Question

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

Answer 1

Hint: In this question circle of radius and revolution per second is given, so with the help of it we will calculate the angular velocity further by applying the formula of angular acceleration in terms of radius of the motion of particle we will get the answer.

Formula used- $\omega = 2\pi n,\alpha = {\omega ^2}r$

Complete step-by-step solution -

Given that radius of circle is $r = 25cm = \dfrac{{25}}{{100}}m = 0.25m$
Revolution per second $n = 2$
As we know that the angular velocity is given as
$\omega = 2\pi n$
Here n is the revolutions per second
So, the angular velocity for this case will be
$\because \omega = 2\pi r \\\ \Rightarrow \omega = 2\pi 2 \\\ \Rightarrow \omega = 4\pi {\text{ rad/}}{{\text{s}}^2} \\\$
As we know that the formula of acceleration is given as
$\alpha = {\omega ^2}r$
Here r is the radius of the circular path of movement.
Substituting all the values as obtained by us in above formula we get
$\because \alpha = {\omega ^2}r \\\ \Rightarrow \alpha = {\left( {4\pi } \right)^2} \times \left( {0.25} \right) \\\ \Rightarrow \alpha = 16{\pi ^2} \times \left( {0.25} \right) \\\ \Rightarrow \alpha = 4{\pi ^2} \\\$
Hence the acceleration of the particle is $4{\pi ^2}$ .

Note- Angular velocity refers to how quickly an object rotates or revolves relative to a different point , i.e. how quickly an object's angular position or orientation changes with time. The angular acceleration relates to the angular velocity transition time scale. It is measured in angle units per squared time unit, and is usually represented by the alpha symbol.