# Why is tension greatest at bottom in circular motion?

• soccer5454
In summary, the object will break when it has the most tension. The string will be most likely to break when it has the most tension at the bottom of the circle.
soccer5454
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.

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 :)

## 1. Why does tension increase at the bottom of circular motion?

Tension in circular motion is caused by centripetal force, which is the force that keeps an object moving in a circular path. At the bottom of the circular path, the centripetal force is at its maximum, causing the tension in the object's path to also be at its greatest.

## 2. Is tension the same at all points in circular motion?

No, tension varies at different points in circular motion. It is greatest at the bottom of the circular path, and decreases as the object moves towards the top. This is because the centripetal force is constantly changing direction, and thus the direction of the tension force also changes.

## 3. Does the mass of the object affect tension in circular motion?

Yes, the mass of the object does affect tension in circular motion. The larger the mass of the object, the greater the amount of centripetal force needed to keep it in circular motion, resulting in a greater tension force.

## 4. What happens if the tension force is too weak in circular motion?

If the tension force is too weak in circular motion, the object will not be able to maintain its circular path and will either fly off in a straight line or collapse towards the center of the circle.

## 5. Can tension in circular motion be completely eliminated?

No, tension is a necessary component of circular motion and cannot be completely eliminated. Without tension, there would be no centripetal force to keep the object in its circular path.

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