Question

If jane and tarzan are initially 8m apart and Jane's mass is 60 kg, what's the maximum tension in the vine that's 25m long, and at what point does it occur?

Answer 1

The tension will be maximum when Jane reaches the bottom point of her swing (i.e Jane is completely upright so that the tension and her weight acts along a straight line)

The centripetal force will act towards the center of the circular path which is equal to

mv^2/r = T - mg

=> T = mv^2/r + mg = m(v^2/r + g)-----------1

We need to find Jane's velocity at the bottom point.

If h is the vertical displacement of Jane,

(2r-h)*h = (L/2)^2

(50 - h)*h = (8/2)^2

h^2 - 50h + 16 = 0

h = 0.322 m (neglecting the other valueof h as h cannot be 49 m)

Now, by conservation of energy, mgh = (1/2)mv^2

=> v^2 = 2gh

Hence 1 becomes,

T = mv^2/r + mg = mg(2h/r + 1) = 60*9.81(2*0.322/25 + 1) = 603.7 N