac{1}{2} \, 	ext{m} $$
Determine the angular velocity, $ \omega $:
The angular velocity $ \omega $ can be calculated using the relation between angular velocity and frequency:
$$ \omega = 2 \pi f $$
where $ f $ is the frequency. Given $ f = 4 \, 	ext{Hz} $,
$$ \omega = 2 \pi 	imes 4 = 8 \pi \, 	ext{rad/s} $$
Calculate the centripetal acceleration, $ a $:
Substitute $ \omega $ and $ r $ into the centripetal acceleration formula:
$$ a = \omega^2 r = (8 \pi)^2 	imes \frac{1}{2} $$
Simplify the expression:
$$ a = 64 \pi^2 	imes \frac{1}{2} = 32 \pi^2 \, 	ext{m/s}^2 $$
Thus, the acceleration of the particle is $ 32 \pi^2 \, 	ext{m/s}^2 $, which corresponds to
Option D
.

Answer

$8 \pi^2$

Answer

$16 \pi^2$

Answer

$24 \pi^2$

Answer

$32 \pi^2$

Work, Energy and Power & Circular Motion

A particle is performing uniform circular motion along the circumference of the circle of diameter 1 m with frequency 4 Hz . The acceleration of the particle in \mathrm{m} / \mathrm{s}^2 is

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