Simple center of mass question - Thanks

[nukeman]

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 :)

Homework Equations

The Attempt at a Solution

I am really confused with this one.

Maybe let's start with the X

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

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

 

[ehild]

Replace every metre stick with a point mass of M at its own centre of mass.

ehild

 

[nukeman]

How do I find each metre sticks own center of mass??

 

[gordonj005]

assuming each stick is of uniform density, it will be in the physical center of the stick.

$x_{c} = \frac{\sum_{i = 1}^{N} m_{i} x_{i}}{\sum_{i = 1}^{N} m_{i}}$

$y_{c} = \frac{\sum_{i = 1}^{N} m_{i} y_{i}}{\sum_{i = 1}^{N} m_{i}}$

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:

$x_{c} = \frac{\sum_{i=1}^{N} x_{i}}{N}$

$y_{c} = \frac{\sum_{i=1}^{N} y_{i}}{N}$

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

$x_{cm} = \frac{x_{cm1} + x_{cm2} + x_{sm3}}{3}$

$y_{cm} = \frac{y_{cm1} + y_{cm2} + y_{sm3}}{3}$

 

[nukeman]

assuming each stick is of uniform density, it will be in the physical center of the stick.

$x_{c} = \frac{\sum_{i = 1}^{N} m_{i} x_{i}}{\sum_{i = 1}^{N} m_{i}}$

$y_{c} = \frac{\sum_{i = 1}^{N} m_{i} y_{i}}{\sum_{i = 1}^{N} m_{i}}$

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:

$x_{c} = \frac{\sum_{i=1}^{N} x_{i}}{N}$

$y_{c} = \frac{\sum_{i=1}^{N} y_{i}}{N}$

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

$x_{cm} = \frac{x_{cm1} + x_{cm2} + x_{sm3}}{3}$

$y_{cm} = \frac{y_{cm1} + y_{cm2} + y_{sm3}}{3}$

What about the stick at the 60 degree angle?

Im actually having trouble with plugging the right info into the formula, let's say for X - what would let's say "Xcm1" be?

 

[gordonj005]

You can resolve the center of mass of the angled stick into its components using trig ratios.

$x = \frac{1}{2} \cos (\frac{\pi}{6})$

$y = \frac{1}{2} \sin (\frac{\pi}{6})$

and then proceed from there.

by $x_{cm1}$ i mean the x component of the center of mass for stick one.

 

[nukeman]

is Xcm1 = .5 m?

 

[gordonj005]

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

 

[nukeman]

Ahhg, I am not getting this. :(

Thanks for trying tho

 

[gordonj005]

keep at it, just try to understand what the formulas are saying.

keep trying, but you can check it against this:

Spoiler

$x_{cm} = \frac{0 + \frac{1}{2}\cos(\frac{\pi}{6}) + 0.5}{3}$

$y_{cm} = \frac{0.5 + \frac{1}{2}\sin(\frac{\pi}{6}) + 0}{3}$

 

[nukeman]

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?

 

[nukeman]

Can anyone confirm this for me please ? :)

 

[gordonj005]

Two words my friend, radian mode.

 

[nukeman]

dammit :)

How about now:

0.31i, 0.25j

Is that correct?

 

[gordonj005]

yup, now i'd suggest going over the problem again or making up your own problem simmilar to this one to practice.