Centripetal acceleration.


by colombo
Tags: acceleration, centripetal
colombo
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#1
May10-06, 12:49 PM
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Iam doing on In this laboratory, you will investigate the centripetal acceleration of a horizontally rotating object. lab report.

on this lab they asking one question ...

Would it be possible to get the string holding the rubber stopper to be PERFECTLY HORIZONTAL during this experiment? Explain.

but, i dont get the solutions for this question. please if any one know the answer can tell me ....or mail me @ ... <email deleted>

thaks for your helping.
NB; i need the answer as soon as possble..coz, i done all the works..but, i dont noe this last quesitons..
i think gravity depent on this quesiotn..but, i dont now how to explane that . pls can any one tell me how to fix this.
thanks again.
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Doc Al
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#2
May10-06, 12:59 PM
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Rather than just asking someone for the answer, why not try to figure it out for yourself. Give it your best shot, explain your thinking, and you'll get plenty of help.

Hint: What supports the weight of the stopper?
colombo
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May10-06, 01:03 PM
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Quote Quote by Doc Al
Rather than just asking someone for the answer, why not try to figure it out for yourself. Give it your best shot, explain your thinking, and you'll get plenty of help.

Hint: What supports the weight of the stopper?
i think the suports the weight of the stopper is gravity ????

colombo
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#4
May10-06, 01:07 PM
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Centripetal acceleration.


i think string's tenstion and stoppers weight with gravity.
using F = Ma
F = Mstpperf * g.
Doc Al
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May10-06, 01:10 PM
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Hint: What component of the string's tension force must balance the stoppers weight?
colombo
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#6
May10-06, 01:12 PM
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Centripetal acceleration.
ac = FC/M.
nrqed
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May10-06, 02:01 PM
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Quote Quote by colombo
Centripetal acceleration.
ac = FC/M.
This is not a force.

Ok, let's go back to the basics.

what are the two forces acting on the stopper as it is rotating?

Draw the situation when the string is not horizontal first. What can you say about the acceleration of the stopper? In what direction is it directed? Now, what can you say about the net force along y? About the net force along x? Are they zero or not?

Now consider the case when the string is strictly horizontal. Do you see any problem you might run into?
colombo
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May10-06, 02:22 PM
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Quote Quote by nrqed
This is not a force.

Ok, let's go back to the basics.

what are the two forces acting on the stopper as it is rotating?

Draw the situation when the string is not horizontal first. What can you say about the acceleration of the stopper? In what direction is it directed? Now, what can you say about the net force along y? About the net force along x? Are they zero or not?

Now consider the case when the string is strictly horizontal. Do you see any problem you might run into?
when the stopper is rotaing FC and Fs aree actiong on the stoper.
FC = M*ac and FS = Mgcosteta ( not horizontal only). the next forc is not zero

when is horizontal FC = M* ac only action also, in this case the net foce is zero ( y and x ) give me some hint on this question..
the question is Would it be possible to get the string holding the rubber stopper to be PERFECTLY HORIZONTAL during this experiment? Explain.
but, i noe its could be Perfectly horizontal. but, i dont noe y.give me some hint...and explean meee
nrqed
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May10-06, 02:36 PM
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Quote Quote by colombo
when the stopper is rotaing FC and Fs aree actiong on the stoper.
FC = M*ac and FS = Mgcosteta ( not horizontal only). the next forc is not zero
[/itex]
The centripetal force is NOT a fundamental force. It is *defined* to be the net force acting toward the center of the circle but it is NEVER one of the forces you put in a free body diagram!!

No, the two forces are : the force of gravity (the weight) which is acting straight down and has a magnitude mg. And the other force is the tenstion in the string, what you called Fs.
when is horizontal FC = M* ac only action also,
[/quote] I am having some trouble following what you wrote ("only action also"??)

When it is horizontal, what is the net force along x? (I know it is Fc but forget about Fc, this is not a fundamental force. You should just think about tension and weight). What is the net force along y?

What is the acceleration along x? what is the acceleration along y?

Now check if it is possible to respect Newton's second law along x and y. What do you notice?



in this case the net foce is zero ( y and x ) give me some hint on this question..
the question is Would it be possible to get the string holding the rubber stopper to be PERFECTLY HORIZONTAL during this experiment? Explain.
but, i noe its could be Perfectly horizontal. but, i dont noe y.give me some hint...and explean meee[/QUOTE]
colombo
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#10
May10-06, 02:48 PM
P: 13
Quote Quote by nrqed
when is horizontal FC = M* ac only action also,
I am having some trouble following what you wrote ("only action also"??)

When it is horizontal, what is the net force along x? (I know it is Fc but forget about Fc, this is not a fundamental force. You should just think about tension and weight). What is the net force along y?

What is the acceleration along x? what is the acceleration along y?

Now check if it is possible to respect Newton's second law along x and y. What do you notice?



in this case the net foce is zero ( y and x ) give me some hint on this question..
the question is Would it be possible to get the string holding the rubber stopper to be PERFECTLY HORIZONTAL during this experiment? Explain.
but, i noe its could be Perfectly horizontal. but, i dont noe y.give me some hint...and explean meee[/QUOTE][/QUOTE]

The net fource along Y is Mg. ( this is could be a tension of the string.)

also, the acceleration along y is zero, and x = ac ( Centripetal acceleration.)


this is lab report abt Centripetal acceleration.

In this laboratory, you will investigate the centripetal acceleration of a horizontally rotating object.
---------------------------------------------------------------
Procedure:
use the apparatus shown in the diagram

Rotate the rubber stopper horizontally until the knot in the spring is stationary. Use a stopwatch to measure the elapsed time for 5 10 revolutions.

Measure the dimensions required to calculate the radius from the centre of the tube to the centre of the rubber stopper.

Measure the mass of the rubber stopper and the steel washers.

Repeat the measurement for different radius and number of washers.

Record your data in a table.


Trial # Mass of stoppers
(g) Mass of washers
Elapsed time
Mass of stoppers Radius
M. washers Revolution
1 11.1g 35.3g 10 5.75s 27.6cm
11.1g 35.3g 10 6.53s 37cm
11.1g 35.3 g 10 7.0s 43.4cm
2 11.1g 21.3g 10 5.4s 23.5cm
11.1 g 21.3 g 10 6.69s 35cm


i did all the questions and i found ac too
i need to anser this quesiton now.
----------------------------------
Que: Would it be possible to get the string holding the rubber stopper to be PERFECTLY HORIZONTAL during this experiment? Explain.
--------------------------
I know it could be horizontal.but, i dont know y


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