# Center of mass/moment

• Dec 19th 2006, 04:12 PM
cyberdx16
Center of mass/moment
A quarter circle of radius 1 foot is cut from sheet metal w/ density of 1lb/ft^2. (This object thus weighs pie/4 lbs). The quarter circle is then attached to a 2ft rod so that the center of the circle is at the middle of the rod and 1/2 of the rod runs along one cut edge of the 1/4 circle. A weight W is attached at the other end of the rod. The whole thing is suspended by a string attached to the middle of the rod and it balances with the horizontal. how heavy is the weight W?
• Dec 19th 2006, 05:15 PM
ThePerfectHacker
Quote:

Originally Posted by cyberdx16
A quarter circle of radius 1 foot is cut from sheet metal w/ density of 1lb/ft^2. (This object thus weighs pie/4 lbs). The quarter circle is then attached to a 2ft rod so that the center of the circle is at the middle of the rod and 1/2 of the rod runs along one cut edge of the 1/4 circle. A weight W is attached at the other end of the rod. The whole thing is suspended by a string attached to the middle of the rod and it balances with the horizontal. how heavy is the weight W?

I do not understand anything. Can you show a picture?
• Dec 19th 2006, 05:34 PM
cyberdx16
• Dec 19th 2006, 07:53 PM
ThePerfectHacker
I kind off see the question.

I think of it like this. If you place a fulctrum at the midpoint it shall balance. That means the sum of the moments is zero. To find the sum of the moments of the quater circle you need to find the first moment of inertia along the x-axis. (Note, in my coordinate system, I let the midpoint be the origin).
That is,
$\displaystyle M_x=\int_D \int y\delta(x,y) dA$
Where,
$\displaystyle \delta(x,y)$
Is the density in every point.
Which is $\displaystyle 1$ (avoiding the units).
Thus, (assuming you know who double integrate),
$\displaystyle \int_0^1 \int_0^{\sqrt{1-x^2}} ydydx$
Thus,
$\displaystyle \int_0^1 \frac{1}{2}-\frac{1}{2}x^2 dx$
Thus,
$\displaystyle \frac{1}{2}x-\frac{1}{6}x^3 \big|_0^1=\frac{1}{3}$
That means the total moment created by the quatercircle is 1/3.

The weight is attach to the end.
Thus, the absolute moment created from the weight matches 1/3. Moment is distance times force. The distance is 1 and weight is w. Thus,
$\displaystyle w=1/3$.

You can thank my engineering professor (assuming I did this right).
• Dec 19th 2006, 08:30 PM
cyberdx16
cool thx a lot