Question

A particle performing $U.C.M$ with radius $3\,m$ and linear speed $9\,m{s^{ - 1}}$. If $\vec r$ is a radius vector, its centripetal acceleration $\vec a$ is:
(A) $\vec a = - 9\vec r$
(B) $\vec a = 81\vec r$
(C) $\vec a = - 27\vec r$
(D) $\vec a = 3\vec r$

Answer 1

Hint: In this question, the distance and velocity are given and we have to determine the acceleration. So, by comparing the formula of acceleration with the formula of velocity, and then by using the given information the centripetal acceleration can be determined.

Formulae Used:
Acceleration of the object is given by,
$a = \dfrac{v}{t}$
Where, $a$ is the acceleration, $v$ is the velocity and $t$ is the time.
Velocity of the object is given by,
$v = \dfrac{d}{t}$
Where, \[v\] is the velocity, $d$ is the distance and $t$ is the time

 Complete step-by-step solution:
Given that,
The radius or distance, $r = d = 3\,m$
The velocity, $v = 9\,m{s^{ - 1}}$
Acceleration of the object is given by,
$a = \dfrac{v}{t}\,...............\left( 1 \right)$
Velocity of the object is given by,
$v = \dfrac{d}{t}\,...............\left( 2 \right)$
From equation (2), the time is written as,
$t = \dfrac{d}{v}\,..............\left( 3 \right)$
Substituting the equation (3) in the equation (1) for the easy calculation, then
$a = \dfrac{v}{{\left( {\dfrac{d}{v}} \right)}}$
By rearranging the terms in the above equation, then the above equation is written as,
$a = \dfrac{{{v^2}}}{d}\,............\left( 4 \right)$
By substituting the velocity and the radius or distance in the above equation (4), then the equation (4) is written as,
$a = \dfrac{{{9^2}}}{3}$
By squaring the terms in numerator, then the above equation is written as,
$a = \dfrac{{81}}{3}$
On dividing the above equation, then the above equation is written as,
$a = 27$
Now, the centripetal acceleration is towards the centre, then
$a = - 27$
Now, the vector form of the acceleration is,
$\vec a = - 27\hat r$
Thus, the above equation shows the centripetal acceleration is towards the centre and the radius $\hat r$.
Hence, the option (C) is the correct answer.

Note:- The negative sign of the acceleration indicates that the acceleration is towards the centre of the uniform circular motion of the particle, then the negative sign is included. The equation (4) can be derived by another form also. It can be derived by comparing the Uniform circular motion of force formula $\left( {F = \dfrac{{m{v^2}}}{r}} \right)$ with Newton’s second law of motion $\left( {F = ma} \right)$.