Physics, Algebra, English, Calculus, Math and Science and Literature, Oh my!

DELETE ME PLEASE


Hey all. I have a BS of Science in Physics (not and BS in the appreviation for Bull-) and I'd like to pass that skill on to you! (I like helping people )

Things I am good at:
-Physics (All levels, including nuclear)
-Math (Algebra and Trig and Calculus, oh my again!)
-English (Essays, papers, dissecting a sentence)
-Politics and political theory
-German (Sprechen Sie Deutsch?)
-Computer Science (including hardware, theory, and scripting)
-Engineering

Things I am alright with:
-Biology
-Chemistry
-Psychology
-Photoshop art

Things you shouldn't ask me about unless you have a way for me to understand the concept:
-Everything else.

Things you're wasting your time with:
-Art.


Leave a message or send a PM. Because I do not check the forums every minute of every day, and because I also have RL issues to deal with, I may not get you help ON THE SPOT. But I can do my best to get it done within a few days. This means please do not ask me for help 10 minutes (or the day before) a major project is due and expect an immediate reply. You may still send your questions to me, however, and I will do my best to get them done.

For free.

 

Sorry to say but you can only REQUEST help in this forum (see the sticky). Don't worry, I made the same mistake awhile ago by making a similar thread to yours.

 

Ahh, alright. Thanks for the heads up!

 

I'm having trouble with center of mass in physics. One problem that is gone over quite a few times and I don't quite understand it is finding the center of mass in a plate in shape of a semicricle. The thing is the same thickness throughout.

 

That's a trick question. If the density is uniform, then the center of mass is located at the geometrical center of the object. For example: if you have a circle that has a uniform density, the center of mass is located at the exact middle of the circle.

Effectively, it's center of gravity is located at:

Draw the semicircle so that the midpoint is at the origin. The axis should look like:

X axis goes from -radius to +radius
Y axis goes from 0 - Radius

Because it is an even distribution, the x axis won't matter. The CoM will be located at x=0 (the semi-circle is symmetric along the y axis).

I know this is a little bit confusing, but lets think of the CoM as an integral in polar coordinates, explained here. (I typed up a 30 sentence explanation, these guys do it better).

Things get very messy when you're changing coordinates. For information on polar coordinates, check this out:

http://en.wikipedia.org/wiki/Polar_coordinates

That probably wasn't as helpful as you'd have liked, but center of mass isn't pretty. There's always an integral involved. In this case, two integrals (polar). It could have been worse, it could have been spherical polar.

 

My teacher was able to do it without polar coordinates as that seems to be more complicated.

He used the equation Rcm=int rdm / M

What exactly is dm?

 

dm is a tiny portion of M. When integrating rdm, if r does not depend on mass, then dm just becomes mass, making the question rm.

However, since your formula involves integral(r/m)dm, the r is constant and can be pulled out, making it the integral:

r(integral(1/m)dm.

This makes it radius x ln(M) since the integral of 1/m dm is Ln(M)

Your answer is the radius times the natural log of the mass.

Do you understand how I got that through the calculus I did? If not, I can break it down further.

 

Looking at the equation again, something seems off. Rdm/M. Are the M's different cases? Is the dm a lower case m and the M an upper case m? If so, that has a much different effect on the integral.

 

No... that's not right.

1/M is the constant that isn't part of the integral, rdm is. M is the total mass of the whole object, the lowercase m, I'm not so sure.

 

little m would have to be a segment of something else then.

What you have then is (1/M)*integral(rdm), correct? dm will have limits of integration that will probably not matter at all. you should then end up with (rm)/M assuming r is radius.

For a fixed object of uniform density, this should help you:
http://en.wikipedia.org/wiki/Centroid

If you set your limits of integration from 0 to M, the result of the equation will be 1/M(rM) which just = r. Which is incorrect. (This right here tells me we need factors of pi and polar gives us just that.) The centroid method would be the easiest to use conceptually, but not mathematically.

I still believe you have to convert to polar coordinates to get anywhere, since the semi-circle isn't getting you anything. Doing a bit more digging, I've found a guide that could help you understand that:

http://www.heisingart.com/dvc/ch 09 center of mass.pdf

 

Nevermind, I don't see that you have sense of what's going on, I'll try to figure it out on my own.

 

Best of luck! I worked it out, my answer came out to be 4R/3pi using polar.

When you figure out how to use it in non polar, would you post your method? I'mm sure there is a way using dm in terms of r and an angle theta, using the small angle approximation to eliminate the arc and turn it into a triangle centered on the radius, but I'm unsure of how to approach that.

Actually, flash of insight! If you do turn it into a small angle using the Radius R as the center of the one wall of the triangle, you can mark the halfway point of the other triangles and use the centroid method for a triangle. The dividing line for the base of the triangle must be the radius, however.