# Energy in circular path with spring

1. Oct 29, 2008

### edowuks

We have a spring that has equilibrium at distance R. the spring is attached at horizontal distance R from circle (R is distance from edge of circle, distance from circles origo is 2R) that has radius R. Mass attached to spring can move freely in that circle path. In circle we have 3 points a=point where spring is at rest. B=at angle pi/2 C=opposite side of spring than the point a. (sorry I dont have picture and my english is bad).
A) calculate velocity at point a when spring is released from point b (at picture we have on paper we can see that at point b it has gravitational potential energy of mgR)
B) How hight should the spring constant be that mass never reachess point c

Ok I have banged my head to wall two days, I cannot figure what should I calculate here. Can I just calculate the A part by 'potential of spring'+mgR=½mv² (can potential of spring be calculated from lenght of pring-R). B-part I dont understand what should I calculate, but with my poor explanation I think no-one can give any hint?

I dont need anykind of solution just hint what I should be calculating, can't picture the problem in my head.

2. Oct 29, 2008

### LowlyPion

The potential in your spring is = ½*k*Δx2 and as you point out that plus drop in height m*g*R will result in your ½*mv2.

Your Δx here can be found by ordinary geometry.

For part 2 you know the KE at b so for the mass to make it to c, which by your description is at the same level as b then there is no need to account for any change in potential energy, so ½*k*(2R)2 = the KE when it was at b.

Last edited: Oct 29, 2008
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