center of mass

Jul 2008
179
1
Three fine bars in length L are arranged in the shape of inverted U as the figure below. The two sidebars have mass M and the central bar mass 3M. What is the location of the center of mass of the set?
 

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arbolis

PHF Hall of Fame
Apr 2008
815
287
I do it intuitionally because I never learned to solve this. It's the same to find the barycenter of the system of bars.
The center of mass is situated at the middle of the "3M" bar on the left-right axis and at \(\displaystyle \frac{2}{10}L=\frac{1}{5}L\) in the bottom direction (following the up/bottom axis).
In other words, if you draw a Cartesian graph with its origin in the middle of the "3M" bars and with its x axis following the 3M bar and its y axis following the direction of the M bars, then the coordinates would be \(\displaystyle (0,-\frac{1}{5})\).
But I'm not sure of my result, so you're maybe better to wait until someone else answer to you better than I.
 

topsquark

Forum Staff
Apr 2008
3,115
661
On the dance floor, baby!
Three fine bars in length L are arranged in the shape of inverted U as the figure below. The two sidebars have mass M and the central bar mass 3M. What is the location of the center of mass of the set?
The CM of the system will be the CM of the three separate pieces. So set up a coordinate system (for reference) at the lower left corner of the diagram. the CM of the left leg is at (0, L/2), the CM of the top bar is at (L/2, L), and the CM of the right leg is at (L, L/2). Now use the CM formula:
\(\displaystyle r_{CM} = \frac{\sum_{n = 1}^N r_im_i}{\sum_{n = 0}^N m_i}\)
in both coordinate directions.

So I'm getting
\(\displaystyle x_{CM} = \frac{0 \cdot M + \frac{L}{2} \cdot 3M + L \cdot M}{M + 3M + M} = \frac{L}{2}\)
as you might have expected and
\(\displaystyle y_{CM} = \frac{\frac{L}{2} \cdot M + L \cdot 3M + \frac{L}{2} \cdot M}{M + 3M + M} = \frac{4L}{5}\)

This is what arbolis told you.

-Dan
 
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