#1

A water molecule consists of an oxygen atom with two hydrogen atoms bound to it. The angle between the two bonds is 106°. If the bonds are l = 0.100 nm long, where is the center of mass of the molecule? (Assume a standard coordinate system with the origin at the center of the oxygen molecule.)

The thing is that i don't understand how you're supposed to find that. Just knowing the length and angles can give you the point? Help? thanks!

The thing is that i don't understand how you're supposed to find that. Just knowing the length and angles can give you the point? Help? thanks!

#2

42

#3

they might form a triangle or something

#4

They for a triangle, center of triangle.

#5

it gives me this...

#6

Alright, first step. Turn that picture 90 degrees clockwise in your minds eye, so that both H atoms are on the bottom. Got it?

Now, you need to resolve the centre of mass in both the x direction and the y direction. Let's start with th x direction (horizontally).

The H atoms are the same distance from the Oxy atom, and equidistant from the middle. the Oxy atom is on the middle. Hence, the Centre of mass of the whole system will also be in the middle.

So now, we can simplify the model of the system.

Imagine that both hydrogen atoms are replaced with a single particle with weight= 2 hydrogen atoms. (i.e. Oxygen weighs 16, H weighs 1, and this new atom which we will call A weighs 2.).

Now, only the two atoms are joined two each other. Oxygen (O) and this new atom (A). The length between these two atoms is the length between the oxygen atoms and the line that both Hydrogen atoms are on.

so your model looks like this:

O------A.

Where O is 6 times heavier than A. Now If the two were exactly the same weight, A would be slap bang in the middle of this. Unfortunately, they aren't. Still with me?

Now, you need to resolve the centre of mass in both the x direction and the y direction. Let's start with th x direction (horizontally).

The H atoms are the same distance from the Oxy atom, and equidistant from the middle. the Oxy atom is on the middle. Hence, the Centre of mass of the whole system will also be in the middle.

So now, we can simplify the model of the system.

Imagine that both hydrogen atoms are replaced with a single particle with weight= 2 hydrogen atoms. (i.e. Oxygen weighs 16, H weighs 1, and this new atom which we will call A weighs 2.).

Now, only the two atoms are joined two each other. Oxygen (O) and this new atom (A). The length between these two atoms is the length between the oxygen atoms and the line that both Hydrogen atoms are on.

so your model looks like this:

O------A.

Where O is 6 times heavier than A. Now If the two were exactly the same weight, A would be slap bang in the middle of this. Unfortunately, they aren't. Still with me?

*Last edited by LordBishek at Jan 6, 2009,*

#7

This sounds suspiciously like chemistry...

#8

Imagine the atoms are just balls of different weights connected by string, and it won't.

#9

Imagine the atoms are just balls of different weights connected by string, and it won't.

Oh, I gotcha. My brain turned off as soon as I read the word "molecule"

#10

make a triangle using the three atoms. you must find the centroid of the triangle, since the centroid is the center of gravity of a triangle. now to find the centroid you begin by drawing a segment from an angle to the midpoint of the side opposite of it. do this to all sides. the point where all lines intersect is the centroid. graph the process. (i would make the two congruent sides each an even number of units when i graph it so i could easily find the midpoints. the third side i would then just measure and divide by two.)

*Last edited by corcmfndr at Jan 6, 2009,*

#11

wouldn't it be 8 times the weight?

and i'm really confused...btw i got y = 0 (friend told me :\ )

and i'm really confused...btw i got y = 0 (friend told me :\ )

#12

make a triangle using the three atoms. you must find the centroid of the triangle, since the centroid is the center of gravity of a triangle. now to find the centroid you begin by drawing a segment from an angle to the midpoint of the side opposite of it. do this to all sides. the point where all lines intersect is the centroid. graph the process. (i would make the two congruent sides each an even number of units when i graph it so i could easily find the midpoints. the third side i would then just measure and divide by two.)

Nice, but I wouldn't recommend graphing, if it's done wrong, it can be innaccurate. I'm explaining it using moments.

#13

Alright, I'm not going to wait, because it's coming up to 3AM here, but I'll be around a while.

Now, we have our system O--------------A

I am aware that it looks like a broken dick with one solitary testicle, but I can't do much about it.

Anyway, we are going to take:

*drumroll*

Alright.

If you can't follow the working up to this point, get out a pencil and paper and draw the diagrams. If you can, get them out anyway, because this bit is hard to explain without the aid of diagrams, and I can't keep uploading pictures.

Now, let's call the distance between O and A something. let's call it L.

(btw, it may actually help if we call A 2H instead because that's what it is, but for simplicity's sake, I'll keep using A)

Now somewhere along this line L is our centre of mass. It will be a certain distance from the Oxygen atom, and this distance we will call Y. So it will be (L-Y) from A.

Draw this out, because even I can't visualise it (I'm slow though).

Alright, so we need to find Y.

Now, a moment about a point is the force exerted by an object multiplied by the distance from that point ASSUMING that force is at right angles to the distance:

|-------------------------------o-----------------|

the o is the point, the two lines are forces, and the broken lines are the two distances between the forces - just like a seesaw.

Now let's say A is the left vertical line, and O is the right one. Y is the right broken line, (L-Y) is the left one. o is the centre of mass (M)

Redraw the diagram:

A--------------(Y-L)----------------M--------Y--------O

Now, we have our system O--------------A

I am aware that it looks like a broken dick with one solitary testicle, but I can't do much about it.

Anyway, we are going to take:

*drumroll*

**Moments**

Alright.

If you can't follow the working up to this point, get out a pencil and paper and draw the diagrams. If you can, get them out anyway, because this bit is hard to explain without the aid of diagrams, and I can't keep uploading pictures.

Now, let's call the distance between O and A something. let's call it L.

(btw, it may actually help if we call A 2H instead because that's what it is, but for simplicity's sake, I'll keep using A)

Now somewhere along this line L is our centre of mass. It will be a certain distance from the Oxygen atom, and this distance we will call Y. So it will be (L-Y) from A.

Draw this out, because even I can't visualise it (I'm slow though).

Alright, so we need to find Y.

Now, a moment about a point is the force exerted by an object multiplied by the distance from that point ASSUMING that force is at right angles to the distance:

|-------------------------------o-----------------|

the o is the point, the two lines are forces, and the broken lines are the two distances between the forces - just like a seesaw.

Now let's say A is the left vertical line, and O is the right one. Y is the right broken line, (L-Y) is the left one. o is the centre of mass (M)

Redraw the diagram:

A--------------(Y-L)----------------M--------Y--------O

*Last edited by LordBishek at Jan 6, 2009,*

#14

O_O

i'm following...but not. i drew it out and stuff but like what about the "seesaw"? where do i go from there? how do you actually find that point?

i'm following...but not. i drew it out and stuff but like what about the "seesaw"? where do i go from there? how do you actually find that point?

#15

Hold your horses dammit, I'm getting to that.

Alright, I've edited the above post.

Right, ONWARD TO GLORY.

While I'm typing up this next bit, go play some videogames or watch some porn. Watch this space.

Alright, I've edited the above post.

Right, ONWARD TO GLORY.

While I'm typing up this next bit, go play some videogames or watch some porn. Watch this space.

*Last edited by LordBishek at Jan 6, 2009,*

#16

Alright. Now we need to take moments.

Remember MOMENTS ABOUT A CENTRE OF MASS ARE EQUAL.

Now, remember how we defined a moment?

Perpendicular Force times distance to a point.

= F x D.

Now our moments are equal, so:

The weight of O is Fo and the weight of A=FA

Now a weight is simply the mass of an object times g the acceleration due to gravity (g)

Fo=Mog = Weight of O

FA=MAg = Weight of A

The moments will be

Mo x g x Y

MA x g x (L-Y)

Moments about a Centre of Mass are equal, therefore

MogY = MAg(L-Y)

The g's are on both sides of the equation, they will cancel:

MoY = MA(L-Y)

Now, remember how much those things weighed? Mo=16, MA= M2H= 2 x 1 =2.

so

let's put that bizzle fo' shizzle back into the equation:

2Y=16(L-Y)

2Y=16L-16Y.

OK, we need to know L to find Y.

Right, go back to your first diagram from your book. I'm going to assume you know simple trig for the next bit, if you don't, I'll explain further, but only briefly here:

we know l (that is a little L) = 0.1 nm

Cos 53 = Adj/Hyp

Cos53 = L/little l

L=(little l) x Cos 53

= 0.1 x Cos 53

=0.0602 nm (check this on your own calculator to make sure this is right)

Stick it back into our Moment equation.

2Y = 16L - 16 Y

18Y = 16L

Y = 16L/18

= (16 x 0.0602)/18

= 0.0535

Your centre of mass is therefore bang between the two H atoms, and (1-0.0535 =) 0.0464 nm from the Oxy atom.

Remember MOMENTS ABOUT A CENTRE OF MASS ARE EQUAL.

Now, remember how we defined a moment?

Perpendicular Force times distance to a point.

= F x D.

Now our moments are equal, so:

The weight of O is Fo and the weight of A=FA

Now a weight is simply the mass of an object times g the acceleration due to gravity (g)

Fo=Mog = Weight of O

FA=MAg = Weight of A

The moments will be

Mo x g x Y

MA x g x (L-Y)

Moments about a Centre of Mass are equal, therefore

MogY = MAg(L-Y)

The g's are on both sides of the equation, they will cancel:

MoY = MA(L-Y)

Now, remember how much those things weighed? Mo=16, MA= M2H= 2 x 1 =2.

so

let's put that bizzle fo' shizzle back into the equation:

2Y=16(L-Y)

2Y=16L-16Y.

OK, we need to know L to find Y.

Right, go back to your first diagram from your book. I'm going to assume you know simple trig for the next bit, if you don't, I'll explain further, but only briefly here:

we know l (that is a little L) = 0.1 nm

Cos 53 = Adj/Hyp

Cos53 = L/little l

L=(little l) x Cos 53

= 0.1 x Cos 53

=0.0602 nm (check this on your own calculator to make sure this is right)

Stick it back into our Moment equation.

2Y = 16L - 16 Y

18Y = 16L

Y = 16L/18

= (16 x 0.0602)/18

= 0.0535

Your centre of mass is therefore bang between the two H atoms, and (1-0.0535 =) 0.0464 nm from the Oxy atom.

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