Question

A particle is moving along a circular path of radius $5m$ with uniform speed $5m{s^{ - 1}}$. What is the magnitude of average acceleration during the interval in which the particle completes half revolution?

Answer 1

Hint
When the particle completes half a revolution, the magnitude of the particle will remain the same but the direction changes. The time can be calculated by the dividing circumference of the semi circle by the velocity of the particle. So by dividing the change in velocity with the time we will get the acceleration.
Formula Used: In this solution we will be using the following formula,
$a = \dfrac{{{v_1} - {v_2}}}{t}$
where $a$ is the acceleration,
${v_1}$ and ${v_2}$ are the velocities and $t$ is the time.

Complete step by step answer
In this question we are given that the velocity of the particle is given as $5m{s^{ - 1}}$. So this is the velocity of the particle at the initial position. After the particle completes half of the revolution, then the magnitude of velocity of the particle remains the same but the direction of the particle changes. Therefore, the velocity of the particle at that position will be $- 5m{s^{ - 1}}$.
Now the total distance travelled by the particle in half a revolution will be equal to half of the circumference of the circle that the particle is following.
The circumference of a circle is given by the formula, $2\pi r$
So half the circumference will be,
$d = \dfrac{{2\pi r}}{2}$
So we get the distance travelled by the particle as $d = \pi r$
Now this distance is travelled by the particle with a velocity of $5m{s^{ - 1}}$. So we can get the time taken by the particle to travel this distance as the distance divided by the speed.
$t = \dfrac{{\pi r}}{5}$
Now in the question we are given the radius of the circle is $5m$
So substituting we get,
$t = \dfrac{{\pi \times 5}}{5}$
Hence we get the time as,
$t = \pi s$
Now we have the initial velocity as, ${v_1} = 5m{s^{ - 1}}$ and the final velocity as ${v_2} = - 5m{s^{ - 1}}$
So we have the acceleration as,
$a = \dfrac{{5 - \left( { - 5} \right)}}{\pi }$
On calculating we get,
$a = \dfrac{{10}}{\pi }m/{s^2}$
Now substituting the value of $\pi$ we get,
$a = 3.183m/{s^2}$
This is the acceleration of the particle.

Note
The motion of a particle, when it is travelling a circular path at a constant speed is called the uniform circular motion. The distance of the particle from the axis of rotation is constant at all times. Though the particle is travelling with constant speed, the velocity is not constant as the velocity not only depends on the magnitude but also the direction of travel.