The grabbing of Jane at the lowest point of the swing is a momentum-conserving event, but some kinetic energy is lost. Then they swing to the highest point with energy conserved again.
Let Tarzan's speed before grabbing Jane = V1:
Tarzan's mass = Mt
Jane's mass = Mj
Mt*g*(1-cos25.9) = (Mt/2)*V1^2
V1 = sqrt[2*g*(1-cos25.9)] = 1.403 m/s
After grabbing Jane,
V1*Mt = V2(Mt+Mj)
V2 = V1*[Mt/(Mt+Mj)]= 0.641V1 = 0.899 m/s
Final maximum angle A is given by
g(1-cosA) = (1/2)V2^2
1 - cosA = 0.04126
cosA = 0.9587
A = 16.5 degrees
Tarzan, King of the Jungle (mass = 73.2 kg), grabs a vine of length 14.1 m hanging from a tree branch. The angle of the vine was 25.9° with respect to the vertical when he grabbed it. At the lowest point of his trajectory, he picks up Jane (mass = 41.0 kg) and continues his swinging motion. What angle relative to the vertical will the vine have when Tarzan and Jane reach the highest point of their trajectory?
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