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 63.0 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. (Figure 1) 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. b) Calculate your rotational kinetic energy.

Explanation / Answer

After that, I'd calculate the moment of inertia for each body part.
If the head is a sphere, then I_head = (2/5)m(d/2)²
If the torso is a cylinder, then I_torso = ½m(d/2)²
If the arms are rods of length L that begin distance r from the axis of rotation,

then I_arms = 2*mL²/3 + mr²
by the parallel axis theorem.

I_arms = 2*mL²/3 + mr²

=(2*63*0.24^2)/3+63*0.30^2=8.08 kg m^2


B)

KE_rot = ½*I*²
where = 72rpm * 2rad/rev * 1min/60s = 7.54 rad/s

KE_rot = ½*I*²

=0.5*8.08*7.54^2

=229.68 j

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