# Uniform circular motion problem

• Femme_physics
In summary: You calculated the maximum reaction force at the shaft supports A and B. In this calculation you must use the self weight of the shaft. Given: 1 meter length of the shaft equals 38.5 [N]The Attempt at a SolutionOkay, So I'm pretty new to uniform circular motion. I think I made a half-decent attempt though. I doubt I got it right though since I'm so new to this. Your method looks fine! :smile:A few things though.You fillled in 15 N for the mass? But 15 N is a force, not a mass.How did you get from v2/r to 2pi
Femme_physics
Gold Member

## Homework Statement

http://img200.imageshack.us/img200/2524/drawingsave.jpg

Weight W that's 15 [N] does circular motion around shaft AB in a frequency of 300 turns per minute

A) calculate the max tension force on shaft CD

[there's another clause, but I rather just try A for now]

## The Attempt at a Solution

Okay,
So I'm pretty new to uniform circular motion. I think I made a half-decent attempt though. I doubt I got it right though since I'm so new to this.

http://img14.imageshack.us/img14/4880/save1h.jpg http://img7.imageshack.us/img7/6522/correctlast.jpg

EDITED: Changed the attempt

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Femme_physics said:
Okay,
So I'm pretty new to uniform circular motion. I think I made a half-decent attempt though. I doubt I got it right though since I'm so new to this.

A few things though.

You fillled in 15 N for the mass? But 15 N is a force, not a mass.

How did you get from v2/r to 2pi 0.1 300 / 60?
The symbols don't seem to match the numbers...EDIT: Just saw your new attempt.
Almost! :)
But I basically have the same comments.

(And what happened to your efforts to give $\pi$ a hat to show off its legs? )

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You fillled in 15 N for the mass? But 15 N is a force, not a mass.

Ohhhhhhhhhh! I get it ;)

How did you get from v2/r to 2pi 0.1 300 / 60?
The symbols don't seem to match the numbers...

Hmmm... well, it's 300 revolutions per minute. So, minute = 60 secs. So, 300/60

And 2 x pi x 0.1 just comes from the formula

V = 2 x Pi x R

The R being 0.1m

How's that wrong?

Femme_physics said:
Ohhhhhhhhhh! I get it ;)

Hmmm... well, it's 300 revolutions per minute. So, minute = 60 secs. So, 300/60

And 2 x pi x 0.1 just comes from the formula

V = 2 x Pi x R

The R being 0.1m

How's that wrong?

Well, as I said you "almost" have it.

And V is not what you wrote.
The proper formula is:

V = 2 x Pi x R x f

What is that?

Of course, f is 1/T, that's why I added 60 seconds to the bottom because it's 300/60 no?

Femme_physics said:
Of course, f is 1/T, that's why I added 60 seconds to the bottom because it's 300/60 no?

Yes, that was right! :)

Hmm... Oh, 60 seconds are also squared. I'll get it fixed :)

Femme_physics said:
Hmm... Oh, 60 seconds are also squared. I'll get it fixed :)

Be warned! Next time I'll string you along a little longer!

Femme_physics said:
Feel free ;)

And yes, I realize I didn't use kg again, but did I get the parenthesis stuff right?

Yes, and I see your π already has nicer legs and looks more like a $\pi$!

LOL.

Okay, I think I ready to solve A now :) I'll try B afterwards.

Thank you so much ILS! You're awesome.

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Femme_physics said:
First off-> *gulps*

Secondly, okay...

EDIT: NEW ATTEMPT!

Is that right?

First off-> you seem to have no trouble!

Except...

Being a little bit sloppy with the numbers. ;)
I didn't check all you substitutions and calculations yet, but I can see the distance of By is wrong...

Have to run now!
(Back to a couple of minutes each time, that I steal from my employer specially for you! )

EDIT: Oh, and you're using the symbol W for 2 different things. You really shouldn't do that!

Thanks, ILS! You're totally right. But if my equations are correct then I'll redo it myself and consider it solved for now (until I compare results, even though I trust you more than any comparison!) ;)

Ah...what'd I do without you... ^^

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Femme_physics said:
Was I correct in saying N is positive (i.e. vector point up) while calculating sum of all moments to A?

You calculated the maximum tension in CD.
At what position is D when the tension is maximal?

What direction does the force on the shaft have when D is in the position of maximum tension?

Ah, good thinking. :) That what "max tension" means! *smacks forehead*.-- Yes, then it must be pointing down then.

Femme_physics said:
Ah, good thinking. :) That what "max tension" means! *smacks forehead*.-- Yes, then it must be pointing down then.

I guess you were already aware of the fact that with the force pointing down, the forces at A and B would be greater (Sum Fy = 0)?
That is, even regardless of maximum tension?

I guess you were already aware of the fact that with the force pointing down, the forces at A and B would be greater (Sum Fy = 0)?
That is, even regardless of maximum tension?

I'm not following. Maximum tension means force pointing down. What do you mean "regardless"? When one happens (max tension) the other must ("force pointing down"). Or so I thought?

Femme_physics said:
I'm not following. Maximum tension means force pointing down. What do you mean "regardless"? When one happens (max tension) the other must ("force pointing down"). Or so I thought?

The pendulum mass at D pulls the shaft down when it is at its lowest point.
This is also the situation where the tensile force is greatest, since gravity pulls it extra down.

The pendulum mass at D pulls the shaft up when it is at its highest point.

It depends on the other forces when the reactive forces at A and B will be greatest.

In this case that will be when the tensile force is downward, which is the same direction as the weight of the shaft which is also down.
Furthermore this is the situation where the tensile force will be greatest.
So everything fits perfectly together to make the reactive forces at A and B greatest.EDIT: And for the record. Your previous calculation is off, since the tensile force should indeed be downward, but I think you already knew that.

What you're saying makes perfect sense.

This is also the situation where the tensile force is greatest, since gravity pulls it extra down.

Yep, that's it :) Brilliant! As always.

EDIT: And for the record. Your previous calculation is off, since the tensile force should indeed be downward, but I think you already knew that.
Yes, thanks to you. :)

http://img98.imageshack.us/img98/9292/1998sol.jpg

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Femme_physics said:
What you're saying makes perfect sense.

Yep, that's it :) Brilliant! As always.

Yes, thanks to you. :)

Thanks! You always make me feel happy when you show your gratitude.

And your calculations look perfectly fine too!

Thanks for saying that!... does that mean I have you Uber seal of Serena's approval? ;)

Femme_physics said:
Thanks for saying that!... does that mean I have you Uber seal of Serena's approval? ;)

Yes!
You get the Seal of ILSe!

I like Serena said:
Yes!
You get the Seal of ILSe!

w00t :)

## 1. What is uniform circular motion?

Uniform circular motion is a type of motion in which an object moves in a circular path at a constant speed. The object's velocity is always tangent to the circular path, and its acceleration is directed towards the center of the circle.

## 2. How is uniform circular motion different from non-uniform circular motion?

Uniform circular motion is characterized by a constant speed and a changing direction, while non-uniform circular motion involves a changing speed in addition to a changing direction. In other words, the object in uniform circular motion moves at a constant rate along a circular path, while the object in non-uniform circular motion experiences a changing rate of motion along the same path.

## 3. What is the centripetal force in uniform circular motion?

The centripetal force in uniform circular motion is the force that acts towards the center of the circle, keeping the object moving in a circular path. This force is necessary to maintain the object's acceleration towards the center of the circle, in accordance with Newton's second law of motion.

## 4. How is centripetal force related to centripetal acceleration?

Centripetal force and centripetal acceleration are directly related. According to Newton's second law of motion, the net force on an object is equal to its mass multiplied by its acceleration. In uniform circular motion, the centripetal force is the net force acting on the object, and the centripetal acceleration is the acceleration towards the center of the circle. Therefore, the centripetal force is equal to the mass of the object multiplied by the centripetal acceleration.

## 5. What is the role of velocity in uniform circular motion?

Velocity plays a crucial role in uniform circular motion. The object's velocity is always tangent to the circular path, meaning it is perpendicular to the object's acceleration towards the center of the circle. The magnitude of the velocity remains constant, while the direction changes continuously, resulting in the object's motion along the circular path.

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