Question

Tarzan has foolishly gotten himself into another scrape with the animals and must be rescued once again by Jane. The 60.0 kg Jane starts from rest at a height of 4.50 m in the trees and swings down to the ground using a thin, but very rigid, 28.0 kg vine 8.50 m long. She arrives just in time to snatch the 72.0 kg Tarzan from the jaws of an angry hippopotamus. What is Jane's (and the vine's) angular speed Part A just before she grabs Tarzan? ω = rad/s Part B just after she grabs him? ω = rad/s Part C How high will Tarzan and Jane go on their first swing after this daring rescue?

Answer 1

initial potential energy of jane + vine = final translation KE of jane + rotational KE of vine

translation KE of jane = (1/2)*M*v^2 = (1/2)*M*L^2*w^2

v = L*w

rotational KE of vine = (1/2)*I*w^2

I = (1/3)*m*L^2

L = length of the vine

m*g*H/2 + M*g*H = (1/2)*M*L^2*w^2 + ((1/3)*m*L^2*w^2)

(28*9.8*4.5/2) + (60*9.8*4.5) = ((1/2)*60*8.5^2*w^2) + ((1/3)*28*8.5^2*w^2)

angular speed w = 1.1 rad/s   <<<========answer

part (B)

angular momentum before grab = angular momentum after grab

(m/3 + M)*L^2*w = (m/3 + M + M1 )*L^2*wf

(m/3 + M )*w = (m/3 + M + M1)*wf

(28/3 + 60)*1.1 = (28/3 + 60 + 72)*wf

wf = 0.54 rad/s <<<==========answer

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part(C)

KE after grab = final PE

(1/2)*(m/3 + M + M1)*L^2*wf = (m*g*h/2) + ((M+M1)*g*h)

(1/2)*(28/3 + 60 + 72)*8.5^2*0.54^2 = (28*9.8*h/2) + ((60+72)*9.8*h)

h = 1.04 m <<<===answer