Why is tension greatest at bottom in circular motion?

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SUMMARY

The tension in a rope during vertical circular motion is greatest at the bottom of the circle due to the combined effects of gravitational force and centripetal acceleration. At the bottom, the tension (T) is calculated using the formula T = mv²/r + W, where W is the weight of the object. In contrast, at the top of the circle, the tension is T = mv²/r - W, indicating that the tension is lower. Understanding the direction of forces is crucial; tension acts upward at the bottom while gravity acts downward, leading to a net increase in tension at that point.

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  • Understanding of Newton's laws of motion
  • Familiarity with centripetal force concepts
  • Knowledge of free-body diagrams
  • Basic algebra for manipulating equations
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  • Study the principles of centripetal acceleration in circular motion
  • Learn how to draw and analyze free-body diagrams for various forces
  • Explore the effects of varying mass and speed on tension in circular motion
  • Investigate real-world applications of circular motion, such as roller coasters and pendulums
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Physics students, educators, and anyone interested in understanding the dynamics of circular motion and the forces involved in vertical motion scenarios.

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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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