# Transfer of Angular Velocity?

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I'm having a little trouble simulating this so I was wondering if someone may know the answer to this problem.

Pretend you're standing on a giant spinning table. If you were standing in the center, and you jumped perfectly straight up, would your angular velocity transfer from the turntable causing you to spin in the air, or would you stop rotating witht the table?

Now pretend your standing on the same turn table near the edge of it. Assuming the friction is high enough for you not to fly off, would any of the angular velocity transfer if you did a perfect jump straight in the air?

It's really hard to tell, because the linear velocity is so fast, the angular velocity goes relatively unnoticed.

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Pretend you're standing on a giant spinning table. If you were standing in the center, and you jumped perfectly straight up, would your angular velocity transfer from the turntable causing you to spin in the air, or would you stop rotating witht the table?

Now pretend your standing on the same turn table near the edge of it. Assuming the friction is high enough for you not to fly off, would any of the angular velocity transfer if you did a perfect jump straight in the air?

In these examples, angular velocity isn't "transferred": you are spinning on your own before jumping, you continue spinning in the air at the same angular velocity (assuming the jump is an impulse pointing straight up), and you still spin when you fall back on the rotating platform. Air resistance, of course, would cause you to slow down your spinning a bit while in the air.

Note that if you jump from an off-center point of the platform you drift outwards because centrifugal force ceases to be balanced by friction as soon as you break contact.

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Note that if you jump from an off-center point of the platform you drift outwards because centrifugal force ceases to be balanced by friction as soon as you break contact.

Correction for OP:

If you jump from an off-centre point then you will move at a tangent to the centre (whilst spinning at same rate as before) because the centripetal force which was provided by the friction (being the force that keeps you moving in a circle) is no longer present.

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So if luca-deltodesco's post is correct, wouldn't that mean I would spin at the same rate as the table when contact breaks regardless of where I'm standing? (Ignoring air resistance)

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that is true yes.

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That's all I need to know,Thanks!

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So if luca-deltodesco's post is correct, wouldn't that mean I would spin at the same rate as the table when contact breaks regardless of where I'm standing? (Ignoring air resistance)

It's been a while since I've done physics, but based on my rigorous thought experiment, I'd say it depends. Your rate of spin is dependant on the difference in velocity on your contact surface. If you touch the surface at only a single point, it is impossible for you to spin.

At the outer edge, your rate of spin would be dependant on the difference in velocity between the foot that is outermost on the wheel (and thus fastest) and the foot that is innermost (and thus slowest). It's greatest when you're standing on the center of the wheel, because one foot is going in one direction, while the other foot is going in the opposite direction at the same speed. Only at the center would your spin be equal to the rate of the wheel.

Your right foot wants to go faster than your left foot, but they are tied together. The end result is a little like what happens when you smack a tether ball, but flying through space with what I'd presume is their average velocity.

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slavik81, that is completely wrong as 'standing' on the wheel, we're assuming there is sufficient friction to prevent sliding, any point on a rigid body has the same angular velocity, as you are standing on the wheel with sufficient friction to prevent sliding you will act like a point on that rigid body and have the same angular velocity as the wheel aswell as any linear tangentenial velocity.

if you didn't have the same angular velocity, then you're feet would be sliding around the centre of the wheel as your linear velocity at that point would be different than the wheel's.

if the wheel is stationary at it's centre, and has an angular velocity w, then if yuo're standing on the wheel you will also have angular velocity w, aswell as a linear velocity v = w cross x, where x is your displacement from the centre of the wheel

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If it helps to convince you, consider the wheel rotating with velocity w, and that the person is standing at a distance x from the wheel, and without loss of generality that his feet lie at distances (x-r) and (x+r) from the centre of the wheel.

given that he and his feet are stationary w.r.t to the wheel as he is not sliding, the linear velocity of the person at his centre is (wx), and the linear velocity of his feet are w(x-r) and w(x+r).
assume the person has an angular velocity of W, then the linear velocity of his feet are wx - Wr = w(x-r) and wx + Wr = w(x+r), which is iff. W = w.

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Good point. That certainly convinces me more than saying I'm completely wrong. I'm pretty sure that all but my conclusions were correct. As you point out, R factors back out when you convert back to angular velocity. So even if you took what I said, when all the calculations are done, you'd end up always getting the same angular velocity. Unless there's some other difference I'm missing.

I'd kind of expected the change of axis of rotation to have some impact, but I guess not if the axes are parallel.

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