Tension in string when ball is at the top of a circle

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Homework Statement


A child is swinging a .325 kg ball at the end of a .74 m long string in a vertical circle. string can withstand a tension of 12 N before breaking. What is the tension in the string when the ball is at the top of the circle if its speed at that point is 3.4 m/s?

Homework Equations


f=ma

The Attempt at a Solution



So I made a free body diagram for the ball...

since it is at the top of the circle, it has no forces acting on it at that time in the x direction. It's velocity is in the positive x direction, but no forces are acting on it in that direction.. right?

Since its acceleration is going towards the center of the circle, I use a = v^2 / r which is = 15.6216 m/s^2, then multiply by mass to get centipetal force, = 5.0772 N

mg gives me force of gravity = 3.18825 N,

Since both of the forces point towards the center of the circle, I would add them up right, and say tension is = to the opposite direction but same magnitude, right?

That would be around a little more then 8 Newtons, but my books answer is 1.89 N, which is the result you get from subtracting Fg from Fcentripetal.

Can anyone explain to me where my logic is going wrong??
 
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Answers and Replies

  • #2
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I know exactly what you're thinking. It's been awhile since I've solved one of these problems, and I did the same thing as you at first. What you have to realize is that when the ball is at the top of the circle, ƒg and ƒTension are both contributing to the centripetal acceleration because both forces are pointing in the direction of the center of the circle. In this case then, we can disregard the sign (+/-) because both forces should essentially act positive.

So, ƒTensiong=mac.

When you solve that out, ƒTension=mv2/r-ƒg

Does that make sense?
 
  • #3
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I found the following doing a Google search:
"It is important to understand that the centripetal force is not a fundamental force, but just a label given to the net force which causes an object to move in a circular path."

Personally, I find it less confusing if I just sum the individual forces and set them equal to (mass) x (acceleration) instead of using centripetal force. For this problem there are only 2 forces - weight and tension. Those should have been the only 2 forces in your free body diagram, true?
 
  • #4
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Those should have been the only 2 forces in your free body diagram, true?
I think Rijad understands this. What he is having trouble with is the sign (+/-) for ƒg and ƒTension.

Maybe the best way to explain it is that any forces pointing in the direction of the center of the circle contribute to the centripetal acceleration. Therefore, if they contribute towards it, they are given a positive sign. That may clear up my initial comment too. It's definitely an awkward thing to try and understand.
 
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  • #5
haruspex
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I think Rijad understands this.
No, I agree with Tom. Rijad was making the very common mistake of taking centripetal force to be an applied force, additional to other forces.
 
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Yeah, you guys are right. My bad. I thought he was saying that he did it like this:

Σƒy=mac
tensiong=mac

If you solve out that equation, you get an answer that is a little greater than 8N. He said that his answer was a bit over 8N, so I figured this was how he did it.

Thanks for clearing up my misunderstanding.
 
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  • #7
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I understand it now guys, thanks.

So the Centripetal force is just the the total of all the forces acting towards the centre, correct?
 
  • #8
haruspex
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I understand it now guys, thanks.

So the Centripetal force is just the the total of all the forces acting towards the centre, correct?
In this case, yes. More generally, it is the component of the net force which is perpendicular to the velocity.
 

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