0
$\begingroup$

I have been working on a particular physics problem but am not sure if I have analyzed it correctly. Here is the problem:

"A bead of mass m is threaded on a metal hoop of radius $R$. There is a coefficient of kinetic friction $\mu_k$ between the bead and the hoop. It is given a push to start it sliding around the hoop with initial speed $v_0$. The hoop is located on the space station, so you can ignore gravity.

...

Find its speed as a function of time. This involves using the frictional force on the bead in Newton’s second law, finding its tangential acceleration on the hoop (which is the time rate of change of its speed) and solving the equation of motion."

Here is my reasoning:

Because the bead is moving in a circle of radius $R$, there must be a centripetal force of magnitude $mv^2/R$, which is given by the normal force on the bead, pointing towards the center of the circle. Thus, $$N=mv^2/R.$$ Now, we also know that there is dynamical friction on the bead as it moves around the hoop. The magnitude of the frictional force is $$f_k=\mu_k|N|=\mu_k mv^2/R.$$ The frictional force imparts a tangential force on the bead, in the opposite direction of its velocity. From Newton's second law applied to the bead, I obtain $$-\mu_k mv^2/R=m\frac{dv}{dt}\tag{1}$$ I then solved the equation of motion by separation of variables, obtaining $$v(t)=\frac{v_0}{1+\mu_k v_0 t/R}$$ The function at least seems to provide a qualitatively correct description of the motion, with $\lim_{t\to\infty} v(t)=0$.

Questions:

1) I think that the negative sign is correct in $(1)$ (otherwise, I would get a singularity in the velocity function at $t=\frac{R}{\mu_k v_0}$) but I am not sure how to justify it. I know the friction opposes the velocity, but $dv/dt$ is the acceleration, not the velocity.

2) Are there any (other) errors in my argument?

$\endgroup$
  • $\begingroup$ Is the hoop anchored to the space station? $\endgroup$ – Cesareo Jun 26 at 16:22
0
$\begingroup$

The velocity decreases so the time derivative is negative. Negative acceleration requires a negative sign for your force as all the quantities in the expression of force are positive.

$\endgroup$

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.