Uniform circular motion problem

[Femme_physics]

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

 

[I like Serena]

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.

Your method looks fine!

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? )

 

[Femme_physics]

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?

 

[I like Serena]

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?

 

[Femme_physics]

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

 

[I like Serena]

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! :)

What about the parentheses?

 

[Femme_physics]

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

 

[I like Serena]

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]

Feel free ;)

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

http://img580.imageshack.us/img580/3733/1515o.jpg

 

[I like Serena]

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!

 

[Femme_physics]

LOL.

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

Thank you so much ILS! You're awesome.

 

[Femme_physics]

http://img857.imageshack.us/img857/4655/finalstuff.jpg

Now that I got A I am carrying on to clause B

B) Calculate the max reaction forces 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]First off-> *gulps*

Secondly, okay...

EDIT: NEW ATTEMPT!

http://img851.imageshack.us/img851/817/rightmethod2.jpg

Is that right?

 

[I like Serena]

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!

 

[Femme_physics]

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... ^^

 

[Femme_physics]

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

FBD on all the forces acting on A:

http://img808.imageshack.us/img808/2780/rcdg.jpg

 

[I like Serena]

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?

 

[Femme_physics]

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

 

[I like Serena]

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 F_y = 0)?
That is, even regardless of maximum tension?

 

[Femme_physics]

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?

 

[I like Serena]

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.

 

[Femme_physics]

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

 

[I like Serena]

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!

 

[Femme_physics]

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

 

[I like Serena]

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

Yes!
You get the Seal of ILSe!

 

[Femme_physics]

Yes!
You get the Seal of ILSe!

w00t :)