Question

(a) A 22.0-kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force is exerted if he is 1.25 m from its center? (b) What centripetal force is exerted if the merry-go-round rotates at 3.00 rev/min and he is 8.00 m from its center? (c) Compare each force with his weight.

Answer 1

a) F₁ = 483 N, b) F₂ = 176 N and c) F₁ / W = 2.2 , F₂ / W = 0.82

Step-by-step explanation:

To find force we use Newton's second law, where acceleration is centripetal

F = m a

The centripetal acceleration is

a = v² / r

The relationship between linear and angular velocity is

v = w r

Substituting Newton's law

F = m (w r)² / r

F = m w² r

Let's calculate for each case

a) Let's reduce the units to the SI system

w = 40.0 rev / min (2pi rad / rev) (1min / 60s)

w = 4,189 rad/s

F₁ = 22.0 4.189² 1.25

F₁ = 482.6 N

F₁ = 483 N

b) w = 3.00 rev / min = 0.314 rad / s

F₂ = 22.0 0.314 2 8.00

F₂ = 176 N

c) the child's weight is

W = m g

W = 22.0 9.8

W = 215.6 N

To make this comparison we can divide the two quantities

F₁ / W = 483 / 215.6

F₁ / W = 2.2

F₂ / W = 176 / 215.6

F₂ / W = 0.82