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Ball in Circular Path

  • Thread starter Mthees08
  • Start date
35
0
1. Homework Statement
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. Homework 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

PLEASE PLEASE PLEASE HELP
 

Answers and Replies

35
0
a) Determine the magnitude and direction of the net force on the ball at the top of the path
The net force is equal to the tension in the string.

b) Determine the speed V(i) of the ball at the top
The tension in the string is the centripetal force. Since you know it, you can get the tangential velocity.

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

d) Determine the horizontal distance the ball travels before hitting the ground
One more element of the horizontal shot. The starting velocity is the tangential velocity.
 
Last edited:
rock.freak667
Homework Helper
6,230
31
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.
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 [itex]\frac{Mv^2}{L}=T+Mg[/itex] and since [itex]T=2Mg[/itex],[itex]\frac{Mv^2}{L}= 3Mg[/itex]
 
35
0
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.
Yes, that's right.
 

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