Centripetal Force Lab Question

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Homework Statement


Basically, we did a centripetal acceleration lab where a hanging mass(s) was connected to a string, run through a tube, and connected to a rubber stopper. The force of gravity of the hanging mass is what supplied the force of tension to the rubber stopper for its centripetal acceleration. This is a picture of the lab:
http://dev.physicslab.org/img/1efa676f-ede0-44dd-ae03-ad06a6e14b18.gif

Homework Equations


mass of hanging mass x force of gravity=mass of rubber stopper x 4(pi)^2(r)(f)^2

The Attempt at a Solution


I understand that because the force of gravity was acting on the rubber stopper, the actual radius of the circle swung would be less than the assumed radius, which is supposed to be constant, and therefore the experimental frequency would be greater than the theoretical frequency. However, why would increasing the weight of the hanging mass reduce this problem?
 

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  • #2
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I understand that because the force of gravity was acting on the rubber stopper, the actual radius of the circle swung would be less ... (snip)

Explain this, and you should see the rest of it.
 
  • #3
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Explain this, and you should see the rest of it.
Causes stopper to swing more perpendicular path?
 
  • #4
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I'll give you a conditional "yes" for that.
 
  • #5
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I don't really understand why this increases the radius though. Because even though the string moves down and causes the upper portion to be more perpendicular, once you start swinging it again doesn't it become less perpendicular?
 
  • #6
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What is the source of tension in the string below the vertical tube? What is(are) the source(s) of tension in the string between the tube and the stopper that oppose the tension applied below the tube?
 
  • #7
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What is the source of tension in the string below the vertical tube? What is(are) the source(s) of tension in the string between the tube and the stopper that oppose the tension applied below the tube?
The source of tension in the string below the vertical tube is equal to the weight of the hanging mass. This is the same force on the rubber stopper...but how does this increase the radius?
 
  • #8
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This is the same force on the rubber stopper ....

In what direction?
 
  • #9
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In what direction?
Horizontally, or radially inwards if we are going to be more specific to the context.
 
  • #10
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I understand that because the force of gravity was acting on the rubber stopper ....

In what direction does this force act?
 
  • #11
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In what direction does this force act?
Well, it would pull the string down, thus exerting a downwards force.
 
  • #12
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Now you have forces in two directions, radial, and downward acting on the stopper. What direction is the resultant on the string? If the gravitational force is constant, and the radial force changes (increase or decrease), what effects will you see?
 
  • #13
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Now you have forces in two directions, radial, and downward acting on the stopper. What direction is the resultant on the string? If the gravitational force is constant, and the radial force changes (increase or decrease), what effects will you see?
If the force of tension is increasing, and the gravitational force is constant, the resultant force on the stopper is towards the tube, which would decrease the radius.
 
  • #14
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Not necessarily, you have the option of changing "f", or "r" in the experiment to change the tension on the string to match that of the increase in the hanging weight.

You have a constant gravitational force acting downward on the swinging stopper in both cases (lighter and heavier hanging weights), and a lower and greater radial force resulting from either decreasing or increasing radius or from decreasing or increasing frequency of revolution. Remember the man named Pythagoras you met back in 7th grade?

You have been through a discussion of the resultant of two forces acting in different directions?
 
  • #15
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Not necessarily, you have the option of changing "f", or "r" in the experiment to change the tension on the string to match that of the increase in the hanging weight.

You have a constant gravitational force acting downward on the swinging stopper in both cases (lighter and heavier hanging weights), and a lower and greater radial force resulting from either decreasing or increasing radius or from decreasing or increasing frequency of revolution. Remember the man named Pythagoras you met back in 7th grade?

You have been through a discussion of the resultant of two forces acting in different directions?
When two forces are acting in different directions, you do a ∑ of forces. Doesn't the greater radial force decrease the radius though? In regards to frequency, are you saying that I start swinging it faster with the hanging masses, thereby increases the centripetal acceleration?
 
  • #16
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You have not been instructed in how to determine the resultant of two forces acting at right angles to one another. Trying to jog your memory when you haven't seen the material obviously won't work.

The resultant of two forces acting perpendicularly to one another is the vector sum of the components of the two forces; can be also regarded as the hypotenuse of a right triangle the base and height of which represent the two forces magnitudes and directions. You have for the hypotenuse the force of the hanging weight which is redirected outward from the top of the tube and slightly downward due to gravitational force on the swinging stopper; you have for "height," the gravitational force downward on the swinging stopper. You are interested in the radial force. Enter Pythagoras, and
FHanging2 - FStopper, grav2 = FRadial2.

I'll let you cook with this for a bit.
 
  • #17
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You have not been instructed in how to determine the resultant of two forces acting at right angles to one another. Trying to jog your memory when you haven't seen the material obviously won't work.

The resultant of two forces acting perpendicularly to one another is the vector sum of the components of the two forces; can be also regarded as the hypotenuse of a right triangle the base and height of which represent the two forces magnitudes and directions. You have for the hypotenuse the force of the hanging weight which is redirected outward from the top of the tube and slightly downward due to gravitational force on the swinging stopper; you have for "height," the gravitational force downward on the swinging stopper. You are interested in the radial force. Enter Pythagoras, and
FHanging2 - FStopper, grav2 = FRadial2.

I'll let you cook with this for a bit.
So, in accordance with the above explanation, increasing the weight from the hanging mass, while the gravitational force acting on the stopper is a constant, increases the radial force. This makes complete and logical sense. However, what exactly do you mean by the radial force? When you say this, I think of the force of tension acting on the stopper, acting radially inwards. Thus, if the force of tension acting on the stopper increases, the centripetal acceleration of the stopper would increase, resulting in a greater frequency. Yet, does an increase in radius have anything to do with this?
 
  • #18
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However, what exactly do you mean by the radial force? When you say this, I think of the force of tension acting on the stopper, acting radially inwards.
Good.

... and, what other force is acting on the stopper? And must be opposed to keep it rotating in a horizontal plane?
 
  • #19
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Good.

... and, what other force is acting on the stopper? And must be opposed to keep it rotating in a horizontal plane?
I believe that the force of tension is the only force that has a horizontal component contributing to the centripetal motion. Besides this, there is the vertical component of the force of tension from Fhanging and the force of gravity. Are you referring to a centrifugal force? The other forces are the only ones we drew on the FBD.
 
  • #20
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The other forces are the only ones we drew on the FBD.

Can you list them, please? There may be one missing.
 
  • #21
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Can you list them, please? There may be one missing.
Force of tension, from the hanging mass, and the force of gravity. A "radial force" was never identified.
 
  • #22
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Can you list them, please? There may be one missing.
Is the radial force, from the resultant force triangle, an actual force?
 
  • #23
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Yes.

You should have two forces acting on the hanging weight: 1) the gravitational force; 2) the tension in the string which is equal to and opposite in direction from the gravitational force. This gives you a net force on the hanging weight of zero, and it doesn't move up or down if everything is working perfectly for you.

The tension in the vertical section of the string is redirected from a vertical direction to a direction with a radial component in the horizontal plane through the top of the tube (the radial component of the resultant force in the string) and slightly downward due to the mass of the stopper being acted upon by the gravitational force (the vertical component in the resultant force in the string). The resultant equals the tension in the string.

If the tension in the string is increased by using a larger hanging weight, and the radius is kept constant, you have to spin the stopper at a higher rate to oppose that tension (resultant). How does that affect the distance below the top of the tube of the plane in which the stopper is being rotated?
 
  • #24
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Yes.

You should have two forces acting on the hanging weight: 1) the gravitational force; 2) the tension in the string which is equal to and opposite in direction from the gravitational force. This gives you a net force on the hanging weight of zero, and it doesn't move up or down if everything is working perfectly for you.

The tension in the vertical section of the string is redirected from a vertical direction to a direction with a radial component in the horizontal plane through the top of the tube (the radial component of the resultant force in the string) and slightly downward due to the mass of the stopper being acted upon by the gravitational force (the vertical component in the resultant force in the string). The resultant equals the tension in the string.

If the tension in the string is increased by using a larger hanging weight, and the radius is kept constant, you have to spin the stopper at a higher rate to oppose that tension (resultant). How does that affect the distance below the top of the tube of the plane in which the stopper is being rotated?
The distance below the top of the tube of the plane in which the stopper is being rotated would increase. Is that what you referred to as the radial force?
 
  • #25
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Now I'm getting sloppy.

Let's define a couple terms: "radius" will mean the radius of the circle describing the movement of the stopper; radius is NOT equal to the length of the string which is to be held constant (refer to the diagram of the experiment).

Try it again.
 

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