Question

Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their center. A rider sits at the outer edge of the disk and holds onto a metal bar while someone pushes on the ride to make it rotate. Suppose that a typical time for one rotation is 4.0 s and the diameter of the ride is 10 ft.A. For this typical time, what is the speed of the rider in m/s?B. What is the rider's radial acceleration, in m/s?C. What is the rider's radial acceleration if the time for one rotation is halved?

Answer 1

Answer:

A = 2.36m/s

B = 3.71m/s²

C = 29.61m/s2

Explanation:

First, we convert the diameter of the ride from ft to m

10ft = 3m

Speed of the rider is the

v = circumference of the circle divided by time of rotation

v = [2π(D/2)]/T

v = [2π(3/2)]/4

v = 3π/4

v = 2.36m/s

Radial acceleration can also be found as a = v²/r

Where v = speed of the rider

r = radius of the ride

a = 2.36²/1.5

a = 3.71m/s²

If the time of revolution is halved, then radial acceleration is

A = 4π²R/T²

A = (4 * π² * 3)/2²

A = 118.44/4

A = 29.61m/s²