Tarzan and Jane in the 21st Century. 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 5.00 m in the trees and swings down to the ground using a thin, but very rigid, 30.0-kg vine 8.00 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 (a) just before she grabs Tarzan and (b) just after she grabs him? (c) How high will Tarzan and Jane go on their first swing after this daring rescue?

Solution 90 P Step 1: Data given Mass of Jane m = 60 kg j Height h = 5.0 m Mass of vine m = 30 kg v Length of vine L = 8.0 m Mass of Tarzan m =t72 kg We angular speed of Jane and vine before she grabs Tarzan We know KE = PE PE = mgh KE = 1/2 mv 2 Thus we can write as mgh = 1/2 × mv 2 Using this we shall find the velocity We get v = gh Substituting values we get v = 2 × 9.8 m/s × 5.0 m v = 98 m /s 2 v = 9.9 m/s We need to find the angular velocity This is given by v = r Here r = L = 8.0 m Thus we get = v/r Substituting values we get = 9.9 m/s/8.0 m = 1.23 rad/s Hence we have the angular velocity of the Jane and wine as she about to get Tarzan as 1.23 rad/s Step 2: We shall law of conservation of momentum we have (m + m )v v + m v = (m + m + m )vv J J t t j t f Substitute the values we get (60 kg + 30 kg ) × 9.9 m/s + 72 × 0 = (60 kg + 30 kg + 72 kg )v f 891 m/s = 162 v f v = 5.5 m/s f In terms of angular velocity it is given by f v /Rf Substitute the values we get f 5.5 m/s/8.0 m f 0.68 rad/s Hence the angular velocity after rescuing Tarzan will be 0.68 rad/s