Question

A 20 kg child is on a swing that hangs from 3.0-m-long chains. what is her maximum speed if she swings out to a 49 ∘ angle?

Answer 1

L = length of the chain = 3 m

θ = angle with the vertical = 49 deg

h = height gained

In triangle ABD

Cosθ = AB/AD

Cosθ = AB/L

AB = L Cosθ

BC = AC - AB = L - L Cosθ

so h = L - L Cosθ

inserting the values

h = (3) - (3) Cos49 = 1.032 m

v = maximum speed at the bottom of the swing

Using conservation of energy

kinetic energy at the bottom of the swing = potential energy at the maximum height

(0.5) m v² = mg h

v = sqrt(2gh)

v = sqrt(2 x 9.8 x 1.032)

v = 4.5 m/s