Question

A particle moves in a circle of radius 25 cm at two revolutions per second. The acceleration of the particle is
$\left( A \right){\pi ^2}{\text{ m}}{{\text{s}}^{ - 2}}$
$\left( B \right)8{\pi ^2}{\text{ m}}{{\text{s}}^{ - 2}}$
$\left( C \right)4{\pi ^2}{\text{ m}}{{\text{s}}^{ - 2}}$
$\left( D \right)2{\pi ^2}{\text{ m}}{{\text{s}}^{ - 2}}$

Answer 1

Hint: In this question use the property that net acceleration working on the particle is the sum of tangential acceleration and the centripetal acceleration, and for a constant angular velocity the tangential acceleration becomes zero so use this concept to reach the solution of the question.

Complete Step-by-Step solution:
The net acceleration working on the particle is the sum of tangential acceleration and the centripetal acceleration.

$ \Rightarrow {a_{net}} = {a_{\tan }} + {a_{cen}}$
Now it is given that the particle moves at the rate of two revolutions per second (Rps).
$ \Rightarrow N = 2$Rps
Now the angular speed of the particle is
$ \Rightarrow \omega = 2\pi N$ rad/sec.
Now substitute the value we have,
$ \Rightarrow \omega = 2\pi \times 2 = 4\pi $ rad/sec.
So as we see that the angular velocity is constant so the tangential acceleration of the particle is zero, as tangential acceleration is the measure of how quickly a tangential velocity changes.
So the net acceleration working on the particle becomes the centripetal acceleration.
$ \Rightarrow {a_{net}} = 0 + {a_{cen}} = {a_{cen}}$
Now as we know that the centripetal acceleration is the square of the linear velocity (v) divided by the radius (R) of the circle.
$ \Rightarrow {a_{cen}} = \dfrac{{{v^2}}}{R}$$ m/s^2$...................... (1)
Now as we know that the linear velocity is the product of angular velocity and the radius of the circle.
$ \Rightarrow v = \omega R$
Now substitute this value in equation (1) we have,
$ \Rightarrow {a_{cen}} = \dfrac{{{{\left( {\omega R} \right)}^2}}}{R} = {\omega ^2}R $ $ m/s^2$.
Now substitute the values we have,
$ \Rightarrow {a_{cen}} = {\left( {4\pi } \right)^2}\left( {\dfrac{{25}}{{100}}} \right)$, $\left[ {\because 25{\text{ cm}} = \dfrac{{25}}{{100}}{\text{m}}} \right]$
$ \Rightarrow {a_{cen}} = {\left( {4\pi } \right)^2}\left( {\dfrac{{25}}{{100}}} \right) = 16{\pi ^2}\left( {\dfrac{1}{4}} \right) = 4{\pi ^2}$ $ m/s^2$.
So the net acceleration working on the particle is
$ \Rightarrow {a_{net}} = {a_{cen}} = 4{\pi ^2}{\text{ m}}{{\text{s}}^{ - 2}}$
Hence option (C) is the correct answer.

Note – Whenever we face such types of questions always recall the formula of centripetal acceleration working on the particle if the particle rotates in a circle having radius (R) which is stated above, then change the linear velocity into angular velocity as linear velocity is the product of angular velocity and the radius of the circle then substitute the values in this equation and simplify, we will get the required answer.