In the middle of a dive, a diver tucks herself into a ball-like shape to execute

Question

In the middle of a dive, a diver tucks herself into a ball-like shape to execute turns at 19.0 rad/s. Her rotational inertia in the tuck position is 1.30 kg · m2 . Then, a moment before entering the water, the diver straightens her body. She is 1.60 m tall, and her mass is 50.0 kg. (a) What is the final rotational inertia of the diver in the straightened position about her center of mass. Treat the diver as a thin uniform rod rotating about the center. (b) How fast is the diver rotating after she extends her body? (c) Calculate a change in her rotational kinetic energy during straightening of her body.

Explanation / Answer

a)

given that w1 = 19 rad/s

I1 = 1.3 kg-m^2

a) moment of inertia when entering water is I = (1/12)*M*L^2 = (1/12)*50*1.6^2 = 10.67 kg-m^2

b) using law of conservation of angular momentum

I1*w1 = I2*w2

1.3*19 = 10.67*w2

w2 = 2.32 rad/s

c) initial kinetic energy is KEi = 0.5*I1*w1^2 = 0.5*1.3*19^2 = 234.65 J

Final kinetic energy is KEf = 0.5*I2*w2^2 = 0.5*10.67*2.32^2 = 28.72 J

change in kinetic energy is KEf - KEi = 28.7-234.65 = -205.95 J

negative sign shows that loss in rotational kinetic energy

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