Question

A child sits at a distance of 5.0 m from the axis of merry-go-round in an amusement park. If the merry-go-round makes 2 rev

in 10.0s, find the (a) tangential speed and (b) centripetal acceleration of the child.

Answer 1

(a) tangential speed

$$v=\frac{2\pi R}{T}=\frac{2\pi*5.0}{5.0}=6.28\:\rm m/s$$

(b) centripetal acceleration

$$a=\frac{v^2}{R}=\frac{6.28^2}{5.0}=7.9\:\rm m/s^2$$