Question

A child sits on the edge of a spinning merry-go-round that has a radius of 1.2 m. The child's speed is 5 m/s. What is the child's acceleration?

Answer 1

Answer:

$20.8 m/s^2$

Explanation:

The centripetal acceleration of the child, which is moving by uniform circular motion, is given by

$a=\frac{v^2}{r}$

where

v = 5 m/s is the tangential speed of the child

r = 1.2 m is the radius of the circular path

Substituting the numbers into the equation, we find

$a=\frac{(5 m/s)^2}{1.2 m}=20.8 m/s^2$