Physics Homework Help Circular Motion

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.

answered Sep 25 '15 at 0:18

I'm having trouble wihh this problem I got in an exam and I got it wrong. In the problem is you have a frictionless table and a nail in the centre where a spring is attached. At the end of the spring there is a mass attached to it. The problem doesn't have numbers, $K$ is the springs constant, $L$ is its natural length, $m$ is the mass of the object attached.

The question is: where is the position of the mass so that the movement is a uniform circular motion?

I feel like this problem is very vague. I have trouble understanding and imagining how can such movement occur. So far i have this:

\begin{equation} K(L-r)=mv^2/r \end{equation}

I'm almost sure this can't be the solution because I never specify the uniformity of the motion, but I don't know how to either. Still, if I consider that the elastic force only acts in the $r$ axes, it's like I'm saying it's uniform, right? Because there's no other force component that can change velocity's magnitude...

Still I'm almost sure I got it wrong because it seems too simple.


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