What is the gravity force on a unit mass inside an homogeneous disc of density d and radius r?

What is the gravity force on a unit mass inside an homogeneous disc of density d and radius r?
That's what I am asking for. The force would be directed also towards the centre.
Correct. Here is a second hint. Pick a point inside the disc, this is where you are calculating the force. Draw the diameter that passes through that point. Draw the cord perpendicular to the diameter that passes through that point. You will get the disc divided into 4 "slices". Now, you can compute the force, I reduced the problem to your computing the 4 areas (there are only 2 distinct areas, due to problem symmetry).
Finally I found that:
1) we draw the diameter to the point (P) under consideration, that is considered as the xaxis.
2) we find the centre of mass of the upper semicircle (0,4*R/(3*π)), where R: radius of the disc.
3) we find the gravitation force that acts to the point P from this upper semicircle. F'=G*m*π*d/2 , where m:unit mass of the point P, d:surface density (e.g. gr/cm^{2})
4) we do the same for the down semicircle. The centre of mass: (0,4*R/(3*π)).
5) we compute the vector sum of the two forces on xaxis (they are cancel out on yaxis). Net force: F=G*m*π*d*cos(atan(4*R/(3*π*r)) , where r:distance of point P from the centre of disc.
Thanks for the help.
I still feel that there is a mistake somewhere. (F' should be: F'=G*m*π*d/(2*cos(atan(4*R/(3*π*r)), and F=G*m*π*d. But this does not make sense)
!) π: is 3.14...
2) https://math.stackexchange.com/quest...alcoordinates
3) yes (not just double due to the projection to xaxis)
4) the denominator x^2 is cancelled by the area of semicircle (π*x^2/2).
OK with 1,2,3,4.
Your error:
The semicircle is divided by the vertical cord into TWO areas, you forgot to consider the contribution of the area between the vertical cord and the circle. This is why the center of gravity of the figure is not at x=0 but somewhere left of it.
The centre of mass of the upper semicircle is at (x=0, y=4*R/(3*π)), and the centre of mass of the down semicircle is at (x=0, y=4*R/(3*π)).
I did not use the vertical cord.
PS. It is a pity that I did not manage to insert a graph
You HAVE to, you are MISSING a force component that operates in the opposite direction along the x axis. This is why you are getting the wrong result. I did this drawing for you to understand where you are making the error.
I used the centre of mass in order not to use integrations (not an expert on this...).
According to the drawing the force F'=G*M*m/(r/cos(φ))^{2}=G*m*π*d*R^{2}*cos^{2}(φ)/(2*r^{2}), where M=π*R^{2}*d/2 and φ=atan(4*R/(3*π*r)=atan(0.4244*R/r).
Hence
F=2*F'*cos(φ)=(G*m*π*d*R^{2}*cos^{3}(atan(0.4244*R/r)))/r^{2}.
The graph of F vs r is quite reasonable.
However, centre of mass could only be used where the gravity field is homogeneous, that is not probably the case.
So I am afraid we HAVE to use integrations ... . Any help on this???
I am grateful for your time spent so far on this post.
Last edited by narsep; 02142018 at 12:33 PM.
Doesn't look right, you should have a subtraction between two expressions , one relating to the section of circle left to the vertical chord and one relating to the section right of the vertical chord.
Sure, you can adapt this very nice solution. All you need to do is to modify the expression for "dM". As a bonus, you will get the force both inside and outside the disc.So I am afraid we HAVE to use integrations ... . Any help on this???
Last edited by AndrewC; 02142018 at 05:18 PM.
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