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 55.0 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

(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. Express your answer to two significant figures and include the appropriate units.

I = ?

(B) Calculate your rotational kinetic energy. Express your answer to two significant figures and include the appropriate units.

K = ?

Clear explanations needed, I have no idea where to begin and I want to learn this. Thanks!

Explanation / Answer

Given

Mass of the dance m = 55 kg

Diameter of the head D = 16 cm = 0.16 m

Width of her body W = 24 cm = 0.24 m

Length of arms L = 60 cm = 0.60 m

Solution

Mass of the head mh = 0.07 x 55 = 3.85 kg

Mass of the arms ma = 0.13 x 55 = 7.15 kg

Mass of the trunk and legs mt = 0.80 x 55 = 44 kg

i) Moment of Inertia of the head

Let us consider the head as a sphere of diameter 0.16 m

The radius of the spherical head Rh = 0.16/2 = 0.08 m

Moment of Inertia of the sphere about it’s diameter

Ih = 2mhRh2/5

Ih = 2 x 3.85 x (0.08)2/5

Ih = 0.009856 Kgm2

ii) Moment of inertia of the arms

Let us consider arms as uniform rods of mass 7.15 kg

Mass of one rod is mr = 7.15/2 = 3.575 Kg

Moment of inertia of a rod about an axis passing through one of its side

Is = mrL2/3

Is = 3.575 x (0.60)2 / 3

Is =0.429 Kgm2

The arm is at a distance d = w/2 from the center of the body

d = 0.24/2

d = 0.12 m

So the moment of inertia of the rod about and axis at a distance 0.12 m is obtained by

Ia = Is + mr(d)2

Ia = 0.429 + 3.575 x 0.122

Ia = 0.48048 kgm2

This is the moment of inertia of one arm, for two arms

I2a = 2 x 0.48048

I2a = 0.96096 kgm2

iii) Moment of inertia of the body

The body is in a shape of cylinder whose diameter is w = 0.24 m

its radius Rt = w/2 = 0.12 m

moment of inertia of a cylinder about its own axis

It = mt(Rt)2/2

It = 44 x 0.122 /2

It = 0.3168 kgm2

Total Inertia

I = Ih + I2a + It

I = 0.009856 + 0.96096 + 0.3168

I = 1.29 kgm2

Angular velocity

= 72 rpm

= 72 x 2 / 60 rad/s

= 7.5 rad/s

Rotational Kinetic energy

K = ½ I 2

K = ½ x 1.29 x 7.52

K = 36.3 J

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