A 65 kg diver is spinning forward in a tuck position at 2.7 rev/s, as shown in F

Question

A 65 kg diver is spinning forward in a tuck position at 2.7 rev/s, as shown in Fig. 13-38a. In this position, the diver's rotational inertia is 3.3 kg·m2. At the peak of her trajectory, the diver suddenly straightens out in a horizontal position as shown in Fig. 13-38b. In this new position, the diver has an overall length of 2.2 m with center of mass midway along this length, and a rotational inertia of 21 kg·m2 about the rotation axis. What is the minimum height of the diver above the water if she is to enter the water hands first? Neglect the diver's horizontal translational motion.

Explanation / Answer

Conservation of angular momentum for the change in position

3.3*2.7=21*?

solve for ?

?=0.4242rev/s

the diver must now revolve 1/4 rev which means the fall time is

.589seconds for her center of mass to be 1.1 m above the water

h=.5*9.81*0.589^2

1.70 m

and 1.1

2.80 m

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