AP Physics CH 5: Centripetal Motion

AI Thread Summary
To derive the minimum speed (Vmin) for a ball at point Z on a circular path, it is essential to understand that point Z is at the top of the loop. The ball must maintain a sufficient speed to counteract gravitational forces, ensuring it does not fall off the track. The relevant equations include V≤(2∏r)/(t) and V=√(GM1)/(r), which relate speed to radius and gravitational force. The radius of the ball's path must equal the radius of the track for stability. Clarifying the exact position of point Z is crucial for accurate calculations.
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Homework Statement



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


Homework Equations


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)
then i had (2∏r)/(t)=√(GM1)/(r)
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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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:
90 degrees? from where? from horizontal? then that is the TOP. are you trying to hide important aspects of the question from yourself?
 
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