Physics Question:Circular Motion

[Paratodoc]

At point A in Experiment 2, the tension in the string is measured to be three times the weight of the ball. What is the speed of the ball?

Point A is basically the ball at 90 degrees (Pi/2) moving in a clockwise fashion.The string is a length L tethered to the center of the circle (e.g. the hand of the person spinning the string).

I set up the equation as follows: 3mg= m(v^2/L). My final answer was the square root of 3gL.

The answer from the book is 2 times the square root of gl. The book's theory is that the tension in the line is equal to 3 mg + mg (4mg), but according to Newton's Third Law the tension by itself should be entirely equal to centripetal force, so why add the weight of the ball too? Isn't is assumed that the tension would inherently include any force from the ball's weight since it is "three times the weight of the ball?"

Thoughts?

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[Paratodoc]

From looking at another question, it appears the book is assuming that the ball will also be engaged in some form of free fall with a force of mg, which is why it would be added. The ball is being spun vertically, so there is a change in the vertical plane. Basically, they're saying that the force of gravity is in the same direction as the tension. I had assumed, I guess wrongly, that because mg was center seeking and the tension was "three times mg" that it had been accounted for. Can someone confirm this interpretation?

 

[V5RED]

At point A in Experiment 2, the tension in the string is measured to be three times the weight of the ball. What is the speed of the ball?

Point A is basically the ball at 90 degrees (Pi/2) moving in a clockwise fashion.The string is a length L tethered to the center of the circle (e.g. the hand of the person spinning the string).

I set up the equation as follows: 3mg= m(v^2/L). My final answer was the square root of 3gL.

The answer from the book is 2 times the square root of gl. The book's theory is that the tension in the line is equal to 3 mg + mg (4mg), but according to Newton's Third Law the tension by itself should be entirely equal to centripetal force, so why add the weight of the ball too? Isn't is assumed that the tension would inherently include any force from the ball's weight since it is "three times the weight of the ball?"

Thoughts?

From looking at another question, it appears the book is assuming that the ball will also be engaged in some form of free fall with a force of mg, which is why it would be added. The ball is being spun vertically, so there is a change in the vertical plane. Basically, they're saying that the force of gravity is in the same direction as the tension. I had assumed, I guess wrongly, that because mg was center seeking and the tension was "three times mg" that it had been accounted for. Can someone confirm this interpretation?

Assuming you are referring to the TBR problem in which a ball is being swung on a string in a vertical circle and point A refers to the apex of the swing....

If you draw a free body diagram you will see that the only forces acting on the ball are tension from the string and gravity. The net force points toward the center of the circle and causes circular motion, so we can write F=ma=mv^2/L=T+mg=3mg+mg=4mg therefore mv^2/L=4mg therefore v^2/L=4g therefore v=sqrt(4Lg).

Tension and weight are separate forces and don't always point in the same direction. Sometimes they have no effect on each other or even oppose each other,. Don't ever assume they are the same force unless you are told that some amount of force refers to them combined.

 

[Paratodoc]

Assuming you are referring to the TBR problem in which a ball is being swung on a string in a vertical circle and point A refers to the apex of the swing....

If you draw a free body diagram you will see that the only forces acting on the ball are tension from the string and gravity. The net force points toward the center of the circle and causes circular motion, so we can write F=ma=mv^2/L=T+mg=3mg+mg=4mg therefore mv^2/L=4mg therefore v^2/L=4g therefore v=sqrt(4Lg).

Tension and weight are separate forces and don't always point in the same direction. Sometimes they have no effect on each other or even oppose each other,. Don't ever assume they are the same force unless you are told that some amount of force refers to them combined.

I had considered gravity, but because TBR said that the tension was "equal to three times the weight" I basically said, "Ok, they are saying that the tension is 3 x the weight. Tension=centripetal force." I never thought to take into account the addition of gravity, because the weight really doesn't add to the tension of a string. If you were to take that same scenario, a ball on a string without the application of any circular motion, and drop the ball it certainly wouldn't add tension to the string. My error was one of reading: They want the velocity of the ball, and certainly the ball is subject to gravitational forces. My mistake. Clearly this is implied when Ta +mg.


Thanks for the input.

 

[sciencebooks]

I had considered gravity, but because TBR said that the tension was "equal to three times the weight" I basically said, "Ok, they are saying that the tension is 3 x the weight. Tension=centripetal force." I never thought to take into account the addition of gravity, because the weight really doesn't add to the tension of a string. If you were to take that same scenario, a ball on a string without the application of any circular motion, and drop the ball it certainly wouldn't add tension to the string. My error was one of reading: They want the velocity of the ball, and certainly the ball is subject to gravitational forces. My mistake. Clearly this is implied when Ta +mg.


Thanks for the input.

From what I'm envisioning, it's not suggesting that the weight adds to the tension, just to the NET force (since net force is what needs to be used in calculating acceleration) and in this case, both forces happen to be in the same direction.

(I don't have the book though so it's hard to say.)

 

[Singhp03]

At point A in Experiment 2, the tension in the string is measured to be three times the weight of the ball. What is the speed of the ball?

Point A is basically the ball at 90 degrees (Pi/2) moving in a clockwise fashion.The string is a length L tethered to the center of the circle (e.g. the hand of the person spinning the string).

I set up the equation as follows: 3mg= m(v^2/L). My final answer was the square root of 3gL.

The answer from the book is 2 times the square root of gl. The book's theory is that the tension in the line is equal to 3 mg + mg (4mg), but according to Newton's Third Law the tension by itself should be entirely equal to centripetal force, so why add the weight of the ball too? Isn't is assumed that the tension would inherently include any force from the ball's weight since it is "three times the weight of the ball?"

Thoughts?

check out this video: http://www.youtube.com/watch?v=32X5Udm57bU&feature=plcp

 

[milski]

Where exactly is the ball, at the bottom of the circle? The answer does not makes sense in that case.

The centripetal force will be the same as the net force, the tension force will not - it's only one of the forces acting on the ball.

At the bottom of the circle the tension is pointed up and is 3mg, the weight is mg, pointed down. The net force ends up 2mg, pointed up.