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


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.

  • $\begingroup$ 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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