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Gareth56

Who's right??

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Can someone setting a friendly argument?

 

The conservation of angular momentum says the angular momentum of a system before must equal the angular momentum after. That is Lbefore = Lafter

 

So, if a runner has linear momentum just before they jump onto a rotating circular platform does this have to be taken into account when calculating the Lbefore ?

 

One argument is that the Lbefore only comprises the rotating circular platform because the runner who only has linear momentum hasn't landed on the platform and the Lafter should include the runners angular momentum using their l converted linear speed to angular speed and their Moment of Inertia at the edge of the rotating platform.

 

So in essence the Lbefore = Lplatform and Lafter = Lplatform + Lrunner

 

 

The alternative argument is that Lbefore = Lplatform + Lrunner and Lafter = Lplatform + runner

 

 

I'm inclined to opt for the first argument because the only thing that has angular momentum before is the rotating platform, the runner doesn't have any angular momentum just linear momentum. It's only when they jump onto the platform do they then possess angular momentum.

 

Sorry to be a tad long winded but if someone could adjudicate I'd be very grateful.

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Angular momentum is defined about an axis - unless a runner is heading directly towards or away from the axis she will have a mass, velocity, and perpendicular distance to the axis (shortest distance between path of travel and axis) . In scalar terms L=mvr ... so yes there can be an ang mom.

 

Remember that angular momentum is conserved in the absence of an external torque - the last bit is worth remembering. It becomes important when you set up your system - if the system is just the platform then the runner would be external and there would be no conservation. If it is the platform and the runner then you need to calculate both bodies both before and after and equate

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Linear momentum and angular momentum are separately conserved, limited to the instant just before and just after the interaction. It's basically a collision.

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So if I've understand the above replies correctly the runner only posses L when they land on the edge of the rotating platform and not before. Therefore only the angular momentum of the platform should be considered in Lbefore.

 

 

Thanks

Edited by Gareth56

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So if I've understand the above replies correctly the runner only posses L when they land on the edge of the rotating platform and not before. Therefore only the angular momentum of the platform should be considered in Lbefore.

 

 

Thanks

 

No, the runner has angular momentum before landing on the platform, unless the hit it directly on-axis. L = r x p

 

if the platform wasn't moving before, it wouldn't start if the runner had no angular momentum.

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No, the runner has angular momentum before landing on the platform, unless the hit it directly on-axis. L = r x p

 

if the platform wasn't moving before, it wouldn't start if the runner had no angular momentum.

 

 

Oh right. Even more confused now as to how someone/something can have Angular Momentum when moving in a straight line. :confused:

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Oh right. Even more confused now as to how someone/something can have Angular Momentum when moving in a straight line. :confused:

 

If you hit something off-axis, it will cause rotation. So it must have angular momentum with respect to that axis.

 

An object moving in a straight line that passes at a distance r0 when it's at right angles has an angular momentum of mvr0. At any other distance r increases by 1/cos(theta) but the cross product has cos(theta) in it, so they cancel. Constant angular momentum, as it must be.

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