Why is tension greatest at bottom in circular motion?

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Homework Help Overview

The discussion revolves around the concept of tension in circular motion, specifically focusing on why tension is greatest at the bottom of a vertical circular path. The original poster is exploring the relationship between tension and gravitational force in this context.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on an object in circular motion, questioning how tension can be greatest at the bottom when gravity is also acting on the object. There are attempts to clarify the directions of these forces and their implications on tension.

Discussion Status

Participants are actively engaging with the concepts, with some providing clarifications about the directions of forces and the conditions under which tension is maximized. There is a recognition of the need to visualize the forces through Free-Body Diagrams, indicating a productive exploration of the topic.

Contextual Notes

Some participants mention the importance of understanding the conditions under which the string may go slack, particularly at the top of the circular path, which introduces additional considerations into the discussion.

soccer5454
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Hi. I am having some difficulty with circular motion.An object is spun vertically on a rope a) when would the string be most likely to break?

The object would most likely to break when it has the most tension, so at the bottom of the circle it is Ft-Fg=Mv^2/r which is (Ft=Mv^2/r + Fg) while the top is Ft=Mv^2/r - Fg. But I don't understand the concept of this, because if both Tenstion and Gravity are going in the same direction how can it be possible? Thanks.
 
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Tension and gravity are NOT "in the same direction". The tension in the rope is always directed toward the middle of the circle. At the bottom, that means tension is UPWARD while gravitational force is directed DOWNWARD. The two are in opposite directions, not the same direction.
 
But if bottom one goes up other goes down how can bottom have the greatest tension?? I thought it would be top because in that case both the force tension and force gravity would go down
 
If you swing a bunch of keys on a keychain around so they trace out a circle in the vertical plane, you'll quickly discover that it is impossible to have the keychain go slack near the bottom of the circle, but it's easy to cause it to go slack near the top of the circle. When the chain goes slack, this indicates there is no nett force acting through the chain.

How to account for the chain going slack at the top of this circular orbit? That's what you need to address.
 
The tension force will always be directed along the string, which will always be pointing to the center.

Gravity always points downwards.

When the object is at the bottom, tension points up and gravity points down. When balancing the forces, T - W = mv^2/r, and T = mv^2/r + W.

When the object is at the bottom, both the tension and gravity point downwards. T + W = mv^2/r and T = mv^2/r - W.

Clearly, T at the bottom is greater than T at the top by twice the weight.

I encourage you to draw the Free-Body Diagrams and balance the forces yourself if what I wrote didn't make sense. You can also try this when the string is at some angle to the horizontal, you should find that the tension will range from T = mv^2/r - W to T = mv^2/r + W.
 
Got it ty so much for help :)
 

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