Question

A particle is moving along a circular path of radius 5m with uniform speed 5m/s. What will be the average acceleration when the particle completes half revolution???

Answer 1

`10/pi ms^-2` see below for direction

Explanation:

The magnitude of acceleration is given by `a=v^2/r= 25/5=5ms^-2`

Acceleration is a vector so we need to take account of its direction. We know it is always acting towards the centre of the circle. The diagram above shows it at various stages in the half revolution.
As for all vectors, we can break the acceleration down into horizontal and vertical components. By looking at the diagram, we can see that vertical components will cancel each other out in a half revolution
.
The horizontal component of the black acceleration vector is `5sintheta`

In a half revolution (anticlockwise) `theta` will go from 0 to 180 degrees (`pi` radians). We need to add up all the possible values for `theta` hence we need to integrate and take an average over the possible angles (i.e `180^0 or pi`)

`5/piint_0^pisintheta=5/pi[-costheta]_o^180=-5/pi[-1-1]=10/pi=3.18ms^-2`
This is in the horizontal direction (left to right on the diagram)