Rotation Problem: determine the number of revolutions

In summary: F = rFsin\theta = 1.2 * 250 * sin\theta = 180 sin\theta But now I'm stuck. I don't know how to convert this into revolutions. I tried using energy, but that didn't work either. I would appreciate any help getting me closer to the correct answer.In summary, the 150 lb man on the diving board will rotate 0.934 revolutions before hitting the water.
  • #1
cehen2
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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 [tex]\theta[/tex] 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
[tex]\sum[/tex][tex]\tau[/tex] = I[tex]\alpha[/tex]
[tex]\tau[/tex] = r x F = rFsin[tex]\theta[/tex]

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.

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

[tex]\sum[/tex][tex]\tau[/tex] = I[tex]\alpha[/tex]
180 sin[tex]\theta[/tex] = 6.71 [tex]\alpha[/tex]
26.8 sin[tex]\theta[/tex] =[tex]\alpha[/tex]

Now I know what I want to do. I want to find the alpha and use this to find [tex]\omega[/tex]. Once the diver leaves the diving board, I assume that [tex]\alpha[/tex] 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 [tex]\pi[/tex]. 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
 
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  • #2
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.
 
  • #3
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, [tex]\theta[/tex]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!
 

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  • #4
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 = ho + vot + 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 [tex]\theta[/tex] in my previous post to [tex]\pi[/tex]/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 [tex]\alpha[/tex] = 26.8 rad/s2. If I plug this into [tex]\theta[/tex] = [tex]\omega[/tex]ot + 1/2 [tex]\alpha[/tex]t2 and solve for t, I get 0.342 seconds for the time it takes to rotate from [tex]\theta[/tex] = 0 to [tex]\pi[/tex]/2. This assumes that alpha is constant, which I do not think is true... I can then plug this into [tex]\omega[/tex] = [tex]\omega[/tex]o + [tex]\alpha[/tex] t . This gives me [tex]\omega[/tex] = 9.17 rad/s. If I then go back and use [tex]\theta[/tex] = [tex]\omega[/tex]t + 1/2 [tex]\alpha[/tex]t2 with the total time to fall t = 1.366 and [tex]\alpha[/tex] = 0 (no more external torque), I get [tex]\theta[/tex] =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 [tex]\alpha[/tex] as a function of [tex]\theta[/tex](t) and integrate instead of assuming [tex]\alpha[/tex] is constant.

Thanks in advance!
 
  • #5


Hello,

Thank you for sharing your attempt at solving this problem. It seems like you have a good understanding of the relevant equations and concepts, but may have made some incorrect assumptions or missed some key information in the problem.

First, it's important to note that the problem does not specify the initial angular velocity of the diver when he leaves the diving board. This means that you cannot assume that his angular acceleration (alpha) is constant throughout the motion. In fact, it is likely that his angular velocity will change as he falls and rotates due to the changing forces acting on him.

Additionally, you mentioned using energy to solve this problem. This could be a valid approach, but it's important to consider all forms of energy involved in the system, including potential and kinetic energy. It may also be helpful to consider the conservation of angular momentum in this problem.

Without giving away the solution, I would suggest revisiting your approach and considering all forms of energy and forces acting on the diver as he falls and rotates. Also, make sure to carefully consider the initial conditions and any assumptions you may be making. I hope this helps and good luck with your solution!
 

FAQ: Rotation Problem: determine the number of revolutions

What is the definition of rotation?

Rotation is the circular movement of an object around a central axis or point.

How is the number of revolutions determined?

The number of revolutions is determined by dividing the total angle of rotation by 360 degrees. This gives the number of full revolutions and any remaining degrees can be converted to partial revolutions.

What units are used to measure rotation?

Rotation is typically measured in degrees, radians, or revolutions. The choice of unit depends on the context and application of the rotation problem.

Can rotation be positive or negative?

Yes, rotation can be either positive or negative. Positive rotation is when an object rotates in a counterclockwise direction, while negative rotation is when an object rotates in a clockwise direction.

How does the direction of rotation affect the number of revolutions?

The direction of rotation does not affect the number of revolutions. The number of revolutions is solely determined by the total angle of rotation, regardless of the direction of rotation.

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