Kleppner and Kolenkow (block sliding in a ring)

1. Dec 2, 2014

geoffrey159

1. The problem statement, all variables and given/known data
A block of mass $m$ slides on a frictionless table. It is constrained to move inside a ring of radius $l$ which is fixed on the table. At $t=0$, the block is moving along the inside of the ring with tangential velocity $v_0$. The coefficient of friction between the ring and the block is$\mu$. Find the velocity and position of the block at later times.

2. Relevant equations
Newton's second law of motion.

3. The attempt at a solution
Since I'm not sure what is meant by later times, I consider a small interval of time where the block sticks to the ring, say $I=[0,T]$, and the block is under circular motion in that interval.
Since we want to know the speed which will have only tangential component in circular motion, we have to calculate $\dot\theta$.

As initial conditions, I get $\dot\theta(0) = \frac{v(0)}{r(0)} = \frac{v_0}{l}$ and $\theta(0) = 0$

Since the block is under ring reaction force radially, and under friction force tangentially, I get by Newton's second law that $a_{\theta} = \mu a_r$, only for time $t$ in $I$. Therefore I get the differential equation
$\frac{\ddot\theta}{\dot\theta} = -\mu \dot\theta$
which thanks to the initial condition gives
$\dot\theta =\frac{v_0}{l} e^{-\mu\theta}$
That equation is equivalent to
$\frac{d}{dt} (e^{\mu\theta}) = \mu\frac{v_0}{l}$
Solving it gives
$\theta(t) = \frac{1}{\mu} ln(1+\frac{\mu v_0}{l}t)$
So that
$\dot\theta(t) = \frac{v_0}{l}\frac{1}{1+\frac{\mu v_0}{l}t}$
Therefore
$v(t) = r(t) \dot\theta(t) =\frac{v_0}{1+\frac{\mu v_0}{l}t}$
and
$\vec r(t) = l (cos(\theta) \vec i + sin(\theta) \vec j )$

3. Question
I begin in physics and I'm unsure about the solution, the question is not very clear either and I find the proof more complicated than other exercises in the book. Does it look good to you?

Last edited: Dec 2, 2014
2. Dec 2, 2014

voko

The differential equation looks good. But the solution you found is not correct.

3. Dec 2, 2014

geoffrey159

Thanks for replying.
What part do you disagree with ? I have done calculations again and I find the same result.

4. Dec 2, 2014

voko

Sorry, I misread your solution. Somehow it seemed to me that you had $e^{-\mu t}$, whereas you had $e^{-\mu \theta}$. I will check the rest of your work now.

5. Dec 2, 2014

geoffrey159

Yes, no problem. Let me know what you think afterwards.

6. Dec 2, 2014

haruspex

I get the same answer, a slightly different way. I integrated $\frac{d\omega}{\omega^2} = -\mu dt$.

7. Dec 2, 2014

geoffrey159

Your way is much better since you get to the result in a single pass while I have to do it in two passes to get to the same result.
Thanks !

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