Question

Tarzan, who has a mass of 85 kg, holds onto the end of a vine that is at a 13 ∘ angle from the vertical. He steps off his branch and, just at the bottom of his swing, he grabs onto his chimp friend Cheetah, whose mass is 30 kg.

What is the maximum angle the rope reaches as Tarzan swings to the other side?

Answer 1

m = mass of tarzan = 85 kg

m' = mass of friend = 30 kg

L = length of the vine

_i = initial angle of vine with the vertical = 13

h_i = initial height from where tarzan steps off = (L - L Cos_i ) = L - L Cos13

_f = final angle of vine with the vertical =

h_f = final height to which tarzan and his friend together rise = (L- L Cos_f ) = L - L Cos

v_i = speed of tarzan at the bottom just before grabbing his friend = ?

v_f = speed of tarzan and his friend after grabbing at he bottom = ?

consider the motion of tarzan from Top to bottom

Using conservation of energy between initial position of tarzan and bottom point

potential energy at the top = kinetic energy at the bottom just before grabbing the friend

mgh_i = (0.5) m v_i²

v_i = sqrt(2gh_i)

v_i = sqrt(2g(L - L Cos13))                                                                            eq-1

m'= mass of friend = 30 kg

similarly , using conservation of energy between bottom point and Top point after grabbing

kinetic energy at the bottom after grabbing = potential energy at top

(m + m') gh_f = (0.5) (m + m') v_f²

v_f= sqrt(2gh_f)

v_f = sqrt(2g(L - L Cos))                                                                                 eq-2

using conservation of momentum at the bottom

initial total momentum before grabbing = final momentum after grabbing

m v_i = (m + m') v_f

using eq-1 and eq-2

m (sqrt(2g(L - L Cos13)) )= (m + m') (sqrt(2g(L - L Cos)) )

inserting the values

(85) (sqrt((1 - Cos13)) )= (85 + 30) (sqrt((1 - Cos)) )

Cos = 0.986

= Cos^-1(0.986)

= 9.6 deg