Question

A merry go round has a radius of 4m and completes a revolution in 2s. Then acceleration of a point on its rim will be:
A) $4{\pi ^2}$
B) $2{\pi ^2}$
C) ${\pi ^2}$
D) $Zero$

Answer 1

In this question we have to find the acceleration of a point on the rim of the merry go round. The radius and the time of one revolution are given. To find acceleration, first we will find the velocity and then further we will move with the calculation.

Complete step by step solution:
Given,
$R = 4m$
$t = 2s$
$Speed = \dfrac{{Distance}}{{Time}}$
The distance in this question is the rim or circumference of the path. So, the distance will be $2\pi R$. Now, we will put the value of distance and time
$\Rightarrow Speed = \dfrac{{2\pi R}}{2}$
$\Rightarrow Speed = \dfrac{{2\pi \times 4}}{2}$
$\Rightarrow Speed = 4\pi$
Acceleration $a = \dfrac{{{v^2}}}{R}$
$\Rightarrow a = \dfrac{{{{\left( {4\pi } \right)}^2}}}{4}$
$\Rightarrow a = 4{\pi ^2}{\text{ m/}}{{\text{s}}^2}$

Hence, from above calculation we have seen that the value of acceleration is $a = 4{\pi ^2}{\text{ m/}}{{\text{s}}^2}$.

Note: In this question we have to find the value of acceleration; so we have used the formula of acceleration. Merry go round is a type of amusement ride. It is circular shaped. The motion of a merry go round is circular motion. In this motion the merry go round exerts a centripetal force on the person riding it. The more we move away from the center of circular motion the more centripetal force will be applied on the person riding it. To keep moving in this motion the merry go round must exert more force on the person.
The centripetal force is the force that makes the body follow a curved path. It acts in the perpendicular direction to the body and towards a fixed point. The centripetal force on the person will be given by following formula;
${F_C} = \dfrac{{m{v^2}}}{R}$
Where,
${F_C}$ is the centripetal force in newton (N)
$m$ is mass in kg
$v$ is velocity in m/s
$R$ is the radius of motion in m