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I am trying to understand how circular motion problems can be solved when we introduce friction as a variable in them. As i understand this problem, the centripetal force, which points towards the center of the ring, equals the normal force. In that case, the friction would be equal to a coefficient times the centripetal force. When the speed changes (due to friction), the centripetal force will change (mv^2/r) and so will the normal. However since the speed is dependent on the friction which in turn is dependent on the speed, I seem to be stuck in a circular situation. Would any one have advice to help me find the relation that will allow me to express v?

Problem description

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Set up Newton's 2nd law. Always set up Newton's 2nd law. Let us pick our positive axis direction as the radial direction (towards the center):

$$ \sum F_{rad}=ma_{rad}\quad\Leftrightarrow\quad n=ma_{rad}=m\frac{v^2}{r} $$

Remember that $v$ varies with time. This still holds at any instant in time. Now, we look for a relation to express $n$ with. That relation could be the kinetic friction expression $f_k=\mu_k n$:

$$ \frac{f_k}{\mu_k}=m\frac{v^2}{r} $$

So far so good. Let's find a relation to express the friction $f_k$ with. We can use Newton's 2nd law in the tangential direction also:

$$\sum F_{tan}=ma_{tan}\quad\Leftrightarrow\quad f_k=ma_{tan}$$

which we just plug in:

$$ \frac{ma_{tan}}{\mu_k}=m\frac{v^2}{r}\quad\Leftrightarrow\quad\frac{a_{tan}}{\mu_k}=\frac{v^2}{r} $$

You might not have thought of putting in this with the $a_{tan}$ still there. But if you can't figure out where to continue, putting what you have together is step no. 1.

Let's think it over for a bit; Hey, $a_{tan}$ and $v$ are along the same path! $a_{tan}$ is just the acceleration that causes a change in the speed $v$. You know the definiton of acceleration:

$$a_{tan}=\frac{dv}{dt}=\dot v$$

(where the $\dot v$ is a short-hand writing for the derivative to time). So let's simply plug in this expression for $a_{tan}$:

$$ \frac{\dot v}{\mu_k}=\frac{v^2}{r}\quad\Leftrightarrow\quad \dot v=\frac{\mu_k v^2}{r} $$

Finally we have our nice differential equation that your question talks about. Solve this and I am sure you'll reach something beautiful.

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  • $\begingroup$ Perfect! Thanks for the help! I was so focused on the angular part of the problem that I forgot to consider the tangential acceleration $\endgroup$ – user1354784 Sep 25 '15 at 0:39

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