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 typical size is a diameter of 12 ft. 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. Estimate a typical time for one rotation. (a) For your estimated time, what is the speed of the rider, in m/s? (b) What is the rider's radial acceleration, in m/s2? (c) What is the rider's radial acceleration if the time for one rotation is halved?

Answer 1

The answers to the question are;

(a) The speed of the rider for your estimated time of 6 s in m/s is 1.91 m/s

(b) The rider's radial acceleration, in m/s² is 2.006 m/s²

(c) The rider's radial acceleration = 8.03 m/s².

Step-by-step explanation:

To solve the question, we note that

Angular velocity ω is the time to make a given change in the angle of an object per unit time.

Therefore ω =
`(\theta)/(t) = (S)/(r*t) = (v)/(r)`

Where:

θ = Position angle

t = Time

S = Arc length

v = Linear velocity

r = Circle radius

Where the Diameter of the merry-go-round is 12 ft the circumference is given by

Circumference of merry-go-round = π×D = π × 12 ft = 37.7 ft

If it takes 6 seconds make one revolution, then we have

ω₁
`= (37.7)/(6*6)` = 1.047 rad/s

(a) The speed is given by

v = r×ω₁ = 6 ft × 1.047 rad/s = 6.28 ft/s = ‪1.91 m/s

The speed of the rider for your estimated time of 6 s in m/s = 1.91 m/s

(b) The radial acceleration in m/s² is given by

a
`_r` = ω₁²×r = where r = 6 ft = 1.83 m

(1.047 rad/s)² × 1.83 m= 2.006 m/s²

The rider's radial acceleration, in m/s² = 2.006 m/s²

(c) If the time for one rotation is halved then the angular velocity is doubled and we have

ω₂ =
`(\theta)/(t)` =
`(2\pi )/((t)/(2) )` =
`2*(2\pi )/(t)` = 2×ω₁ = 2×1.047 rad/s = 2.094 rad/s

The radial acceleration in m/s² is then given by

ω₂²× r = (2.094 rad/s)² × 1.83 m = 8.03 m/s²

The rider's radial acceleration = 8.03 m/s²