Simple center of mass question - Thanks

  • Thread starter nukeman
  • Start date
  • #1
655
0

Homework Statement



Hey all. Below is the question and picture related to it.

"Three uniform metre sticks each of mass M are attached together as follows: Stick 1 lies along the y axis from y = 0 to y = 1.0m. Stick 2 lies along the x axis from x = 0 to x = 1.0m. Stick 3 makes an angle of 60 degrees with the y axis. Find the x and Y coordinates of the centre of mass of the system of three metre sticks, and marks its approximate location on the diagram of the three sticks."



----Sorry, its kinda upside down :)
j1251k.jpg


Homework Equations




The Attempt at a Solution



I am really confused with this one.

Maybe lets start with the X

The formula I know is: Xcm = m1x1 + m2x2 + m3x3 / m1+m2+m3

But how do I input into this formula... ??
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,909
Replace every metre stick with a point mass of M at its own centre of mass.

ehild
 
  • #3
655
0
How do I find each metre sticks own center of mass??
 
  • #4
56
0
assuming each stick is of uniform density, it will be in the physical center of the stick.

[itex] x_{c} = \frac{\sum_{i = 1}^{N} m_{i} x_{i}}{\sum_{i = 1}^{N} m_{i}} [/itex]

[itex] y_{c} = \frac{\sum_{i = 1}^{N} m_{i} y_{i}}{\sum_{i = 1}^{N} m_{i}} [/itex]

these describe the center of mass, but assuming it is uniform, the masses are taken out and you are left with just the physical center:

[itex]x_{c} = \frac{\sum_{i=1}^{N} x_{i}}{N}[/itex]

[itex]y_{c} = \frac{\sum_{i=1}^{N} y_{i}}{N}[/itex]

now plug that into the same equation for all three sticks:

[itex]x_{cm} = \frac{x_{cm1} + x_{cm2} + x_{sm3}}{3}[/itex]

[itex]y_{cm} = \frac{y_{cm1} + y_{cm2} + y_{sm3}}{3}[/itex]
 
Last edited:
  • #5
655
0
assuming each stick is of uniform density, it will be in the physical center of the stick.

[itex] x_{c} = \frac{\sum_{i = 1}^{N} m_{i} x_{i}}{\sum_{i = 1}^{N} m_{i}} [/itex]

[itex] y_{c} = \frac{\sum_{i = 1}^{N} m_{i} y_{i}}{\sum_{i = 1}^{N} m_{i}} [/itex]

these describe the center of mass, but assuming it is uniform, the masses are taken out and you are left with just the physical center:

[itex]x_{c} = \frac{\sum_{i=1}^{N} x_{i}}{N}[/itex]

[itex]y_{c} = \frac{\sum_{i=1}^{N} y_{i}}{N}[/itex]

now plug that into the same equation for all three sticks:

[itex]x_{cm} = \frac{x_{cm1} + x_{cm2} + x_{sm3}}{3}[/itex]

[itex]y_{cm} = \frac{y_{cm1} + y_{cm2} + y_{sm3}}{3}[/itex]

What about the stick at the 60 degree angle?

Im actually having trouble with plugging the right info into the formula, lets say for X - what would lets say "Xcm1" be?
 
  • #6
56
0
You can resolve the center of mass of the angled stick into its components using trig ratios.

[itex] x = \frac{1}{2} \cos (\frac{\pi}{6})[/itex]

[itex] y = \frac{1}{2} \sin (\frac{\pi}{6})[/itex]

and then proceed from there.

by [itex]x_{cm1}[/itex] i mean the x component of the center of mass for stick one.
 
  • #7
655
0
is Xcm1 = .5 m?
 
  • #8
56
0
is Xcm1 = .5 m?
well that depends on what you choose to be your first stick. if you're talking about the bottom one then you'd be correct
 
  • #9
655
0
Ahhg, im not getting this. :(

Thanks for trying tho
 
  • #10
56
0
keep at it, just try to understand what the formulas are saying.

keep trying, but you can check it against this:

[itex]x_{cm} = \frac{0 + \frac{1}{2}\cos(\frac{\pi}{6}) + 0.5}{3}[/itex]

[itex]y_{cm} = \frac{0.5 + \frac{1}{2}\sin(\frac{\pi}{6}) + 0}{3}[/itex]
 
  • #11
655
0
Gordon or anyone, here is what I got. Can you tell me if the answer I got to this question is correct, including image (posted below)

(.333 i + .168 j) m

Image of where centre of mass roughly would be?

55s2v7.jpg
 
Last edited:
  • #12
655
0
Can anyone confirm this for me please ? :)
 
  • #13
56
0
Two words my friend, radian mode.
 
  • #14
655
0
dammit :)

How about now:

0.31i, 0.25j

Is that correct?
 
  • #15
56
0
yup, now i'd suggest going over the problem again or making up your own problem simmilar to this one to practice.
 

Related Threads on Simple center of mass question - Thanks

  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
4
Views
7K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
9
Views
3K
  • Last Post
Replies
3
Views
972
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
4
Views
2K
Top