# centripetal force.

• October 27th 2008, 04:43 AM
john haas
centripetal force.
I need to know what formula is required to solve this problem

A pail of water is swung by a cord tied to its handle in a vertical circle
What must be its minimum speed at the top of the circle if no water is
to spill out?
• October 27th 2008, 06:35 AM
Rafael Almeida
john,

The formula for centripetal force is $F = \frac{mv^2}{R}$, where m is the mass, v the the velocity and R is the radius of the movement.
• October 27th 2008, 06:46 AM
Chop Suey
Quote:

Originally Posted by john haas
I need to know what formula is required to solve this problem

A pail of water is swung by a cord tied to its handle in a vertical circle
What must be its minimum speed at the top of the circle if no water is
to spill out?

Also, note that at top of the circle, the force that maintains circular motion is the gravitational force.
• October 27th 2008, 07:00 AM
john haas
need mass
Rafael
Thanks for supplying the formula.
What if I am not given mass.
All that is given is the radius and acceleration as gravity(9.81m/s^2).

from the above I was able to solve velocity as 3.09 m/s

I am thinking at the top of the vertical would I use a negative acceleration.

• October 27th 2008, 07:15 AM
Rafael Almeida
john,

As Chop Suey correctly pointed out, at the top of the circle, the force that mantains the motion is ONLY the gravitational force (weight). This happens because this force will be applied no matter what, and a stronger force would imply a higher velocity by the formula I gave you.

Since you know the force is the weight:

$F = \rlap{//}mg = \frac{\rlap{//}mv^2}{R} \Rightarrow g = \frac{v^2}{R} \Rightarrow v = \sqrt{Rg} \ \ (I)$

Also notice that, considering the centripetal force to be constant and using Newton's Second Law:

$F = ma = \frac{mv^2}{R} \Rightarrow a = \frac{v^2}{R}$

What gives you the modulus of the centripetal acceleration for any point in the circle, which does not depend on mass. So, another argument to solve the question is to see that, since no other force other than weight is applied to the pail, its acceleration must be equal to gravity. Setting $a = g$ in the above formula will lead you to the same result as in (I).

Hope this helps,