Rotation Problem: determine the number of revolutions

[cehen2]

Homework Statement

It's been a while since I've done a problem like this, and I wanted to see if anyone could spot my errors. I know that I'm making an incorrect assumption somewhere, but can't quite figure it out.

The problem states:

A 150 lb man is on the edge of a diving board, 30 feet above the surface of the water. He curls up into a ball and leans forward at an angle \theta until he falls while rotating to the surface of the water. If his center of gravity is 1.2 feet above his feet on the diving board, approximately how many revolutions will he complete before he hits the water?

Given:
Weight = 150 lbs; Mass = 150 lbs / 32.2 = 4.66 slugs
h = 30 feet

Find:
Number of revolutions

Homework Equations

I = MR²
\sum\tau = I\alpha
\tau = r x F = rFsin\theta

The Attempt at a Solution

My thought is that when the person leans forward far enough over the board, his center of gravity will create a torque that will start the spinning. Once he tips off of the board, he continues to spin while falling into the water.

I = MR² = 4.66 * (1.2)² = 6.71 *I'm assuming you just treat the center of gravity as a point mass and not like a sphere or ring.

\tau = r x F = rFsin\theta = 1.2 * 250 * sin\theta = 180 sin\theta *This is one of the areas where I don't quite know what theta to use..

\sum\tau = I\alpha
180 sin\theta = 6.71 \alpha
26.8 sin\theta =\alpha

Now I know what I want to do. I want to find the alpha and use this to find \omega. Once the diver leaves the diving board, I assume that \alpha goes to 0 since no more external torques are applied to the system. But I do not think I can assume that alpha is constant throughout the initial motion of rolling off of the diving board. I can find the time for the diver to hit the water since acceleration should be constant and equal to 32.2 ft/s². Once I know this, I can figure out the total radians that the diver rotates through and convert to revolutions by dividing by 2 \pi. The answer is supposed to be 0.934 revolutions, but I don't quite know how to get there. I tried using energy as well, but I wasn't able to get that to work either.

I would appreciate any tips to get me moving in the correct direction.

Thanks

 

[rock.freak667]

You should use projectile motion equations to get the time it takes for him to reach the water then use that as the time to get the number of revolutions.

 

[cehen2]

I think I should have included a picture initially. Please see the quick sketch (not to scale) showing what I am describing. I understand the use of projectile motion equations when items are launched off of a ledge with a significant velocity in the x direction. I don't think this is the case for this particular problem, but definitely correct me if I'm wrong. The way I'm interpreting it is that the person curled up into a ball leans forward until they fall straight down while rotating. I can find the time to fall fairly easily; however, my question is about how to handle the torque providing the initial rotation.

Edit: Also attached is a FBD. Initially when the person is curled up without leaning forward, the only forces he experiences are his weight and the normal force of the platform. When the person leans forward far enough, if you take his center of mass as the pivot point, the normal force creates a torque. At some critical angle, \theta_c, the normal force goes to 0 and there is no longer a torque on the person. My issue is finding this critical value.

Thanks for any additional help!

 

[cehen2]

Just wondering if anyone else had any ideas.

I tried rock.freak667's suggestion to use projectile motion equations to get the time. I solved h = 1/2 gt² for t and got t = 1.366 seconds. This is the same time I got when I just used the equations for constant acceleration (i.e. h = h_o + v_ot + 1/2 gt²).

Here is the closest I could get to the correct answer. I know I'm missing some fundamental assumption, but I can't quite place it.

Let's say I set \theta in my previous post to \pi/2. This assumes that when the person's center of gravity is parallel to the diving board, this is the moment he loses contact with the board. This gives me \alpha = 26.8 rad/s². If I plug this into \theta = \omega_ot + 1/2 \alphat² and solve for t, I get 0.342 seconds for the time it takes to rotate from \theta = 0 to \pi/2. This assumes that alpha is constant, which I do not think is true... I can then plug this into \omega = \omega_o + \alpha t . This gives me \omega = 9.17 rad/s. If I then go back and use \theta = \omegat + 1/2 \alphat² with the total time to fall t = 1.366 and \alpha = 0 (no more external torque), I get \theta =12.53 radians = 1.99 revolutions. This is about 1 revolution too many based on what the answer in the book says. I'm sure I am making an incorrect assumption somewhere. I'm not sure if I'm supposed to treat \alpha as a function of \theta(t) and integrate instead of assuming \alpha is constant.

Thanks in advance!