# AP Physics CH 5: Centripetal Motion

## Homework Statement

Derive an expression for Vmin, the minimum speed the ball can have at point Z without leaving the circular path.

V≤(x/t)
V≤(2∏r)/(t)
V=√(GM1)/(r)

## The Attempt at a Solution

I set 2∏r = to x since its in a circle. i then had V≤(2∏r)/(t)
in the end i got t=2∏√(r3)/(Gm1)
it didnt make sense to me so i came here for help /:

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lightgrav
Homework Helper
The key equation here are v.v=-a.r , and ΣF=ma .
Is "z" the top of a circular loop? if so, then a fast ball would be pushed down by the track (helping mg to accelerate the ball's mass downward). too slow, and the radius of the ball's path, given by r = -v.v/g , would be too small to stay on the track. you want the radius of the ball's path to be the same as the Radius of the track.

They want the speed, not the time.

point "z" is about 90 degrees there are also points "M, P, and Q". "Q" is about 270 degrees. "P" is 180 and m is about 60.

idk if that helps

Last edited:
lightgrav
Homework Helper
90 degrees? from where? from horizontal? then that is the TOP. are you trying to hide important aspects of the question from yourself?