My Math Forum  

Go Back   My Math Forum > Science Forums > Physics

Physics Physics Forum


Thanks Tree2Thanks
  • 1 Post By Benit13
  • 1 Post By skeeter
Reply
 
LinkBack Thread Tools Display Modes
May 4th, 2015, 05:38 PM   #1
Newbie
 
Joined: Apr 2015
From: Washington

Posts: 4
Thanks: 0

Question Center of mass of a rigid body (semi circle)

Hey, so I have a question that s been stumping me for a while. So here it goes.

Find the center of mass of the semicircular disk of radius with length a? Assume it has a constant density of 1.

Basically, point of origin is smack dab center of the semicircle, with -a being to the left, and +a being to the right.

I took a practice exam, and they *explain* the answer to me, but their explanation sucks.

They give me the equation $\displaystyle Y_c = 1/M \int Y dm$
Then they inform me that $\displaystyle dm = 2x * dy$
Where does the 2x come from?

Then they say that $\displaystyle x = \sqrt{a^2 - y^2} $
Where does that come from?

So the equations transfers from $\displaystyle y_cm = 2/(\pi*a^2) \int y(2*\sqrt{a^2 - y^2} dy $

to

$\displaystyle y_cm = 2/(\pi*a^2) \int -2/3*(2*\sqrt{a^2 - y^2}^{3/2}$

I am super confused as to where they are pulling all of these numbers from. Any kind of explanation of what's going on would be fantastic.
Attached Images
File Type: jpg stupidproblem.JPG (39.4 KB, 7 views)

Last edited by skipjack; May 5th, 2015 at 07:25 AM.
SpareLiver is offline  
 
May 5th, 2015, 01:49 AM   #2
Senior Member
 
Joined: Apr 2014
From: Glasgow

Posts: 2,156
Thanks: 731

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
Originally Posted by SpareLiver View Post

They give me the equation $\displaystyle y_c = 1/M \int y dm$
Then they inform me that $\displaystyle dm = 2x * dy$
Where does the 2x come from?
Consider slicing up the semicircle into many thin rectangles with width $\displaystyle dy$. The length of every slice is $\displaystyle 2x$, so the total area is $\displaystyle 2x dy$.

Specifically, it is $\displaystyle dA = 2x dy$, which is the change in area. The reason for the use of area here is that for uniform shapes (homogeneous distribution of mass) the centre of mass can be found by using

$\displaystyle y_{com} = \frac{1}{A} \int y dA$

instead, which is described purely in terms of area. This can be done because the mass of each individual mass packet can be divided through.

i.e.
$\displaystyle y_{com} = \frac{1}{M} \int y dm$
$\displaystyle = \frac{1}{Am} \int y m dA$
$\displaystyle = \frac{m}{Am} \int y dA$
$\displaystyle = \frac{1}{A} \int y dA$

Note that this set of operations is only valid when the mass does not depend on y (or x).

Quote:
Then they say that $\displaystyle x = \sqrt{a^2 - y^2} $
Where does that come from?
The equation of a circle is

$\displaystyle x^2 + y^2 = r^2$

So with a little rearrangement the equation that describes x in terms of y for this particular semicircle ($\displaystyle r=a$) is

$\displaystyle x = \sqrt{a^2 - y^2}$

This needs to be done because x depends on y. As y increases from 0 to a, x decreases. This relationship between x and y can be substituted (for x) so that your integration is over one variable only.

Quote:
So the equations transfers from $\displaystyle y_cm = 2/(\pi*a^2) \int y(2*\sqrt{a^2 - y^2} dy $

to

$\displaystyle Y_cm = 2/(\pi*a^2) \int -2/3*(2*\sqrt{a^2 - y^2}^{3/2}$

I am super confused as to where they are pulling all of these numbers from. Any kind of explanation of what's going on would be fantastic.
The area of a semicircle can be found by halving the area of a circle:

$\displaystyle A = \frac{\pi a^2}{2}$

So you take your integral, substitute for x so there is only 1 independent variable, perform the integral (with limits) and then divide by the total area above to get the centre of mass.
Thanks from SpareLiver

Last edited by skipjack; May 5th, 2015 at 07:28 AM.
Benit13 is offline  
May 5th, 2015, 04:55 AM   #3
Math Team
 
skeeter's Avatar
 
Joined: Jul 2011
From: Texas

Posts: 2,949
Thanks: 1555

nice to know ...

$\displaystyle 2\pi x_c \cdot \frac{\pi a^2}{2} = \frac{4\pi a^3}{3}$

$\displaystyle x_c = \frac{4a}{3 \pi}$

Pappus's Centroid Theorem -- from Wolfram MathWorld
Thanks from SpareLiver
skeeter is offline  
Reply

  My Math Forum > Science Forums > Physics

Tags
body, center, circle, mass, rigid, semi, semicircle



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Rotation of a rigid body Animate Physics 1 November 19th, 2014 06:20 PM
center of mass boaz Algebra 6 November 17th, 2013 11:40 AM
Center of mass slice_of_god Real Analysis 1 June 4th, 2013 02:11 PM
center mass kfarnan Calculus 1 December 1st, 2011 12:14 PM
Question about rigid body analysis cmmcnamara Physics 1 December 6th, 2009 11:53 PM





Copyright © 2019 My Math Forum. All rights reserved.