Force on Stopper in UCM & Hanging Washer Relationship

In summary, the relationship between the weight of the hanging washers and the force acting on the stopper by the string is that the tension in the string, provided by the gravitational force acting on the washers, is the centripetal force that allows for circular motion of the stopper. The washers are in equilibrium due to the equal and opposite forces of tension and weight acting on them. The statement to "let L = R" is taking into account the angle of the string below the horizontal and its effect on the horizontal component of the tension.
  • #1
Klymene15
10
0

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.
 
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  • #2


Klymene15 said:

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.
 
  • #3
  • #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


The Anonymous said:
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]
 

1. What is the force on a stopper in uniform circular motion (UCM)?

The force on a stopper in UCM is known as centripetal force, which is directed towards the center of the circular path and keeps the object moving in a circular motion.

2. How does the radius of the circular path affect the force on the stopper in UCM?

The force on the stopper is directly proportional to the square of the velocity and inversely proportional to the radius of the circular path. This means that as the radius decreases, the force required to keep the stopper in UCM increases.

3. What is the relationship between the speed of the stopper and the force in UCM?

The force on the stopper is directly proportional to the square of the velocity. This means that as the speed of the stopper increases, the force required to keep it in UCM also increases.

4. How does the mass of the stopper affect the force in UCM?

The mass of the stopper does not directly affect the force in UCM. However, a larger mass may require a greater force to keep it in UCM at a given speed and radius.

5. What is the relationship between the force on a hanging washer and the radius of the circular path?

The force on a hanging washer is also known as tension and is directed towards the center of the circular path. The tension is directly proportional to the radius of the circular path, meaning that as the radius increases, the tension in the string supporting the washer also increases.

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