Question

A particle moves in a circular path of radius r with speed v. It then increases its speed to 2v while travelling along the same circular path. The centripetal acceleration of the particle has changed by what factor?
A. 0.25
B. 0.5
C. 2
D. 4

Answer 1

To solve this problem, use the formula for centripetal acceleration for a particle in uniform circular motion in terms of speed of the particle and the radius of the circular path. Substitute the values in the formula and find the centripetal acceleration of the particle moving with speed v. Then, using the same formula find the centripetal acceleration of the particle when it is moving with speed 2v. Divide these two values of centripetal acceleration and find the factor by which centripetal acceleration of the particle has changed.

Formula used:
${a}_{c}= \dfrac {{v}^{2}}{r}$

Complete solution step-by-step:
Centripetal acceleration of the particle is given by,
${a}_{c}= \dfrac {{v}^{2}}{r}$
Where,
${a}_{c}$ is the centripetal acceleration
$v$ is the speed of the particle
$r$ is the radius of the circular path
When the particle is moving with speed v, the centripetal acceleration is given by,
${a}_{1}= \dfrac {{v}^{2}}{r}$ …(1)
When the particle is moving with speed 2v, the centripetal acceleration is given by,
${a}_{2}= \dfrac {\left({2v}\right) ^{2}}{r}$
$\Rightarrow {a}_{2}= \dfrac {4{v}^{2}}{r}$ …(2)
Dividing equation. (1) by (2) we get,
$\dfrac {{a}_{1}}{{a}_{2}}= \dfrac {\dfrac {{v}^{2}}{r}}{\dfrac {4{v}^{2}}{r}}$
$\Rightarrow \dfrac {{a}_{1}}{{a}_{2}}=\dfrac {1}{4}$
$\Rightarrow \dfrac {{a}_{1}}{{a}_{2}}= 0.25$
Thus, the centripetal acceleration of the particle has changed by a factor of 0.25.

So, the correct answer is option A i.e. 0.25.

Note:
Students must not get confused between centripetal acceleration and radial acceleration. Centripetal acceleration is also known as radial acceleration. The direction of the centripetal acceleration is radially towards the centre of the circular path. If the speed of the particle increases or decreases during the circular motion then the net force deviates the particle from its initial path.