Two bodies are attached with a string. One body is rotating on a plane centering a point. A hole is present at the center of rotation and the string is passed through the hole and attached with a hanging body. The plane has friction. I need to find out the graph of the velocity of the hanging body. I have described the problem (with the masses, coefficient of friction, radius of rotation) in the picture. [$m_1=1.5$ kg (rotating body), $m_2=2$ kg (hanging body), $r=1$ m (initial radius of rotation), $\mu=0.2$ (coefficient of kinetic friction)]

For the reason mentioned at the bottom, I wrote the solution on paper and took pictures. I have uploaded the images in this link. The link goes to a folder in my google drive.

Below, I briefly present what I have done on paper:

$m_2g \cdot dr = m_2a_r \cdot dr + m_1a_r\cdot dr + m_1a_x\cdot dx + \mu m_1 g \sqrt{(dx)^2 + (dr)^2}$

Dividing by $dt$, then by $v_r$, then putting $\tan(\theta) = \frac{v_x}{v_r}$, it is possible to obtain:

$v_r' = \frac{m_2}{m_1+m_2}g - \frac{m_1}{m_1+m_2} \mu g \frac{v_r}{\sqrt{v_x^2 + v_r^2}}$

$v_x' = -\mu g \frac{v_x}{\sqrt{v_x^2 + v_r^2}}$

My questions are:

  1. In my solution, I got a system of differential equations. Is the system correct? Does it actually describe what is going on in the problem?
  2. I solved the system numerically. Is it possible solve the system in some other way? Is it possible to get an explicit expression for $v_r$?
  3. Is there an easier way to solve this problem?
  4. In page 9 (the pages are labeled in top right), I obtained another equation for $a_r$. But the equation does not hold when $t=0$. What have I done wrong in that page?

*I unfortunately had to write the solution on paper and upload the pictures. Writing down the solution would have made the question too long. I hope this isn't a huge inconvenience. I also included the tag differential-equations because I am asking if there is a way to solve the system (not numerically).

  • $\begingroup$ Duplicate : physics.stackexchange.com/q/659844/305718 $\endgroup$
    – ACB
    Aug 19 at 9:50
  • $\begingroup$ @ACB Wait, I added a solution, that question doesn't have it. $\endgroup$
    – Hoque
    Aug 19 at 9:52
  • 1
    $\begingroup$ It is not a good thing to present solutions as images through links. Briefly type it here. $\endgroup$
    – ACB
    Aug 19 at 9:57

I do not think your equations of motion are correct. The equations of motion can be expressed in polar coordinates on the flat surface \begin{align} &x = r \, \cos(\theta)\\ &y = r \, \sin(\theta) \end{align}

and I think they look something like this:

\begin{align} (m_1+m_2)\,&\frac{d^2 r}{dt^2} \,=\, m_1\,r\,\left(\frac{d\theta}{dt}\right)^2 \, -\, m_2g\, - \, \mu m_1 g\,\, \frac{\frac{dr}{dt}}{\sqrt{\left(\frac{dr}{dt}\right)^2 \, +\, r^2\left(\frac{d\theta}{dt}\right)^2}} \\ m_1\,r^2 \, &\frac{d^2 \theta}{dt^2} \, = \, - \,2 \, m_1\,r \, \frac{d r}{dt} \left(\frac{d \theta}{dt}\right)^2\, - \, \mu m_1 g\,\, \frac{r\,\frac{d\theta}{dt}}{\sqrt{\left(\frac{dr}{dt}\right)^2 \, +\, r^2\left(\frac{d\theta}{dt}\right)^2}} \\ \end{align}

I have serious doubts that these equations are explicitly solvable, because I do not think that they are integrable in the first place (they do not have conservation laws because the friction breaks them, except for the fact that these equations are not changed by rotation around the center of the plane, i.e. when $\theta \mapsto \theta + c$). Hence I would not expect explicit expression for the solutions.

You could also write the system as a system of first order differential equations as follows:

\begin{align} &\frac{dr}{dt} \, = \,v \\ &\\ &\frac{d \theta}{dt} \, = \,\omega \\ &\\ &\frac{d v}{dt} \,=\, \frac{m_1}{m_1+m_2}\,r\,\omega^2 \, -\, \frac{m_2g}{m_1+m_2}\, - \, \frac{\mu m_1 g}{m_1+m_2}\,\, \frac{v}{\sqrt{v^2 \, +\, r^2\omega^2}} \\ &\\ &\frac{d \omega}{dt} \, = \, - \, 2\, \frac{v \, \omega^2}{r} \, - \, \mu g\,\, \frac{\omega}{r \, \sqrt{v^2 \, +\, r^2\omega^2}} \\ \end{align}

and use this formulation for numeric simulations.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.