Circular motion, find V when only given m and r

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Homework Statement



I don't want anyone to do it for me i am just sortof stuck any hints would be good.
OK so an object is spinning and its mass is x it's path has a radius of z, it is swinging in the vertical plane. What is the slowest it may be swung while maintaining the circular motion

Homework Equations



Fc=(mv^2)/r

The Attempt at a Solution



ok so Fnet=ma so i could just substitute ma=mv^2 / r or fg=mv^2 / r because fg would be pulling it down and the minimum force would be to counteract gravity?

i don't think this is right at all but don't really know where to go form here.
 
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Centripetal force is not a "real" force, like gravity or tension or friction. Centripetal force is a name for another force that is acting to cause circular motion. So look at the motion of the object and identify at certain places what force(s) are contributing to the centripetal force and circular motion. Look at the top specifically. What forces are acting there, and what is the minimum required amount of each force to keep the object moving in a circular path rather than falling. You have the right idea and equations, but the minimum force is not counteracting gravity like you thought, you were right that that isn't correct.
 
wbandersonjr said:
Centripetal force is not a "real" force, like gravity or tension or friction. Centripetal force is a name for another force that is acting to cause circular motion. So look at the motion of the object and identify at certain places what force(s) are contributing to the centripetal force and circular motion. Look at the top specifically. What forces are acting there, and what is the minimum required amount of each force to keep the object moving in a circular path rather than falling. You have the right idea and equations, but the minimum force is not counteracting gravity like you thought, you were right that that isn't correct.

ok so at the top it would be the slowest because it is working against gravity.
therefore at the top the acceleration would be straight down.
M(9.8)=mv^2 / r

and since i have the mass and radius i may solve for v?
 
You are right, at the top the acceleration is straight down. What force(s) are causing that downward acceleration? I think you know the correct answer to that, I just want to make sure.

Also, the speed will be the same at every point in the motion, because in intro physics we learn about uniform circular motion, meaning that the velocity is constant and the angular acceleration is zero.

One last thing, in the equation you wrote M(9.8)=mv[itex]^{2}[/itex]/r, the m's are the same, both referring to the rotating object, so they cancel algebraically. Did they even give you the mass in the problem, it really is not needed.