Ball in Circular Path

1. Mar 31, 2008

Mthees08

1. The problem statement, all variables and given/known data
A ball if mass M attached to a string of length L moves in a circle in a vertical plane as shown above. At the the top of the circular path, the tension in the string is twice the weight of the ball. At the bottom, the ball just clears the ground. Air resistance is negligible. Express all answers in terms of M, L, and g
a) Determine the magnitude and direction of the net force on the ball at the top of the path
b) Determine the speed V(i) of the ball at the top
the string is now cut
c) Determine the time it takes the ball to reach the ground.
d) Determine the horizontal distance the ball travels before hitting the ground

2. Relevant equations
F=ma
W=mg
centripetal A=V(tangental)^2/r
d=rt
Vfinal=Vinitial+at

3. The attempt at a solution
a) i tried to set up an equilibrium of forces equation using tension and weight but got 0 points
b) i got 2 points i used the centripetal acceleration equation and get Vt=sqrt(MgL)
c)0 points... i really messed this up... I should have gotten this...
d) i used the d=rt but had the incorrect time so if i fix c (or someone explains it) then this is easy

2. Apr 1, 2008

Dr. Jekyll

The net force is equal to the tension in the string.

The tension in the string is the centripetal force. Since you know it, you can get the tangential velocity.

If the string is cut when the ball is at the top, then it's just a horizontal shot. You know that $$g=9.81 ms^{-2}$$ and the distance that ball has to travel until it reaches the ground so you can get the time.

One more element of the horizontal shot. The starting velocity is the tangential velocity.

Last edited: Apr 1, 2008
3. Apr 1, 2008

rock.freak667

Actually, at the top of the circle, the forces acting are the Tension,T, and the weight,W. The resultant of those 2 forces provides the centripetal force required to keep the object in a circle.

So that $\frac{Mv^2}{L}=T+Mg$ and since $T=2Mg$,$\frac{Mv^2}{L}= 3Mg$

4. Apr 1, 2008

Dr. Jekyll

Yes, that's right.