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Circular mgomentum - Tension of a string.

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Circular momentum - Tension of a string.

Homework Statement


A body of mass 1kg is tied to a string and rotates on a horizontal frictionless table. If the length of the string is 40.0 cm and the speed of revolution is 2m/s, find the tension in the string.

2. The attempt at a solution

Since the two forces acting on the string is gravity and tension, and the momentum is horizontal, it means that Ty (vertical component of the tension) is equal and opposite to the gravity, and thus, the net force is Tx component (which would thus also be the centripetal force). Hence, we can find the tension of the string by finding the centriputal acceleration, and consequently the magnitude of that force, then use pythagoras theorem to find the tension.

My problem here is that I'm not quite sure about the relation of the length of the string and the radius of the circle around which the mass moves. I tried to hold my hand still whilst drawing a circle with my pencil and noticed that the diameter was very close to the length of the pencil, but I'm not sure if that is enough to assume that the radius must be 20cm. I think that if someone helps me out with this part, I'll most likely be able to solve the question.

EDIT: Another thing which confuses me is that in the book, it says that the magnitude of the tension is 10m. However, As the mass is 1kg, the magnitude of gravity should be 10N, and thus so should the Ty component. That means that the tension has to be at least greater than 10 (as it is the hyoptenuse if we imagine the two components and it as a right-angled triangle).
 
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Answers and Replies

  • #2
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Force of gravity has no role to play here. The circular motion takes places on the horizontal table. The forces acting in the horizontal direction are the tension and the centrifugal force. There is no component of weight in the horizontal direction.
 
  • #3
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Ah! I didn't pay any attention at all to the fact that it says horizontal table. I'm aware of the fact weight has no horizontal direction. But since its on the horizontal table, we can assume that the net force on the vertical direction is 0, correct? Also, what would the centrifugal force be?
 
  • #4
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Ah! I didn't pay any attention at all to the fact that it says horizontal table. I'm aware of the fact weight has no horizontal direction. But since its on the horizontal table, we can assume that the net force on the vertical direction is 0, correct? Also, what would the centrifugal force be?
Yep, the net force in vertical direction is 0.
Centrifugal force=[itex]\frac{mv^2}{r}[/itex]
 
  • #5
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That's centriputal force, isn't it? Also, I was under the impression that centripetal force is more of a description of a force than an actual force. On that basis, isn't the only force the horizontal tension force which ACTS like a centriputal force?
 
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  • #6
rcgldr
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Since the string tension is always perpendicular to the path of the mass (in this case a circular path), then the force the string exerts onto the mass is a centripetal force.
 

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