A 63 kg diver is spinning forward in a tuck position at 1.7 rev/s , as shown in

Question

A 63 kg diver is spinning forward in a tuck position at 1.7 rev/s, as shown in part (a) of the figure. In this position, the diver's rotational inertia is 4.5 kg·m2. At the peak of her trajectory, the diver suddenly straightens out in a horizontal position as shown in part (b) of the figure. 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

4.5 * 1.7 = 21 * w

w = 0.3643 rev/s

The time it takes to make a complete revolution given by the diver must now revolve 1/4 rev which means the fall time is [ (1/4 rev) / (0.3643 rev/s) ] = 0.6863 seconds for her center of mass to be 1.1 m above the water

now we know the time he stays in air from heigh h till he reach water,

h = u*t+0.5*a*t^2

u = 0 for vetical velocity at height h,and a= 9.8

h = 0.5*9*0.6863^2

h = 2.307 m

Therefore Minimum height is

h = 2.307 + 1.1 = 3.407 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.