Question

A child sits on the edge of a spinning merry-go-round that has a radius of 1.5m. The child's speed is 2 m/s. What is the child's acceleration?

Answer 1

Since child is revolving on the merry go round

So here speed of child is 2 m/s and its radius is given as 1.5 m

Now during revolution of wheel the acceleration of child is known as centripetal acceleration

It is given as

`a = (v^2)/(R)`

`a = (2^2)/(1.5)`

`a = (8)/(3) m/s^2`

so here acceleration of child will be 2.67 m/s^2