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

In summary, when solving for the tension in a string in a vertical circle, it is important to take into account both the weight of the object and the centripetal force acting towards the center of the circle. The centripetal force is not a separate force, but rather the net force that causes an object to move in a circular path. It is the sum of all forces acting towards the center of the circle.
  • #1
Rijad Hadzic
321
20

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??
 
  • Like
Likes Truman I
Physics news on Phys.org
  • #2
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
I found the following doing a Google search:
"It is important to understand that the centripetal force is not a https://www.khanacademy.org/science/physics/centripetal-force-and-gravitation/centripetal-forces/a/science/cosmology-and-astronomy/universe-scale-topic/light-fundamental-forces/v/four-fundamental-forces, 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
TomHart said:
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.
 
Last edited:
  • #5
Truman I said:
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.
 
  • Like
Likes TomHart
  • #6
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.
 
  • Like
Likes TomHart
  • #7
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
Rijad Hadzic said:
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.
 

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

1. What is tension in a string?

Tension in a string is the force transmitted through the string when it is pulled at both ends. It is a measure of how much the string resists being pulled apart.

2. How is tension affected when a ball is at the top of a circle?

At the top of a circle, the ball is experiencing a centripetal force towards the center of the circle. This force creates tension in the string, as the string is being pulled towards the center along with the ball.

3. Does the tension in the string change as the ball moves around the circle?

Yes, the tension in the string changes as the ball moves around the circle. At the top of the circle, the tension is at its maximum, while at the bottom of the circle, the tension is at its minimum.

4. What factors affect the tension in the string when the ball is at the top of a circle?

The tension in the string is affected by the mass of the ball, the speed of the ball, and the radius of the circle. The higher the mass and speed of the ball, or the smaller the radius of the circle, the greater the tension in the string will be.

5. How is the tension in the string related to the weight of the ball?

The tension in the string is directly related to the weight of the ball. As the weight of the ball increases, the tension in the string also increases. This is because the weight of the ball creates a downward force, which is counteracted by the tension in the string pulling upwards.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
275
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
2
Replies
38
Views
6K
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
3K
Back
Top