1
$\begingroup$

I was solving this problem:

A ring of mass $M$ hangs from a thread and two beads of mass $m$ slide on it without friction.The beads are released simultaneously from the top of the ring and slides down in the opposite sides.

We are asked to find the condition on $m$ such that the ring will move up during the motion of the beads.

Now I wrote down the equation $$ N + mg\cos\theta =\frac{mv^2}{r} $$ where $N$ is the normal reaction force provided by the ring (I am working in the frame of reference of the bead) and by using the work energy theorem I get $$ \frac{mv^2}{r} = 2mg(1-\cos\theta)$$ After that, by solving for $N$, I take the downward component of $N$ and multiply it by $2$ for the two beads so it becomes $2N\cos\theta$ which provides force to lift the ring up. Now differentiating and finding maximum force for corresponding $\theta$, we get $$F_\text{max} = \frac{2mg}{3}$$ Now my question is, I will get the correct answer which is $m>3M/2$ if I use $F>Mg$ where $F = 2N\cos\theta$, but shouldn't I write it as $F>(M+2m)g$ considering the weight of the other two small beads sliding upon the ring?

$\endgroup$
0
$\begingroup$

Your confusion will be removed if you consider the FBD of the ring itself. enter image description here

(Mg acts from center of mass of the ring.)
I think you can now see why the weight of the beads is inconsequential while determining equilibrium condition of the ring: regardless of the various forces felt by the beads, the ring itself feels only the following forces:
1. Weight
2. Tension of the ring
3. Normal reaction from both the beads
Hence the weight of the beads is not to be used in the comparison: F>Mg
NOTE:
Do, as an additional exercise, check whether the beads will tend to "fly off" from the ring before or after the ring will tend to move upwards(though logic says it WILL happen so).

$\endgroup$

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.