Human Rotational Energy. A dancer is spinning at 72 rpm about an axis through he

Question

Human Rotational Energy. A dancer is spinning at 72 rpm about an axis through her center with her arms outstretched, as shown in the following figure. From biomedical measurements, the typical distribution of mass in a human body is as follows:
Head: 7.0%
Arms: 13%(for both)
Trunk and legs: 80.0%
Suppose the mass of the dancer is 56.5 kg , the diameter of her head is 16 cm, the width of her body is 24 cm, and the length of her arms is 60 cm.

Part a:

Calculate moment of inertia about dancer spin axis. Use the figures in the following table to model reasonable approximations for the pertinent parts of your body.

Part B:

Calculate your rotational kinetic energy.

Explanation / Answer

for head :

mh = mass of head = 0.07 m = 0.07 x 56.5

d = diameter of head = 0.16 m

r = radius of head = 0.08 m

Ih = moment of inertia of head = (2/5) mhr2 = (0.4) (0.07 x 56.5) (0.08)2 = 0.010125

For arms :

mass of each arm , m = 0.13 m/2 = 0.13 x 56.5 x 0.5

L = length of each arm = 0.6 m

r = distance from axis = 12 cm = 0.12

moment of inertai of hands = 2 (mL2/3 + mr2) = 2 (m) (L2/3 + r2) = 2 (0.13 x 56.5 x 0.5 ) (0.62/3 + 0.122) = 0.987

moment of inertia of trunk and leg :

m = 0.8 x 56.5 kg

r = 0.12 m

moment of inertia = (0.5) m r2 = (0.5) ( 0.8 x 56.5) (0.12)2 = 0.325

total moment of inertia = I = 0.010125 + 0.987 + 0.325 = 1.32

b)

w = 72 rpm = 72 (2 x 3.14 / 60) = 7.54

rotational KE is given as

rotational KE = (0.5) I w2 = (0.5) (1.32) (7.54)2 = 37.5 J

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