My Math Forum Center of mass of a rigid body (semi circle)

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May 4th, 2015, 05:38 PM   #1
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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.
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Last edited by skipjack; May 5th, 2015 at 07:25 AM.

May 5th, 2015, 01:49 AM   #2
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Quote:
 Originally Posted by SpareLiver 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.

Last edited by skipjack; May 5th, 2015 at 07:28 AM.

 May 5th, 2015, 04:55 AM #3 Math Team     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

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