Why is tension greatest at bottom in circular motion?

AI Thread Summary
In circular motion, tension is greatest at the bottom of the circle due to the opposing forces of tension and gravity. At the bottom, tension acts upward while gravity acts downward, resulting in a net force that must equal the centripetal force required for circular motion. The equation T = mv^2/r + W illustrates that tension increases at the bottom because it must counteract both the gravitational force and provide the necessary centripetal force. Conversely, at the top of the circle, tension is reduced since both forces act in the same direction, allowing the string to potentially go slack. Understanding these force dynamics is crucial for grasping the principles of circular motion.
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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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