Question

A particle moves in a circle of radius 25 cm at 2 revolutions per second. The acceleration of the particle is _________${\pi ^2}m/{s^2}$.

Answer 1

Hint: We will use the concept of centripetal acceleration to solve that question. Expression of centripetal acceleration is given by: $a = \dfrac{{{\omega ^2}}}{r}$, where $\omega $ is the angular velocity and r is the circular trajectory radius.

Formula used: $a = \dfrac{{{\omega ^2}}}{r}$ and $\omega $= $2\pi f$.

Complete step-by-step solution -

As we know, an acceleration that is guided along the radius towards the middle of the circular path acts on a body undergoing uniform circular motion. That acceleration is called centripetal acceleration.
The magnitude of a particle's acceleration moving in a circular motion is given by,
$a = \dfrac{{{\omega ^2}}}{r}$.
This being so, radius r = 25 cm and frequency = 2 revolutions per second.
We remember, angular speed, $\omega $= $2\pi f$
$ \Rightarrow \omega = 2\pi \times 2 = 4\pi rad/\sec $.
Radius = 25cm = $\dfrac{{25}}{{100}}m = 0.25m$.
Still, Centripetal acceleration,
$ \Rightarrow a = {\omega ^2}r$
Putting all the values given in this equation, we get
$
   \Rightarrow a = {(4\pi )^2} \times 0.25m{s^{ - 2}} \\
   \Rightarrow a = 4{\pi ^2}m{s^{ - 2}} \\
$
Then the particle's acceleration is $4{\pi ^2}m{s^{ - 2}}$.

Note: First we need to remember some basic points of uniform circular motion in this type of problem. We'll then use the relationship between angular velocity and angular acceleration to resolve the problem. Through this relationship, when the frequency and radius of the circular path is given we can easily find both of them. We can get the required response through this.