velocity/acceleration of merry-go round

Question

An 8.0 m radius merry-go round completes one revolutions every 7.0 s.

a) what is the angular velocity of the ride?

b) With what speed are children moving when they ride the ride?

c) what is the centripetal acceleration the kids feel when riding on the merry-go round?

d) what is the average acceleration of each child over the course of a full revolution turn of the merry go round?

Answer 1

The period of motion here is T=7s per one revolution.

The angular velocity is ω = 2π/T = 2 * 3.14 radians / 7s = 0.90 radians/s or just 0.90 1/s (some teachers count radians as units, others do not)

This is the solution to part a.

for part b we can use the relation between tangential velocity and angular velocity to solve it:

v = rω = 8m * 0.90 1/s = 7.2m/s

This is the solution to part b.

part c is simple, the equation for centripetal acceleration resulting from uniform circular motion is

a_c = v²/r = (7.2m/s)²/(8m) = 6.4m/s²

This is the solution to part c.

part d is more difficult and requires consideration for the vector properties of acceleration. Acceleration is a vector which means that it has a magnitude and a direction just like velocity. Start by imagining the centripetal acceleration vector as any one child goes around the merry go round. Now think of just two points, one when they begin and one when they are halfway around the circle. The magnitude of the acceleration is the same at both places, but the directions are opposite. These are equal and opposite acceleration vectors and cancel one another when added together. Now we can do this same process at every point along the circle, that at any point in the rotation there is an opposite point halfway around that will cancel our accelerations. When we add up all of these we get a total net acceleration of 0 for one full rotation. For each child we therefore have an average acceleration of 0m/s² for one full turn of the merry go round. This is the solution to part d.