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Thread: Gravity within disc.

  1. #1 Gravity within disc. 
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    What is the gravity force on a unit mass inside an homogeneous disc of density d and radius r?
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  2. #2  
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    Quote Originally Posted by narsep View Post
    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?
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  3. #3  
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    You may be kind enough to help me.
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  4. #4  
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    Quote Originally Posted by narsep View Post
    You may be kind enough to help me.
    Hint: the net force in the center is zero. How about in an arbitrary point?
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    That's what I am asking for. The force would be directed also towards the centre.
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  6. #6  
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    Quote Originally Posted by narsep View Post
    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).
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  7. #7  
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    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.
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  8. #8  
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    Quote Originally Posted by narsep View Post
    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.
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  9. #9  
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    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)
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  10. #10  
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    Quote Originally Posted by narsep View Post
    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.
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  11. #11  
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    !) 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).
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  12. #12  
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    Quote Originally Posted by narsep View Post
    !) 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.


    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.
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  13. #13  
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    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
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  14. #14  
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    Quote Originally Posted by narsep View Post
    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.
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  15. #15  
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    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.
    Last edited by narsep; 02-14-2018 at 11:33 AM.
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  16. #16  
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    Quote Originally Posted by narsep View Post
    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.
    Last edited by AndrewC; 02-14-2018 at 04:18 PM.
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