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

2. Originally Posted by narsep
What is the gravity force on a unit mass inside an homogeneous disc of density d and radius r?
You can calculate it yourself if you know how to do integration. Do you?

3. You may be kind enough to help me.

4. Originally Posted by narsep
You may be kind enough to help me.
Hint: the net force in the center is zero. How about in an arbitrary point?

5. That's what I am asking for. The force would be directed also towards the centre.

6. Originally Posted by narsep
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).

7. Finally I found that:
1) we draw the diameter to the point (P) under consideration, that is considered as the x-axis.
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/cm2)
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 x-axis (they are cancel out on y-axis). 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.

8. Originally Posted by narsep
Finally I found that:
1) we draw the diameter to the point (P) under consideration, that is considered as the x-axis.
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/cm2)
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 x-axis (they are cancel out on y-axis). 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.
You are welcome.

9. 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)

10. Originally Posted by narsep
I still feel that there is a mistake somewhere. (F' should be: F'=G*m*π*d*cos(atan(4*R/(3*π*r)/2, and F=G*m*π*d. But this does not make sense)
1. What is "n"?
2. Show step by step calculation of the center of mass of the upper semicircle
3. Did you double the force due to the lower semicircle contribution?
4. Force looks like . I don't quite see those dimensions in any of your expressions.

11. !) π: is 3.14...
2) https://math.stackexchange.com/quest...al-coordinates
3) yes (not just double due to the projection to x-axis)
4) the denominator x^2 is cancelled by the area of semicircle (π*x^2/2).

12. Originally Posted by narsep
!) π: is 3.14...
2) https://math.stackexchange.com/quest...al-coordinates
3) yes (not just double due to the projection to x-axis)
4) the denominator x^2 is cancelled by the area of semicircle (π*x^2/2).
OK with 1,2,3,4.

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.

13. 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

14. Originally Posted by narsep
I did not use the vertical cord.
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.

15. 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*R2*cos2(φ)/(2*r2), where M=π*R2*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*R2*cos3(atan(0.4244*R/r)))/r2.
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.

16. Originally Posted by narsep
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*R2*cos2(φ)/(2*r2), where M=π*R2*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*R2*cos3(atan(0.4244*R/r)))/r2.
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.

I am grateful for your time spent so far on this post.
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.

So I am afraid we HAVE to use integrations ... . Any help on this???
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.

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