A guys spinning on a chair has mass dropped on him

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When a mass drops into the lap of a person spinning on a chair, the conservation of angular momentum suggests that the angular velocity (ω) should decrease due to the increased moment of inertia. However, the specific effects depend on where the mass lands relative to the axis of rotation; if it lands at the center, it does not create torque, potentially allowing the spin rate to remain unchanged. The moment of inertia of the mass must be accurately calculated, particularly for a spherical object like a bean bag, which complicates the analysis. The discussion emphasizes the importance of precise calculations and modeling to understand the system's behavior before and after the mass is added. Ultimately, the angular momentum remains conserved, but the angular velocity may change depending on the moment of inertia of the combined system.
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Homework Statement


A guy is spinning on a chair with his hands at rest on his lap. As he is spinning, a large mass drops into his hands/lap. Does the guy continue spinning at the same rate, a slower rate, or a faster rate?


Homework Equations


L = Iω = m(r^2)ω


The Attempt at a Solution


If mass is added, shouldn't ω decrease due to the conservation of angular momentum?
But I did a quick experiment with my brother and I didn't seem to slow down...
 
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wootman23 said:

Homework Statement


A guy is spinning on a chair with his hands at rest on his lap. As he is spinning, a large mass drops into his hands/lap. Does the guy continue spinning at the same rate, a slower rate, or a faster rate?


Homework Equations


L = Iω = m(r^2)ω


The Attempt at a Solution


If mass is added, shouldn't ω decrease due to the conservation of angular momentum?
But I did a quick experiment with my brother and I didn't seem to slow down...

What's the moment of inertia of the mass that was dropped?

Or maybe a better question is what is the change in angular momentum of the dropped mass itself? (Hint: it depends on other variables besides just the value of the mass alone. It also depends on the shape of the mass, exactly where it lands on the guy in the chair [i.e., whether it lands on the center of rotation, or some distance from it], etc.)
 
collinsmark said:
Or maybe a better question is what is the change in angular momentum of the dropped mass itself? (Hint: it depends on other variables besides just the value of the mass alone. It also depends on the shape of the mass, exactly where it lands on the guy in the chair [i.e., whether it lands on the center of rotation, or some distance from it], etc.)

Well the question says the mass is a bean bag...so maybe its spherical?

And it falls in his lap which in basically on the axis of rotation so it doesn't provide any torque.
So, I'm thinking the speed stays the same, but the equation doesn't support that.
L = Iω = m(r^2)ω
If the mass increases the angular speed should decrease.

Am I using the wrong equation?
 
wootman23 said:
Well the question says the mass is a bean bag...so maybe its spherical?

And it falls in his lap which in basically on the axis of rotation so it doesn't provide any torque.
So, I'm thinking the speed stays the same, but the equation doesn't support that.
L = Iω = m(r^2)ω
If the mass increases the angular speed should decrease.

Am I using the wrong equation?

mr2 isn't the right formula for the moment of inertia of a sphere (or spherical beanbag) rotating around its own center.

When solving this problem, you might discover that the moment of inertia of the guy plus chair without the bag doesn't change much compared to the moment of inertia of the guy, chair and bag combination, thus you might be able to ignore the effects of the bag. This depends on how accurate your instructor would like you to be. So what is the angular momentum of the bag before it is dropped? what is the angular momentum of the bag after it is dropped? (Hint: It's not ωmr2. That's not the right formula for a spherical bean bag rotating around its center. Nor is it the right formula for a guy in a chair.)

If you want to be precise, you need to
  • find the angular momentum of the guy on the chair before the mass is dropped on him. As part of this process find the moment of inertia of the guy plus the chair before the mass is dropped on him.
  • Find the moment of inertia and angular momentum of the beanbag before it is dropped on the guy.
  • Then find the moment of inertia of the guy and beanbag on the chair (all combined).

The angular momentum of the system will be the same before and after the beanbag is dropped. But that does not necessarily mean the angular velocity will be the same before and after. For that you must compare the moment of inertia of the guy spinning on the chair before the bag was dropped, and the moment of inertia of the system after the bag was dropped.
 
collinsmark said:
If you want to be precise, you need to
  • find the angular momentum of the guy on the chair before the mass is dropped on him. As part of this process find the moment of inertia of the guy plus the chair before the mass is dropped on him.
  • Find the moment of inertia and angular momentum of the beanbag before it is dropped on the guy.
  • Then find the moment of inertia of the guy and beanbag on the chair (all combined).

I don't think I need to be precise because this is just a concept question. No actual measurements are given.

I have no idea for the find the angular momentum of the guy on the chair either before of after the bean bag is dropped on him.
If we are being precise, then the moment of inertia the spherical bean bag would be (2/5)M(R^2).

I really appreciate the help collinsmark, but I'm just getting more confused. Maybe if I ask a different question?

Say you have a wheel whose center is attached to a friction-less vertical pole and is free to rotate about it; the wheel is parallel with the floor; the wheel is spinning.
Next, say you drop a ball of significant weight (say its weight is half of the wheels weight), right onto the center of the wheel.
Does the wheel still spin with the same angular velocity?
 
wootman23 said:
I don't think I need to be precise because this is just a concept question. No actual measurements are given.

I have no idea for the find the angular momentum of the guy on the chair either before of after the bean bag is dropped on him.
If we are being precise, then the moment of inertia the spherical bean bag would be (2/5)M(R^2).
Great! :smile:

So now what you might want to do is model the moment of inertia of a guy sitting in the chair. (perhaps model it by a couple of cylinders and use the parallel axis theorem). Then plug some ballpark numbers in. Is the moment of inertia of the beanbag significant?

I really appreciate the help collinsmark, but I'm just getting more confused. Maybe if I ask a different question?

Say you have a wheel whose center is attached to a friction-less vertical pole and is free to rotate about it; the wheel is parallel with the floor; the wheel is spinning.
Next, say you drop a ball of significant weight (say its weight is half of the wheels weight), right onto the center of the wheel.
Does the wheel still spin with the same angular velocity?

That's a good idea. Considering the problem by modeling a spinning wheel is easier than modeling a guy on a chair. But you can answer this question for yourself. :wink:

  1. Find the angular momentum of a stationary ball plus the angular momentum of a spinning wheel. Assume the wheel is spinning at angular velocity ω0. Call the angular momentum of the system L0.
  2. Find the moment of inertia of the same wheel plus the spherical ball attached at its center, where the ball's mass is half that of the wheel. (You'll still have to decide on the relationship between the ball's radius and the wheel's radius.) Call the moment of inertia of this new wheel-ball system, I1.
  3. Invoke conservation of angular momentum, such that the new angular momentum is L1 = L0.
  4. Divide L1 by I1 to find the new angular velocity, ω1.

Plug some typical numbers in. How does ω1 compare to ω0 for typical values?

[Edit: And what happens as the ratio of the ball's radius to the wheel's radius approaches zero? (Rball/Rwheel → 0)]
 
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