Determining the center of mass for a particle/molecule...?

Question

In the ammonia (NH_3) molecule, the three hydrogen (H) atoms form an equilateral triangle; the center of the triangle is 9.40*10^-11 m from each hydrogen atom. The nitrogen (N) atom is the apex of a pyramid, with the three hydrogen atoms forming a base. The nitrogen-to-hydrogen atomic mass ratio is 13.9, and the nitrogen-to-hydrogen distance is 10.14*10^-11 m. Locate the center of mass of the molecule relative to the nitrogen atom, and explain how you got your answer.

Answer 1

You can find the center of mass of the 3 Hs first, then find the center of mass between that point and the N. Draw an equilateral triangle with an H at each point. The symmetry of the problem indicates that the center of mass (COM) for the 3 Hs should be in the center of this triangle. Treat this point as having a mass of 3H and find the COM between it and the N atom. If the z axis is pointing out of your paper, they both lie on it. The general formula for finding center of mass of a point mass system is:

r1*m1+r2*m2+... /(m1+m2+...)

Where r is the location of each particle of mass m. (You can just use this equation to solve the problem from the beginning, but you have to deal with 3-D coordinates). Treating the 3 Hs as 1 point of mass 3H, the equation reduces to:

3H*r1 * 13.9H*0/(13.9H+3H) --taking the N atom at the origin.

From the geometry of a "pyramid" you should be able to derive r1=9.4e-11*cos(60)+9.4e-11.

Interestingly enough, this answer is generally wrong (this is the right answer for you, though, unless this question was asked in a quantum mechanics course) . The NH3 molecule is normally in an energy eigenstate where the N "oscillates" rapidly between either side of the H triangle. Technically the center of mass is thus inside the plane of the H molecules, or 9.4e-11cos60+9.4e-11 meters away from the N atom. If you want to piss of your teacher you can tell him that.

Answer 2

The center of mass R of a system of particles is defined as the average of their positions r(i), weighted by their masses m(i):
R=(1/M)Sigma[(m(i) r(i)]

where M is the total mass of the system, equal to the sum of the particle masses.
Accordingly, let centroid of the tetrahedron be the origon of the co-ordinate system or the reference point.
Height of the pyrimid is =(10.14*10^-11 m)^2 - (9.40*10^-11)^2
Position of the centroid from centre of the base is at a distence
[10.14*10^-11 m)^2 - (9.40*10^-11)^2] /3 on the line of height.
Then, r(1), r(2), r(3) = [{(10.14*10^-11 m)^2 - (9.40*10^-11)^2} /3]^2 + (9.40*10^-11)^2
And, R(4) the distance between the nitrogen atom and the centroid is
=[(10.14*10^-11 m)^2 - (9.40*10^-11)^2 ]*(2/3)
Then, R={1/(3amu +13.9 amu)}*[3*r(1) +13.9 r(4)]
Note - convert amu to the relevent mass unit

Answer 3

First upload up each and every of the mass interior the device: M + M + M = 3*M next, upload the lots expanded with their x-coordinate on the grounds that finding midsection of mass alongside the x-axis: M*0 + M*0 + M*L = M*L ultimately, divide step 2 by ability of step a million: (M*L) / (3*M) = (a million/3)*L

Answer 4

In case of a triangle the center of mass is the point of intersection of it's medians i.e. the centroid. I guess that's enough information to calculate the center of mass. If not then contact me on [e-mail] .

Answer 5

Maybe it's good

Answer 6

this is hard i think u can use geometric method
you should also try using your book