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A child jumps off a rotating platform

  • Thread starter yasar1967
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1. A child of mass 25kg stands at the edge of a rotating platform of mass 150kg and radius 4m. The platform with the child on it rotates with an angular speed of 6.2 rad/s. The child jumps off in a radial direction. (a) What happens to the angular speed of the platform? (b) What happens to the platform if, a little later , the child, starting at rest, jumps back onto the platform? (Treat the platform as a uniform disk)



2. I=1/2MR^2(platform) ; I=MR^2(child), L=Iω



3. My Attempt was to calculate the angular speed of the platform by using the fact that angular momentum is conversed. So two different inertias rotating at the same speed and then one of it leaves the system. Therefore the platform now has a different speed. But the solution I see in the book states that due to the fact that the child leaves the system in a radial direction the speed of the platform is NOT affected. How come? how is it important the direction? whether he leaves the platform radial or tangential the inertia is changed so the speed must be different. Why am I wrong, if I am?
 
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Andrew Mason
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2. I=1/2MR^2(platform) ; I=MR^2(child), L=Iω

3. My Attempt was to calculate the angular speed of the platform by using the fact that angular momentum is conversed. So two different inertias rotating at the same speed and then one of it leaves the system. Therefore the platform now has a different speed. But the solution I see in the book states that due to the fact that the child leaves the system in a radial direction the speed of the platform is NOT affected. How come? how is it important the direction? whether he leaves the platform radial or tangential the inertia is changed so the speed must be different. Why am I wrong, if I am?
Consider the angular momentum of the rotating platform and of the child separately. Does the child jumping off provide a torque to the platform? Recall that torque is the rate of change of angular momentum: [itex]\tau = Id\omega/dt = dL/dt[/itex] so if there is no torque, there is no change in angular momentum, hence angular speed, of the platform.

When the child jumps back on, however, what happens? Is there is a torque applied by the child to the platform?

AM
 
PhanthomJay
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When the child jumps off radially, he has still a tangential speed v = wR, where w is the same as his initial angular speed. Thus, his momentum remains the same and so must the momentum of the platform remain the same. No change in the angular speed of either. If he had jumped off tangentially, his tangential speed woul be different, thus there would be a change in the platforms angular speed as well. The situation is different however when he jumps back on from rest in a radial direction. Then there IS a change in the platforms speed, which you should calculate.
 
BobG
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2. I=1/2MR^2(platform) ; I=MR^2(child), L=Iω



3. My Attempt was to calculate the angular speed of the platform by using the fact that angular momentum is conversed. So two different inertias rotating at the same speed and then one of it leaves the system. Therefore the platform now has a different speed. But the solution I see in the book states that due to the fact that the child leaves the system in a radial direction the speed of the platform is NOT affected. How come? how is it important the direction? whether he leaves the platform radial or tangential the inertia is changed so the speed must be different. Why am I wrong, if I am?
You know the moment of inertia of a rotating system is equal to the sum of the moments of inertia of each of the parts (which is why the formula for moment of inertia is different for different shape objects). Once the overall moment of inertia is found, you multiply by the angular velocity to get the angular momentum. In other words, you have:

[tex]L = (\frac{1}{2}MR^2 + mr^2)\omega[/tex]
Using the distributive property,
[tex]L = \frac{1}{2}MR^2\omega + mr^2\omega[/tex]
So you could say that the angular momentum of the system is equal to the sum of the angular momentum of each of the parts. You just removed one part, so you're left with only the angular momentum of the platform, itself.

The radial direction is only important because the child has to impart some force to propel herself off of the platform. You wanted the force to be in a direction that didn't create any torque on the platform. In other words, the direction was chosen to focus in on the concept of removing part of the system.

Jumping back on in a radial direction will have an effect, since she will have no angular momentum. You're adding inertia without adding angular momentum.

Personally, I think she's going to have substantially less mass when she jumps back on. You realize her linear speed is nearly 55 mph when she jumps off, right?
 
Hootenanny
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Personally, I think she's going to have substantially less mass when she jumps back on. You realize her linear speed is nearly 55 mph when she jumps off, right?
:rofl: I hope to god that's just a typo :rofl:
 
Andrew Mason
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:rofl: I hope to god that's just a typo :rofl:
If she has substantially less mass because of the speed at which she jumped off she ain't going to be jumping back on....:frown:

AM:
 

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