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Circular motion and orbits AP Physics

  • #1
A ball of mass M attached to a string of length L moves in a circle in vertical plane. At 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 when it is at the top.
I know that ƩF=T=Fc but that's about all I can get. Can someone please help.


B) Determine the speed v initial of the ball at the top.

I did;
ƩF=T+mg= mv^2/L and since T=2mg+mg=3mg
so
3mg=mv^2/L
the masses cancel out and you take the square root of whats left so
v=the square root of 3lg
 

Answers and Replies

  • #2
A) Look at all of the forces acting on the ball at the top. There is more than tension.
B) Your math looks correct. Just note that the net force on the ball at the top is actually [tex]ƩF = 3mg[/tex], not [tex]T= 3mg[/tex]. It looks like your next line is right, but just make sure you have your terms correct.
 
  • #3
For part A I don't understand at all how to find the magnitude and direction. at the top you have tenstion, net force, and mg (force of gravity). That's all I about know.
 
  • #4
The magnitudes you already know; it's given to you. Force of gravity, as you said in the post before, has a strength of mg. Tension is twice that. So there's the magnitudes!

The directions should be straightforward. Gravity is always acting toward the center of the planet. The tension is a little more tricky to think about. Ask yourself, "What way does the string have to pull to make the net motion circular?"
 
  • #5
Ohhh, duhh. Sometimes when it's soo easy I over think -__- Thaaanks (:
 

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