Kleppner and Kolenkow (block sliding in a ring)

[geoffrey159]

Homework Statement

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.

Homework Equations

Newton's second law of motion.

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?

 

[voko]

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

 

[geoffrey159]

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

 

[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.

 

[geoffrey159]

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

 

[haruspex]

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

 

[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 !