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So I have the following statement.

"A merry-go-round is spinning with a fixed angular speed. As a person is walking towards the edge, the force of static friction must increase in order for the person not to slide off."

I have no clue why is it true. I tried to figure out the problem with the following equations: rω = v (mv^2)/r = Tension but I can't seem to get past the tension. I guess what it causes the person to slide off would be the tangential acceleration (and therefore a friction force is needed), but I can't find a way of getting to this last variable with the previous mentioned equations...

I need help at figuring out specifically what (if I'm not right about the tangential acceleration) causes the person to slide off and how is it affected by an increase in the radius (as the person walk towards the edge.)

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The fact that you've got a "tension" in there suggests to me that you went to the back of the chapter and looked for equations with the right symbols in them. I'd like to discourage you from doing that as it will get in the way of your understanding. Instead start from "If the person is not to fall off, then she must be accelerating (because she is traveling in a curved path)", then draw a free body diagram and identify the force that could be providing the acceleration. –  dmckee Dec 11 '11 at 22:30
    
By Tension T, I was meaning the centripetal force. I just didn't use the name, sorry. –  Yokhen Dec 11 '11 at 22:42

1 Answer 1

For the person not to slip, there must be a centripetal force of $mv^2/r = m r \omega^2$ towards the centre.

Since $v$ varies with $r$ while $\omega$ is fixed ($v=r\omega$), it is probably easier to take the second form, in which case this force has to increase as $r$ increases. This forces comes from friction since there are no other forces in the plane of rotation.

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It seems a little contradictory.... if the Centripetal force Fc = (mv^2)/r ... wouldn't it instead decrease as r increases? In the other hand, there is the other equivalence which makes a little more sense (mrω^2) in which if r increases, the centripetal force increases. It seems like a contradiction... :S –  Yokhen Dec 11 '11 at 22:42
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It's not a contradiction because v is not fixed. As Henry said, v varies with the radius too. –  Arnoques Dec 11 '11 at 23:12
    
to repeat, v^2 increases faster than r –  Nic Dec 12 '11 at 12:23

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