# Centripetal Force Lab Question

1. Nov 29, 2014

### SigmaForce

1. The problem statement, all variables and given/known data
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:

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

3. 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?

2. Nov 29, 2014

### Bystander

Explain this, and you should see the rest of it.

3. Nov 29, 2014

### SigmaForce

Causes stopper to swing more perpendicular path?

4. Nov 29, 2014

### Bystander

I'll give you a conditional "yes" for that.

5. Nov 29, 2014

### SigmaForce

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. Nov 29, 2014

### Bystander

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. Nov 30, 2014

### SigmaForce

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. Nov 30, 2014

### Bystander

In what direction?

9. Nov 30, 2014

### SigmaForce

Horizontally, or radially inwards if we are going to be more specific to the context.

10. Nov 30, 2014

### Bystander

In what direction does this force act?

11. Nov 30, 2014

### SigmaForce

Well, it would pull the string down, thus exerting a downwards force.

12. Nov 30, 2014

### Bystander

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. Nov 30, 2014

### SigmaForce

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. Nov 30, 2014

### Bystander

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. Nov 30, 2014

### SigmaForce

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. Nov 30, 2014

### Bystander

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. Nov 30, 2014

### SigmaForce

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. Nov 30, 2014

### Bystander

Good.

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

19. Nov 30, 2014

### SigmaForce

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. Nov 30, 2014

### Bystander

Can you list them, please? There may be one missing.