# Homework Help: Find the min. velocity for which a particle will remain on the loop

1. Jan 8, 2010

### holezch

1. The problem statement, all variables and given/known data
A particle m is moving in a vertical circle of radius R inside a track. There is no friction. When m is at its lowest position, its speed is v0. What is the lowest value of v0 for which m will go completely around the circle without losing contact of the track?

2. Relevant equations

F= mv^2/r

conservation of energy

3. The attempt at a solution

Is this correct:

The min velocity will be at the top, where 1/2mv^2 + U(max height) = 1/2 mv0^2 ( from the bottom where U(0) = 0 ). So, if the particle is going to "fall off" the track, it should be where there isn't enough kinetic energy / speed. So.. if it DOES fall off, we would have N = 0, mv^2/r = mgsintheta - N = mgsintheta.

From 1/2 mv^2 + U(max height) = 1/2 mv^2 + mg2r = 1/2 mv0^2 we can solve for v from the top

and it turns out to be v = sqrt( v0^2 - 4gr )

then plug that into mv^2/r = mgsintheta and that would be the min. speed of the particle's total movement.. and you can solve for the initial speed from there. so if the min. speed is larger than the value that would have N = 0 , then it won't fall off.

thanks

Last edited: Jan 8, 2010
2. Jan 8, 2010

### JaWiB

I think that should work; however, I don't know what theta represents in this problem.

3. Jan 9, 2010

### holezch

thanks, I just had to split my weight into radial and tangential components