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How to calculate the height of center of gravity?

  1. Mar 9, 2015 #1
    upload_2015-3-9_19-44-38.png

    In the above figure, a small weight is placed at 2 cm (0.02 m) from the center of the rectangular prism. Lets call the mass of this small weight (M1) and the mass of the entire body including the small weight (M) Given the dimensions above, I want to know how to calculate the height of center of gravity from the base of the prism. I know that we have to take moments but I am a little confused. If anyone can help clarify how this can be done, it would be very helpful.

    P.S - Just in case anyone is wondering, COG is not at the center of the object i.e it is not H/2 (Yes I did think that at first). The reason for this, the small weight has a certain mass and placing the weight on the object will alter its center of gravity. With increasing weight the height of center of gravity should increase from the base.

    P.P.S. - The prism is hollow on top and not fully enclosed. And the horizontal black line which is parallel to W represents a rod on which the small weight was placed

    So if anyone knows how to get the center of gravity by taking moments, please explain it to me
     

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  2. jcsd
  3. Mar 9, 2015 #2

    Svein

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    By definition the position of the Center of Gravity is given by [itex] \frac{\sum_{j}^{}m_{j}\vec{r_{j}}}{\sum_{j}^{}m_{j}}[/itex] (when you are calculating with discrete masses and discrete positions). You are allowed to use the total mass of discrete objects placed at the COG for that object.
     
  4. Mar 10, 2015 #3
    But how do we calculate the position of the whole entire object. If the mass of the object is M, how do I take moments around that mass? Like its not located at a particular point, so what position do I consider?
     
  5. Mar 10, 2015 #4

    SteamKing

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    You pick a convenient reference point, like the bottom of the prism, on centerline, say one end of the prism.

    You measure the moment arms based on that set of reference locations, or coordinate system, if you will. After you have summed up the moments about that reference system and divided by the total mass of the prism, the c.o.g. will then be referred to the origin of that same set of coordinate axes.
     
  6. Mar 11, 2015 #5
    But how do I take moments around the mass of the whole object, the small weight is located at a particular point so taking moments around it by picking an origin is easy. But the mass of the whole prism is 'M', so how do I take moments around this mass?
     
  7. Mar 11, 2015 #6

    Svein

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    First find the center of gravity for the prism. Assume that the side walls and bottom are uniform (then you can easily find the center of gravity for each).
     
  8. Mar 13, 2015 #7
    Okay I have another question. Is the mass moment of inertia of a cylindrical shell that is hollow on top and bottom different from the mass moment of inertia of a cylindrical shell that is hollow on top but enclosed at the bottom?
     
  9. Mar 13, 2015 #8

    SteamKing

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    Yes. Why would it not be so?
     
  10. Mar 14, 2015 #9
    What is the mass moment of inertia of a cylinder that is hollow only on to and enclosed at the bottom? The 'handbook of mass moment of inertia' that I am referring to has equations of mass moment of inertia for a solid cylinder and a thin cylindrical shell that is hollow on top and bottom.
     
  11. Mar 14, 2015 #10
    I was thinking that of adding the mass moment of inertia of a cylindrical shell with the mass moment of inertia of a thin circular disk.

    Like, the mass moment of inertia of a cylindrical shell about its longitudinal axis is (M/12) * ((6R^2)+(H^2))

    And the mass moment of inertia of a thin circular disc about its longitudinal axis is (MR^2)/4

    So can I add these two equations to get the equation for mass moment of inertia of hollow cylinder with an enclosed bottom?
     
  12. Mar 14, 2015 #11

    SteamKing

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    You can. If you are interested in calculating the moment of inertia for a cylindrical shell and end cap rotating about the longitudinal axis, then adding the MOIs for the individual pieces will give the MOI for the whole body.

    Usually, you want to find the mass moment of inertia of inertia of a composite body about its center of mass or some other point of reference. In these cases, the parallel axis theorem must also be used to calculate the correct mass moment of inertia.
     
  13. Mar 14, 2015 #12
    Alright. Thank you very much. I am still a little confused about the COG though, I'll try and look into it more.
     
  14. Mar 14, 2015 #13

    SteamKing

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    If you're confused about how to determine the center of gravity, then why are you asking questions about calculating the mass moment of inertia?

    In other words, you should understand the former before worrying about the latter, which, as far as I can determine, has no bearing on the question raised in the OP.

    And I'm confused. Haven't you taken a course in Statics or Dynamics before now? Usually, the concept of moments and c.o.g. or centers of mass are elementary topics covered in these courses.
     
  15. Mar 14, 2015 #14
    No, I don't need to know one to find the other, they are not related, they are simply steps to a calculation for finding another parameter. And as for your second question, yes I have taken courses of Statics and Dynamics. Now for the mass moment of inertia, they taught us the parallel axis theorem and perpendicular axis theorem at the beginning of the course and we learned how to calculate MOI for simple 2D shapes such as a rectangle or an I-Beam, but later on in the course they discouraged us from calculating MOI for every single shape that we encounter while solving a question and instead provided us with a data booklet that contained equations for most of the shapes (not all), and for centre of gravity, I am very familiar with the concept, and we have covered that too just not in this particular manner, it was never taught in an extensive way and also, I have gone to my Dynamics professor with this particular COG question and he was a little confused too. He said he has to think about it and to go back to him after a day or two. So if anything, the system has to be blamed and not me. At least I am trying to figure out what I don't understand. Hope that clears your confusion. Good day.
     
  16. Mar 14, 2015 #15

    SteamKing

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    Tables save time, but only if you are thoroughly familiar with the concept.

    This is rather disconcerting. If your professor has trouble with this basic concept, I am not optimistic that he has a firm grasp of more complicated (or subtle) topics.

    You may find the following article of use:

    http://ruina.tam.cornell.edu/Book/COMRuinaPratap.pdf

    I would pay particular attention to the example problems at the end and to the tabular form method of calculation for the c.o.g. illustrated on page 7.

    If you want to publish an attempt at calculating the c.o.g. for the prism illustrated in the OP, I would be glad to review your calculations and offer comment.
     
  17. Mar 16, 2015 #16
    Hi. I have calculated the COG for a similar situation but with a different shaped object. I have created a thread and posted my attempt there, it is titled 'Please verify COG calculation', please check and let me know if I have done it correctly, if I have then I am on the right track and can do so for this rectangular prism or other shapes.
     
  18. Mar 16, 2015 #17
    I am attaching the word file here, if it helps to view it on word
     

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  19. Aug 18, 2016 #18
    Please delete this thread.
     
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