Question

1. A 31.0 kg child on a swing is displaced by a maximum angle of 20 degrees to the vertical above her rest position. Assuming no loss of energy and that the swing's rope is 2m long:

   a) At what point during the swing will she attain her maximum speed?
   b) What will be her maximum speed through the subsequent swing?
   c)Assuming this maximum height was the result of one push from her parent, what was the average force exerted by the parent if s/he pushed over a distance of 70 cm?

Answer 1

a) At the lowest position of vertical displacement (y=0 )velocity of child is maximum.

b) At this point maimum speed is

$v= \omega\sqrt{A^2-o^2}=\omega A$

So A is amplitude of motion

A = 11 – cos

so,

and

$\omega = \sqrt\frac{l}{g}$

so we get

$\omega = 0.45 rad/s$

so speed of child

$v=\omega A= 0.054m/s$

3) Now restoring force coefficient

$k=\omega^2 m= 6.27 N/m$

so average force

$F=kx= 6.27 \times 0.7=4.39N$