Lets swing a ball on the end of ONE string and divide call it 2 T
You were not told what angle or height so I will say dropped from height h.
m g h = (1/2) m v^2
so at the bottom v = sqrt (2gh)
then the centripetal acceleration = v^2/R
and 2 T = m g + m v^2/R
2 T= m (g + 2 g h/R) = mg (1 +2 h/R) = force up on child if sat has no mass :)
Now if you were given maximum angle theta instead of height h then
h = R - R cos theta = R(1-cos theta)
if theta is small cos theta = 1 -theta^2/2 +.....
so
h = (R /2)theta^2 note theta in radians
A child of mass m swings in a swing supported by two chains, each of length R.
The mass of the swing’s seat is negligible. If the tension in each chain at the
lowest point is T, find:
4.1 the child’s speed at the lowest point;
4.2 the force exerted by the seat on the child at the lowest point.
1 answer
1 answer