Question

A ball is whirled on the end of a string in a horizontal circle of radius R at speed v. By which one of the following means can the centripetal acceleration of the ball be increased by a factor of 1.5 (or 3/2)?

Answer 1

Explanation:

When an object moves in a circular path, it will have circular acceleration. Its magnitude of acceleration is given by :  

$a=\omega^2R$

Since, $\omega=\dfrac{2\pi }{T}R$

$a=(\dfrac{2\pi}{T})^2R$

T is the time period

R is the radius of the circular path

To increase the centripetal acceleration bu a factor of 1.5 or 3/2, radius of circle must be increase by a factor of 6 and T is increased by a factor of 2 such that,

R'=6R and T'=2T

So,

$a'=(\dfrac{2\pi}{T'})^2R'$

$a'=(\dfrac{2\pi}{(2T)})^2(6R)$

$a'=\dfrac{6}{4}(\dfrac{2\pi}{T})^2R$

$a'=\dfrac{3}{2}(\dfrac{2\pi}{T})^2R$

Hence, this is the required solution.