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Centripetal force/motion experiment

by ibwm
Tags: circular, motion, orbit, rotation, tension
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ibwm
#1
Mar22-12, 10:56 PM
P: 4
1. The problem statement, all variables and given/known data

Hi, the aim of this experiment is the investigate the relationship between the centripetal force acting on an object moving in a circle of constant radius and its frequency of revolution. For the experiment, there is a rubber stopper attached to fishing line, that is passed through a glass tube, and an alligator clip is used to keep the radius constant. Hanging on the fishing line are slotted masses, which are altered in order to alter the weight force, and in turn alter the centripetal force supplied by tension.

I have to amend a diagram to include the real rotation of the stopper and draw labelled vectors representing the forces acting on the rubber stopper whilst it is orbiting. I have tried to draw this diagram, but I am unsure as to whether it is correct, and have hence come here in seek of verification.

2. Relevant equations

Nil.

3. The attempt at a solution

The ideal diagram looks like this,


Whilst the amended one looks like this,


But with the use of vector addition, I am unsure as to how this produces the tension resultant vector. Shouldn't there be a horizontal force too? If so, what is it? Please help.

Thank you in advance.
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QuarkCharmer
#2
Mar22-12, 11:25 PM
QuarkCharmer's Avatar
P: 1,035
Seems like if you just take the x component of [itex]F_{c}[/itex] in diagram 2, the problem simplifies to the ideal case. You would have to know the angle between the glass tube and the wire, or the distance from the circular path to the top of the glass tube.
LawrenceC
#3
Mar23-12, 12:20 PM
P: 1,195
Please describe how the experiment is performed. When are the weights added?

the-ever-kid
#4
Mar23-12, 12:44 PM
P: 53
Centripetal force/motion experiment

hey there could you tell me why you hung the mass to the thread down below.........the weight would simply slide down and reduce the radius. i believe you are missing a clip on the top end or are you?....

could you post a detailed procedure of the experiment .

it would make it easier to answer you then.....
the-ever-kid
#5
Mar23-12, 01:09 PM
P: 53
http://www.batesville.k12.in.us/phys..._and_speed.htm
ibwm
#6
Mar23-12, 09:48 PM
P: 4
Quote Quote by LawrenceC View Post
Please describe how the experiment is performed. When are the weights added?
Quote Quote by the-ever-kid View Post
hey there could you tell me why you hung the mass to the thread down below.........the weight would simply slide down and reduce the radius. i believe you are missing a clip on the top end or are you?....

could you post a detailed procedure of the experiment .

it would make it easier to answer you then.....
The experiment was conducted as followed:
Apparatus
- Thin glass tubing (approximately 20cm long, no sharp edges)
- Fishing line (1.5m length)
- Alligator clip
- Large 2-hole rubber stopper
- Mass carrier and slotted masses (50g each)
- Stopwatch(es)
- Metre ruler
- Graph paper

1. Securely tie one end of the fishing line to the rubber stopper. Pass the line through the tube and attach a 50g mass carrier to this end as shown. Attach an alligator clip to the line so as to set the radius of the circular path of the stopper to a value between 80 and 100cm. Add five 50g masses to the carrier making the total mass 300g.
2. Whirl the stopper over your head in a horizontal circular path at such a rate that the alligator clip is just pulled up to, but does not touch, the lower end of the tube.
3. Maintain the rate of revolution so that the alligator clip remains at the same position and measure time taken for 30 revolutions of the stopper, at least 4 times.
4. Add 50g to the mass carrier, keeping the alligator clip in the same position. Repeat steps 2 and 3.
5. Repeat step 4 until you have used all your masses. Enter your results in a table as shown. Calculate the average time taken for 30 revolutions of the stopper for each mass. Then, determine the time taken for one revolution (period).

With the two tables looking like:


So, we vary the mass on the hanger, thus producing different magnitudes of weight. Tension is supplied by this weight. I'm not quite sure about how it actually works and how the stopper can rotate without the radius altering. The alligator clip can move up and down if the experiment is not conducted properly, thus altering the radius. It is the person who is using the apparatus' responsibility to ensure the alligator clip is in relatively the same position throughout the duration of the experiment in order try and maintain a constant radius.

Due to gravity, the stopper is not 90 degrees to the tubing (as shown in the 'amended diagram'). We are to draw labelled vectors of all of the forces acting on the system, and I am unsure as to whether my diagram is completely correct, or not.
PeterO
#7
Mar24-12, 04:05 AM
HW Helper
P: 2,316
Quote Quote by ibwm View Post
1. The problem statement, all variables and given/known data

Hi, the aim of this experiment is the investigate the relationship between the centripetal force acting on an object moving in a circle of constant radius and its frequency of revolution. For the experiment, there is a rubber stopper attached to fishing line, that is passed through a glass tube, and an alligator clip is used to keep the radius constant. Hanging on the fishing line are slotted masses, which are altered in order to alter the weight force, and in turn alter the centripetal force supplied by tension.

I have to amend a diagram to include the real rotation of the stopper and draw labelled vectors representing the forces acting on the rubber stopper whilst it is orbiting. I have tried to draw this diagram, but I am unsure as to whether it is correct, and have hence come here in seek of verification.

2. Relevant equations

Nil.

3. The attempt at a solution

The ideal diagram looks like this,


Whilst the amended one looks like this,


But with the use of vector addition, I am unsure as to how this produces the tension resultant vector. Shouldn't there be a horizontal force too? If so, what is it? Please help.

Thank you in advance.
In the real situation, the Tension and the Weight force combine to make the horizontal Centripetal force.

In the ideal situation, the centripetal force is Tension, while the radius of rotation is L, the length of fishing line out the top of the tube.

In the second, real situation, where the stopper string is "drooping" at angle θ the horizontal, the Centripetal Force is only Tcosθ. However, the radius of the actual circle followed is only Lcosθ and when you finally get the relationship you will see that it is not surprising that those two cosθ factors effectivel compensate for each other / cancel out.
It is thus entirely reasonable to ignore the droop.


For those posters unfamiliar with the experiment, the slotted masses are added before each trial. In earlier editions of this experiment was performed with steel washers of a fixed, but undetermined mass, so the relation was derived in terms of F, 2F, 3F, 4F loads, rather than xx grams.

Once the weights are added, the tube is held above your head [not vertically, your heand operates at aboiut 45 degrees - you have to be able to see your hand holding the glass tube.

WHile holding the hanging weights, you begin spinning the stopper. As you do it you can feel the masses being supported by the tension in the fishing line - the same tension that supplies the centripetal force to the stopper.

When you experience approximate balance, you release the masses and try for balance.
If you rotate too fast, the stopper moves out / the masses move up.
If you rotate too slowly, the stopper moves in / the masses move down.
Get the speed just right and the masses stay in position - and a constant radius is maintained.
The paper clip is there so you can judge the balance situation. You usually try to maintain the paperclip at a chosen distance below the tube - perhaps 1 cm.

It is surprisingly easy to achieve balance after only a few minutes of trialling.

Your partner uses a stop watch to time, say, 10 rotations so that you can then calculate the Period - and thus frequency if you would like.
the-ever-kid
#8
Mar24-12, 02:02 PM
P: 53
okay you wanted to know

But with the use of vector addition, I am unsure as to how this produces the tension resultant vector. Shouldn't there be a horizontal force too? If so, what is it? Please help.
listen don't use vector addition it is a concept created solely to confuse guys like use........with mind boggling algebra....(trigonometry is a concept that rescues physics(mechanics))

let us see................ in the real case the string is inclined right?

yup

let that angle that you have be theta.....
Spoiler

(BTW while doing the experiment you can make sure that there is no inclination by practicing....i mean alot)


back to the explanation

see you measure the side of the cone kay...that is your hypotenuse so the force vector along it is a vector sum of some two vectors that are perpendicular to each other. let them be the perpendicular and horizontal components...

the vertical one is the one that balances F_g and the horizontal one balances the centripetal force of the apparent circular path.
Spoiler

(*BTW the ideal case is really a special case of the real case where the angle theta is 90*)


do your basic trig and find the perpendicular and horizontal components...

you know how to measure frequency i assume...
Spoiler
Your partner uses a stop watch to time, say, 10 rotations so that you can then calculate the Period - and thus frequency if you would like.


now as you know that the centrefugal force surely relates to the mass of the cork the radius and the frequency in some way or the other...

remember these quantities are fundamental as the SI units are fundamental and the force if the only derived unit...... kay.....

now the frequency has a unit second inverse.....mass as kg and length as metre..... in force the si unit is kg*m/s2 right compare the poweres you get a relation force is proportional to mass of the cork the length of the string and inversely proportional to the square of the timeperiod or directly to the square of frequency....

figure the constants using the data given ..... and sams your uncle.....
PeterO
#9
Mar24-12, 06:48 PM
HW Helper
P: 2,316
Quote Quote by the-ever-kid View Post
hey there could you tell me why you hung the mass to the thread down below.........the weight would simply slide down and reduce the radius. i believe you are missing a clip on the top end or are you?....

could you post a detailed procedure of the experiment .

it would make it easier to answer you then.....
I gave you the detailed procedure, but unfortunately you responded with several, very incorrect, assertions.
I trust OP can sift through the correct / incorrect parts of your post.
PeterO
#10
Mar25-12, 01:03 AM
HW Helper
P: 2,316
Quote Quote by the-ever-kid View Post
okay you wanted to know



listen don't use vector addition it is a concept created solely to confuse guys like use........with mind boggling algebra....(trigonometry is a concept that rescues physics(mechanics))

I presume you were trying to say "guys like us" - and even got that wrong.

let us see................ in the real case the string is inclined right?

yup

let that angle that you have be theta.....
Spoiler

(BTW while doing the experiment you can make sure that there is no inclination by practicing....i mean alot)


Practice all you like - that just isn't going to happen. Rotate fast enough and you won't be able to notice the droop, but the masses will be jammed against the bottom of the glass tube meaning a centripetal force much greater than the weight of the masses

back to the explanation

see you measure the side of the cone kay...that is your hypotenuse so the force vector along it is a vector sum of some two vectors that are perpendicular to each other. let them be the perpendicular and horizontal components...

the vertical one is the one that balances F_g and the horizontal one balances the centripetal force of the apparent circular path.
Spoiler

(*BTW the ideal case is really a special case of the real case where the angle theta is 90*)


Some more errors here: The tension IS the hypotenuse, and the horizontal component of the Tension is the centripetal force.

do your basic trig and find the perpendicular and horizontal components...

you know how to measure frequency i assume...
Spoiler
Your partner uses a stop watch to time, say, 10 rotations so that you can then calculate the Period - and thus frequency if you would like.


now as you know that the centrefugal force surely relates to the mass of the cork the radius and the frequency in some way or the other...

Alert! Alert! - there is no such thing as centrifugal Force so anything that follows should be suspected


remember these quantities are fundamental as the SI units are fundamental and the force if the only derived unit...... kay.....

now the frequency has a unit second inverse.....mass as kg and length as metre..... in force the si unit is kg*m/s2 right compare the poweres you get a relation force is proportional to mass of the cork the length of the string and inversely proportional to the square of the timeperiod or directly to the square of frequency....

figure the constants using the data given ..... and sams your uncle.....
The unit comparison in the last paragraph is useful.
the-ever-kid
#11
Mar25-12, 01:39 PM
P: 53
I presume you were trying to say "guys like us" - and even got that wrong.
that was a silly typo......

Alert! Alert! - there is no such thing as centrifugal Force so anything that follows should be suspected
i was merely quoting the name given to the psuedo-force acting in a rotating non-inertial frame.

and the horizontal component of the Tension is the centripetal force
i figured that it was an error later sorry..
The tension IS the hypotenuse
i have never mentioned anything about being the hypotenuse or not.i said that the force of tension was directed along the hypotenuse.... now you're the one assuming things...

The unit comparison in the last paragraph is useful.
Thank you.....


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