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 6.0 s and the diameter of the ride is 16 ft.
For this typical time, what is the speed of the rider in m/s? Express your answer in meters per second. V AEV O ? m /s Submit Request Answer Part B What is the rider's radial acceleration, in m/s27 Express your answer in meters per second squared. V AED + O 2 Grad Submit Request Answer
Part C What is the rider's radial acceleration if the time for one rotation is halved? Express your answer in meters per second squared. tad m/s2 Submit Request Answer

Answer 1

Answer:

2.55348650884 m/s

2.67400481907 m/s²

10.6960192763 m/s²

Explanation:

d = Diameter of the ride = 16 ft=$16\times 0.3048=4.8768\ m$

r = Radius = $\dfrac{4.8768}{2}=2.4384\ m$

t = Time taken = 6 seconds

Velocity is given by

$v=\dfrac{2\pi r}{t}\\\Rightarrow v=\dfrac{2\pi 2.4384}{6}\\\Rightarrow v=2.55348650884\ m/s$

The speed is 2.55348650884 m/s

Acceleration is given by

$a=\dfrac{v^2}{r}\\\Rightarrow a=\dfrac{(\dfrac{2\pi 2.4384}{6})^2}{2.4384}\\\Rightarrow a=2.67400481907\ m/s^2$

The acceleration is 2.67400481907 m/s²

When the speed is halved

$v=\dfrac{2\pi 2.4384}{\dfrac{6}{2}}=5.10697301768\ m/s$

$a=\dfrac{v^2}{r}\\\Rightarrow a=\dfrac{(\dfrac{2\pi 2.4384}{\dfrac{6}{2}})^2}{2.4384}\\\Rightarrow a=10.6960192763\ m/s^2$

The acceleration is 10.6960192763 m/s²