Question

A twisting swing has a plank seat 2ft wide of negligible mass. It takes a force of 20 punds applied to each side of the seat edge to twist a 180 pound person sitting in the seat. That force is pprox. costant as the person is wrapped up 5 twists. After being released from rest it takes 8 seconds to unwind if teh person extended their full height of 6 feet, and 4 sec if they are doubled up to 1/2 their height: a) what is the angular velocity and acceleration of the person on each case? b) what is the moment of inertia of the person in each case? c) how high vertically did the person rise when the swing was being twisted? d) approximate the moment of inertia of the person's body by assuming it is a 6-foot rod spinning about its center. calculate that moment of inertia and compare with the experimental values. e) formulate concllussions

Answer 1

given force F = 20 lbs

weight of person, N = 180 lbs

for 5 twists, work done = W

let the arm of the lever be x = 2/2 = 1 ft

then

W = 5*F*x*2*pi = 10*pi*F*x

a. case 1, t = 8 s

then

5*2*pi = 0.5*alpha*t^2

alpha = 0.981747 rad/s/ds

w = alpha*t = 7.853 rad/s

case 2, t = 4 s

5*2*pi = 0.5*alpha*t^2

alpha = 3.92699 rad/s/s

w = alpha*t = 15.707963 rad/s

b. case 1.

alpha = F*x/I

I = 1.27323954 kg m^2

case 2

alpha = Fx/I

I = 5.0929592 kg m^2

c. vertical height = h

h*N = F*5*2*pi*x

h = 3.490 ft

d. assuming a 6 ft rod spinning about center = I

I = ml^2/12

I = 180*36/12*32 = 16.875 kg m^2

e. as the rod is assumed to be turning about center perpendicular to the axis the values are different

for a rod about axis

I = mr^2/2 = 1.27323954

(180/32)r^2/2 = 1.27323954

r = 0.6728353379 ft

human body is a cylinder of about r = 0.67283533795424270952747302815463ft