Question

A particle moves on a circle of radius r with centripetal acceleration as function of time as ${{a}_{c}}={{k}^{2}}r{{t}^{2}}$ where k is a positive constant. Find the resultant acceleration.

$\begin{aligned} & \text{A}\text{. }k{{t}^{2}} \\\ & \text{B}\text{. }kr \\\ & \text{C}\text{. }kr\sqrt{{{k}^{2}}{{t}^{4}}+1} \\\ & \text{D}\text{. }kr\sqrt{{{k}^{2}}{{t}^{4}}-1} \\\ \end{aligned}$

Answer 1

Hint: The centripetal acceleration should be enough for maintaining the circular motion. Hence, we can compare it with the conventional formula to find the velocity of the particle. Tangential acceleration can be found by differentiating the velocity of the particle. The resultant acceleration is the vector combination of centripetal acceleration and tangential acceleration.

Formula Used:
Centripetal acceleration for a rotating object is given by,
${{a}_{c}}=\dfrac{{{v}^{2}}}{r}$

Where,
$v$ is the velocity of the particle
$r$ is the radius of the particle

Complete step by step answer:

Given, the centripetal acceleration is,
${{a}_{c}}={{k}^{2}}r{{t}^{2}}$..................(1)

If a particle is rotating in a circle of radius r with velocity v, the centripetal acceleration is given by,
${{a}_{c}}=\dfrac{{{v}^{2}}}{r}$.................(2)

Comparing equation (1) and (2) we get,
${{k}^{2}}r{{t}^{2}}=\dfrac{{{v}^{2}}}{r}$
$\Rightarrow {{v}^{2}}={{k}^{2}}{{r}^{2}}{{t}^{2}}$
$\Rightarrow v=krt$

Hence, the velocity of the particle is given by,
$krt$................(3)

As you can see in the expression of velocity, the velocity is also changing with time.

Hence, there must be a tangential acceleration which is responsible for the changing velocity.

We know that acceleration is the rate of change of velocity.

So, we can write,
${{a}_{t}}=\dfrac{dv}{dt}$....................(4)

We can put the value of velocity that we found in equation (3) into equation (4).

So, we have,
${{a}_{t}}=\dfrac{dv}{dt}=\dfrac{d(krt)}{dt}=kr$

So, the tangential acceleration is given by,
${{a}_{t}}=kr$

Tangential acceleration is working along the tangent of the circle and the centripetal acceleration is working along the radius of the circle.

Hence, both these accelerations are perpendicular to each other.

So, we can find the resultant acceleration to be,
$a=\sqrt{a_{t}^{2}+a_{c}^{2}}$
$\Rightarrow a=\sqrt{{{k}^{2}}{{r}^{2}}+{{k}^{4}}{{r}^{2}}{{t}^{4}}}$
$\Rightarrow a=kr\sqrt{{{k}^{2}}{{t}^{4}}+1}$

So, the required value is,
$a=kr\sqrt{{{k}^{2}}{{t}^{4}}+1}$

The correct answer is (C).

Note: As the centripetal acceleration is changing with time, the velocity should also be changing with time to maintain a constant radius of motion. Otherwise, the particle will move away or move towards the centre of the circle.