# Thread: Circular Motion Question

1. ## Circular Motion Question

2 Particles, having mass of $\displaystyle m$ and $\displaystyle 2m$ respectively are connected by a light inextensible string, threaded through a fixed smooth ring. The lighter particle moves uniformly around a horizontal circle radius of radius $\displaystyle r$, while the other particle remains at rest.

Find the speed of the lighter particle?

No idea how start this, I've dealt with a bit of circular motion but not sure how to relate it to the mass of the particles?

Thanks in advance for any help

2. In order for the lighter particle to be moving in a circle, it has to be experiencing acceleration towards the center of the circle with magnitude mv^2/r. That acceleration is supplied by the heavier particle trying to respond to gravity, i.e., (2m)g. These 2 accelerations are equal.

3. Originally Posted by qmech
In order for the lighter particle to be moving in a circle, it has to be experiencing acceleration towards the center of the circle with magnitude mv^2/r. That acceleration is supplied by the heavier particle trying to respond to gravity, i.e., (2m)g. These 2 accelerations are equal.
Thank you, that's surprisingly easy...

Accelerations are equal, therefore $\displaystyle \frac{mv^2}{r} = 2mg$

Cancelling out the masses and multiplying through by r, we get $\displaystyle v^2 = 2gr$, so $\displaystyle v = \sqrt{2gr}$?

Thanks again for the quick reply

4. Would the larger particle have an acceleration though if it remains stationary? Would have thought that this would be zero?