How would I go about solving this problem: "Diver, weight given, in rigid position with radius of gyration of 1.2 ft, radius from ground to CG 1.5 ft. He is initially vertical, and tips forward to 90º. How many revolutions does he make before hitting the water at 30ft?"

Question

anonymous · Answer

I'm thinking that his initial energy is completely potential energy, and his final is completely kinetic and rotational, but this doesn't get me the correct answer. The book's answer is 0.934 revolutions.

anonymous · Answer

The equation I'm starting with is:
$$mgh = mv^2/2+mk^2\omega^2/2$$, where k is the radius of gyration
I'm then figuring that I can transform [v] to $$v = r\omega$$ but this gives me the wrong answer. Any ideas?

kociela · Answer

is there any mass given for moment of inertia ? that previous would be maybe right if the speed would instantly be the ending one.  otherwise "radius from ground to CG 1.5 ft" ? does the diver extend his arms.

kociela · Answer

so in the upper you got the time from potential energy equation and one point is that the rotational energy is different part. diver get it from his muscle-work.

kociela · Answer

from potential energy by y = y_o - v(y)*t - 1/2 * g * t^2 ?

anonymous · Answer

no mass, but i am given weight. mass moment is constant throughout

anonymous · Answer

I got similar answers to one person that replied, and apparently deleted their answer, being 1.3656 seconds and 1.59 revolutions. These are not right. Any ideas on what I'm missing?