Question

You are the technical consultant for an action-adventure film in which a stunt calls for the hero to drop off a 24.9-m tall building and land on the ground safely at a final vertical speed of 3.87 m/s. At the edge of the building's roof, there is a 137-kg drum that is wound with a sufficiently long rope (of negligible mass), has a radius of 0.541 m, and is free to rotate about its cylindrical axis with a moment of inertia I0. The script calls for the 56.1-kg stuntman to tie the rope around his waist and walk off the roof.

a) Determine the required value of the stuntman's acceleration if he is to land safely at a speed of 3.87 m/s, and use this value to calculate the moment of inertia of the drum about its axis. Tries 0/99

b) What is the angular acceleration of the drum? Tries 0/99

c) How many revolutions does the drum make during the fall?

Answer 1

Let T be the tension in the rope..
mg-T=ma....(1)
Tr=I*?=Ia/r or T=Ia/r²

Putting this value of T in (1) and simplifying

                     a = mg/(m+I/r²)

Part b: u = 0,S = 24.9,v = 3.87

             a = (v²-u²)/2S = 0.30 m/s²

part a: acceleration: a = mg/(m+I/r²)

                               0.3 = 56.1*9.8/(56.1+I/0.541^2)

Moment of inertia: I = 480.26 kgm²

Using (1), T=m(g-a) = 56.1(9.8-0.3) = 532.95 N

Angular acceleration: ?=Tr/I = 532.95*0.541/480.26 = 0.6 rad/s^2

h=n * 2?r or n= 24.9/(2*?*0.541)= 7.3 = 7 (approx)