Swinging Child?

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 45 degree angle?

Answer 1

Use the law of conservation of energy. At the bottom potential energy is 0 so energy is equal to mv^2/2. At the top kinetic energy is 0 so energy is equal to mgR(1 - cos(45)) = mgR(1 - 1/sqrt(2)). So mv^2/2 = mgR(1 - 1/sqrt(2)), then v= sqrt(gR(2 - sqrt(2)))

Answer 2

This is a law of conservation of energy problem.

One will switch from Kinetic translational energy (.5*m*v^2) to potential energy (m*g*y -- where y is the height).

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The total energy will be conserved, so it is best to first set the equations equal to zero.

.5*m*v^2 - (m*g*y) =0 => .5*m*v^2 = m*g*y)

The masses cancel.

.5*v^2 = (g*y)

Now, one has all the variables, except y.

We use the height of the chain and the right triangle above to calculate the adjacent side ( the height of the ball off the ground currently), and then we subtract the calculated value from the 3m given to find the height it is elevated from rest position.

3 -(cos45*3) = .88m = y

Plug in y, and now we have the answer:

v= 4.15m/s

Answer 3

At the top of her swing her potential energy is
u = m*g*h
where h = 3 - 3*sin45

At the bottom, all of that potential energy is in form of kinetic energy (1/2 * m * v^2).