twisting swing has a plank seat 2 feet wide of negligible mass. It takes a force

Question

twisting swing has a plank seat 2 feet wide of negligible mass. It takes a force of 20 nds applied to each side of the seat at force is approxi ou ely constant as the rson is wrapped up five twists. After being person is extended their full height of 6 pe released from rest it takes 8 seconds to unwind if the up (to half velocity and acceleration of the person in 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 bein twisted? (d) Approximate the moment of inertia of the person's body by assuming it is a (a) What is the final foot rod spinning about its center. Calculate that moment of inertia and compare with the experimental values. (e) Formulate conclusions

Explanation / Answer

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 = 0.6728353ft

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