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As far as I know, if an object has a speed $s$ and a force is applied to it which generates an acceleration of $s^2/r$, then the object will start moving in a circle of radius $r$. Also, as far as I understand, this circular motion is constant (the radius will remain constant at all times).

However, I've seen physics problems that go something along the lines of a person stands on a merry-go-round, going at a certain tangential speed, find the friction needed for the person to not slip off, and the correct answer would be

$$\mu N m = m \frac{s^2}{r} \implies \mu = \frac{s^2}{N r}$$

Given what I've said in the first paragraph, I don't understand why the person would slip off. If the acceleration is equal to $s^2/r$, why is friction needed?

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1 Answer 1

up vote 5 down vote accepted

To remain on the merry-go-round, the person must be accelerated towards the center. How is the force applied to the person to provide that acceleration? The static friction provides a way for the floor of the merry-go-round to force the person along the circular path. If the floor were to suddenly become frictionless and the person was not otherwise attached, the person would continue moving along a line tangent to the merry-go-round.

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Oh, I see. Simple enough. – Paul Manta Jun 27 '12 at 20:18

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