Problem 7-51 Energy cons. - Part 7

A boy is playing with a rope tied to a tree near his favourite swimming hole. Initially the boy is stationary and the rope (of length \(3.7 \;m\)) makes an angle of \(48^\circ\) with the vertical. He then lifts his feet slightly and starts to swing freely. If air resistance is neglected, use conservation of energy to determine:

(a) his speed at the bottom of the swing
(b) the minimum height, relative to his initial position, to which he can swing.

Accumulated Solution

At point 2, \(E_P = 0\)

At point 2, \(E_K = (1/2)mv^2\)

\(h = 3.7(1 - \cos48) \;m = 1.22 \;m\)

E at point  \(1 = 0 + mgh\)

E at point  \(2 = (1/2)mv{_2}{^2} + 0\)

\(v_2 = 4.9 \;m/s\)    (answer to part (a))

\(mgh = (1/2)mv{_2}{^2} \\ v{_2}{^2} = 2gh = 2(9.8)(1.22) = 23.9 \\ v_2 = 4.9\; m/s  \; \text{  (answer to part (a))}\)

With our chosen origin the boy has no \(E_P\) at the bottom of the swing, just \(E_K\). What happens to this as he swings through the lowest point?

(A)   It disappears at the bottom and he comes to rest.

(B)   It is converted back into \(E_P\) as he rises again.

(C)   It is converted into heat.