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## Homework Statement

Picture-

https://flic.kr/p/zi1ays

The rubber stopper will be spun at a constant speed in a horizontal circle. The hanging mass is only in contact with the string.[/B]

A) For the rubber stopper, state the forces and what they are exerted by and on

B) Does the net force have a horizontal component, a vertical component, both, or neither? Indicate the direction and explain

C) Use newtons second law to write the net force equation in terms of the mass of the stopper, the speed (v) of the stopper, and the radius (r) of it's orbit. Then use the definition of net force to show that Fnet is equal to Tcosθ, where T is the magnitude of the tension force exerted on the rubber stopper by the string.

## Homework Equations

F=m(v^2/r)

## The Attempt at a Solution

A) [/B]For the stopper there is a tension force and a weight force. I believe the tension force points inward toward the center of the circle,while the weight force points downward. I think the tension force is exerted by the stopper and exerted on the string. I believe the weight force is exerted on the stopped and exerted by gravity. But i'm not sure if what I think they are exerted on and by are correct?

**I believe the net force is only horizontal because the forces are being directed inward toward the center of the circle. But, I'm not sure how weight would play in because it would be a force in the vertical direction, though the rubber stopper will be spinning at a constant speed, so the vertical components wouldn't be playing a role. But, i'm not sure what would cancel this force out? Also, I feel like the tension force might have both a horizontal and vertical component since it's at somewhat of an angle...**

B)

B)

Edit: If the net force is only horizontal and tension is, indeed, split in x and y components. Then the vertical tension and force of weight would cancel each other out, making the net force only in the horizontal direction?

C) All I think the first part is asking for is that Fnet=ma

Then the second part would be ΣFx=TcosΘ and ΣFy=TsinΘ-mg=0

which makes the total net force to be ΣF=TcosΘ

I think this all the question asks for and I think this might be right.

C) All I think the first part is asking for is that Fnet=ma

_{r}, which is equal to Fnet=m(v^2/r)Then the second part would be ΣFx=TcosΘ and ΣFy=TsinΘ-mg=0

which makes the total net force to be ΣF=TcosΘ

I think this all the question asks for and I think this might be right.

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