A 75-kg person stands on a tiny rotating platform with arms outstretched. Part A

Question

A 75-kg person stands on a tiny rotating platform with arms outstretched.

Part A

Estimate the moment of inertia of the person using the following approximations: the body (including head and legs) is a 60-kg cylinder, 17 cm in radius and 1.7 m high; and each arm is a 7.5-kg thin rod, 61 cm long, attached to the cylinder.

Express your answer using two significant figures.

Part B

Using the same approximations, estimate the moment of inertia when the arms are at the person's sides.

Express your answer using two significant figures.

Part C

If one rotation takes 1.9 s when the person's arms are outstretched, what is the time for each rotation with arms at the sides? Ignore the moment of inertia of the lightweight platform.

Express your answer using two significant figures.

Part D

Determine the change in kinetic energy when the arms are lifted from the sides to the horizontal position.

Express your answer using two significant figures.

Explanation / Answer

let M = 60 kg

m = 7.5 kg

L = 61 cm = 0.61 m

R = 17 cm = 0.17 m

A) moment of inertia of the person, I = I_body + I_hands

= 0.5*M*R^2 + 2*(m*L^2/12 + m*(L/2 + R)^2)

= 0.5*60*0.17^2 + 2*(7.5*0.61^2/12 + 7.5*(0.61/2 + 0.17)^2)

= 4.72 kg.m^2

B) I = I_body + I_hands

= 0.5*M*R^2 + 2*m*R^2

= 0.5*60*0.17^2 + 2*7.5*0.17^2

= 1.3 kg.m^2

C) initial angular speed, w1 = 2*pi/1.9 = 3.307 rad/s

final angular speed, w2 = ?

use Conservation of angular momentum

I1*w1 = I2*w2

==> w2 = I1*w1/I2

= 4.7165*3.307/1.3005

= 12 rad/s

time taken to complete one revolution, T = 2*pi/w2

= 2*pi/12

= 0.52 s <<<<<<<<----------------Answer

D) Ki = 0.5*I1*w1^2 = 0.5*4.7165*3.307^2 = 25.8 J

Kf = 0.5*I2*w2^2 = 0.5*1.3005*12^2 = 93.6 J

delta_K = kf - ki

= 93.6 - 25.8

= 68 J <<<<<<<<----------------Answer

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