Question

Merry-go-rounds are a common ride in park play-grounds. 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 6.0 s and the diameter of the ride is 16 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

A)The speed of the rider is 2.6m/s

B)The rider's radial acceleration is  $2.75m/s^2$

C)The radial acceleration when the time for one rotation is halved is $10.7 m/s^2$

Explanation:

Time for one rotation t=6s

$diamter\ of\ the\ ride=16\ feet=16\times 0.3048=4.9 m\\radius\ r=diameter/2=4.9/2=2.45 m$

speed is the rate of change of distance with time.

to calculate the distance we have to find the circumference of the ride

A)$circumference=2\pi r=2\times3.14\times 2.45\\=15.4\\speed\ of\ the\ ride\ v=circumference/time=15.4/6=2.6m/s$

B)

$radial\ acceleration=v^2/r\\=2.6^2/2.45=2.75 m/s^2$

C)If the time of rotation is halved, speed/velocity as well as the radial acceleration changes.

new time t'=3s

new velocity v'=circumference/t'=15.4/3=5.13 m/s$new\ radial\ acceleration=v'^2/r=5.13^2/2.45=10.7m/s^2$