Relationship between force on a stopper in Uniform Circular Motion and Hanging washer

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Homework Statement



What is the relationship between the weight of the hanging washers and the force acting on the stopper by the string?

Here is the diagram provided by the book: http://www.goodreads.com/photo/user/5034346-kylaia-formerly-known-as-klymene?photo=454140

It also says to simply assume L=R

Homework Equations



ƩF=ma, where the sum of the forces equals mass times acceleration.

ƩF=(mv2)/r, where the sum of the forces on an object in rotation equals mass times velocity squared, all divided by the radius

W=mg, where the force of the weight equals mass times the acceleration due to gravity (9.8)

The Attempt at a Solution



The only force acting on the stopper, since it is in rotation, is the tension of the string.
ƩFstopper=(mstoppervstopper)/r

There is tension and weight acting on the washers.
ƩFwashers=Twashers-Wwashers=0

Therefore, Twashers=Wwashers=mwashers*9.8

I'm guessing that somehow, the system has to be in equilibrium, because that's the only way anything could be put in a relationship. But I don't see how it could possibly be in equilibrium. And I don't understand how to relate the forces on the washers to the force acting on the stopper.

It's all purely theoretical, so there are no given numbers.
 

Answers and Replies

  • #2
PeterO
Homework Helper
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Homework Statement



What is the relationship between the weight of the hanging washers and the force acting on the stopper by the string?

Here is the diagram provided by the book: http://www.goodreads.com/photo/user/5034346-kylaia-formerly-known-as-klymene?photo=454140

It also says to simply assume L=R

Homework Equations



ƩF=ma, where the sum of the forces equals mass times acceleration.

ƩF=(mv2)/r, where the sum of the forces on an object in rotation equals mass times velocity squared, all divided by the radius

W=mg, where the force of the weight equals mass times the acceleration due to gravity (9.8)

The Attempt at a Solution



The only force acting on the stopper, since it is in rotation, is the tension of the string.
ƩFstopper=(mstoppervstopper)/r

There is tension and weight acting on the washers.
ƩFwashers=Twashers-Wwashers=0

Therefore, Twashers=Wwashers=mwashers*9.8

I'm guessing that somehow, the system has to be in equilibrium, because that's the only way anything could be put in a relationship. But I don't see how it could possibly be in equilibrium. And I don't understand how to relate the forces on the washers to the force acting on the stopper.

It's all purely theoretical, so there are no given numbers.

It is far from theoretical!! I can't see the picture you ahve attached but this reads like a standard PSSC [Physical Science Study Committee] Practical exercise.
PSSC was a Senior High Text and Prac manual developed/written in the 1960s and used in schools for many years from then.
 
  • #4


Well, the system is not in equilibrium in the sense that there needs to be a net centripetal force acting on the rotating object in order to enable circular motion.
However, the sum of the forces on the washers will be zero. The tension in the string and the gravitational force acting on the washers means that the washers are in equilibrium - which makes sense since they aren't accelerating.
What provides the centripetal force on the rotating stopper?
The tension in the string.
What provides the tension in string?
The force of gravity acting on the washers.

cheers
 
  • #5
PeterO
Homework Helper
2,426
48


Well, the system is not in equilibrium in the sense that there needs to be a net centripetal force acting on the rotating object in order to enable circular motion.
However, the sum of the forces on the washers will be zero. The tension in the string and the gravitational force acting on the washers means that the washers are in equilibrium - which makes sense since they aren't accelerating.
What provides the centripetal force on the rotating stopper?
The tension in the string.
What provides the tension in string?
The force of gravity acting on the washers.

cheers

All good.

ANd the reference to "let L = R" is allowing for the fact that the string from tube [I assume the string goes through a tube - still have not seen the picture] to the stopper is not exactly horizontal.
The tension must be providing an upward component to balance the (small) weight of the stopper.
Geometry of the situation will show that the actual radius of the circular motion is L.cosθ, but then the actual horizontal component of the tension is T.cosθ so the "droop factor" is rendered irrelevant. [θ was the angle below the horizontal that the string is angled down due to the weight of the stopper]
 

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