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A question related to Center of Mass and Linear Momentum

  1. Dec 1, 2014 #1
    A uniform rod of length 2L and mass 2m is bent on its center at a right angle. The rod hangs down from the ceiling using a thread, so that one part, AB, is horizontal, and the other part, BC, is vertical.
    The question is what's the distance (x) between the suspension point D to the point A (the end of the bar).

    My problem is in building the equations. Shall I refer to the vertical part as a seperate load of weight on the center point B? Which points should I pick to build the equations from? Should the equations be related to the point A?

    zn4n08.png

    Thanks in advance.
     
  2. jcsd
  3. Dec 1, 2014 #2

    SteamKing

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    The concept you are looking for here is indeed based on calculating the location of the center of mass of the angle. You want the point of suspension to pass thru vertical line connecting the string or wire suspending the angle and the c.o.m.

    Use point A as your reference, since it seems the problem is looking for you to calculate this distance. You should be able to determine the centers of mass for rods AB and BC by inspection. All you have to do is write a moment equation for each rod about point A and divide by the total mass of both rods. (IOW, because the two rods are joined, their combined moment will be the sum of the moments of each rod about point A.)
     
  4. Dec 1, 2014 #3
    Ok, I'll start by writing Newton's second law for the forces applied on the suspension point D. That is,
    [itex] \sum{F}\mathrm{{=}}{T}\mathrm{{-}}{2}{mg}\mathrm{{=}}{0} [/itex]
    (2m)*g is becuse all of the rod's mass is applied as a downward force due to gravity. So
    [itex]T=2mg[/itex]
    since the system is in equilibrium.

    Now as for the moment equation, when you say "point of reference", do you mean I should take the displacement vector's magnitude as the distance from where the force is applied to point A or to point D (which is the point where the rotation axis is)? And does the upward force which is applied by the thread is taken into account if the distance to the rotation axis (point of suspension) is 0?
     
  5. Dec 1, 2014 #4

    SteamKing

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    If you write your moment equations with the moment arms measured from point A, you will automatically calculate the distance 'x' when you divide the total moment by the total mass. Also, the tension force in the thread is not included in the moment calculation to determine the c.o.m. of the angle.
     
  6. Dec 1, 2014 #5
    Ok, then one force that is applied here (apart from the tension caused by the thread) is in the point B. This force is caused by BC's gravity, and is operating exactly on point B. The other force is AB's gravity, applied on the c.o.m (point D). And still I can't get the moment equation right, since
    [itex]\sum{\mathit{\tau}}\mathrm{{=}}{\mathrm{(}}{mg}{\mathrm{)}}\mathrm{\cdot}{L}\mathrm{{+}}{\mathrm{(}}{mg}{\mathrm{)}}\mathrm{\cdot}{x}\mathrm{{=}}{0}[/itex]
    both forces are gravity forces pointed downwards, hence taken in "+" sign, where one is the AB's gravity from c.o.m point D (distanced x from point A), and the other force is similar, caused by BC and distanced L from point A. By the way, when I say "distanced" I refer to the moment arm, of course.
     
  7. Dec 1, 2014 #6

    SteamKing

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    I think point D is the location of the thread, not necessarily the c.o.m. of rod AB.
     
  8. Dec 1, 2014 #7
    So that's where I have a misunderstanding: On the one hand, the whole rod AC is placed perfectly so that the angle between AB and the vertical line (the line of the thread, if you want) is 0 degrees. On the other hand, point D cannot be in the middle because of BC's existance, which forces the c.o.m to be located closer to point B, rather than to point A.

    Anyways, the equation I wrote above is still valid though, because D is distanced x from A, and B is distanced L from A, although you made a great point here. Still I can't anywere from that equation, whether because it is wrong somewhere or because I miss a point.
     
  9. Dec 1, 2014 #8

    SteamKing

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    What is the c.o.m. of rod AB only, measured from point A?
    What is the c.o.m. of rod BC only, measured from point A?

    If you can also calculate the masses of rods AB and BC, you're all set to calculate the moment of the angle ABC about point A.
     
  10. Dec 1, 2014 #9
    Ignoring BC, the c.o.m of rod AB is L/2 distant from A. Ignoring AB, the c.o.m of rod AB is [itex]
    \raise0.7ex\hbox{${\sqrt{5}L}$}\!\!\left/{}\right.\!\!\lower0.7ex\hbox{${2}$}[/itex] distant from A. Why? I would say because both rods are two halves of one uniform rod, one dimensional rod, and in such a rod the c.o.m must be centered, isn't it?

    I don't need to calculate the masses since they're already given (2m for the whole rode, m for each half of it, i.e AB=BC=m [kg]).
     
  11. Dec 1, 2014 #10

    SteamKing

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    You don't need to calculate the straight line distance from A to the c.o.m. of rod BC, only the horizontal distance. Since you want to line up the thread with the c.o.m. horizontal position from point A, the vertical location of the c.o.m. of the angle ABC is irrelevant in this case.
     
  12. Dec 2, 2014 #11
    Horizontal distance is from c.o.m of BC to A is L, Horizontal distance from c.om of AB is L/2. Now trying to build the equation, I get:
    [itex] \sum{\mathit{\tau}}\mathrm{{=}}{\mathrm{(}}{mg}{\mathrm{)}}\mathrm{\cdot}\frac{L}{2}\mathrm{{+}}{\mathrm{(}}{mg}{\mathrm{)}}\mathrm{\cdot}{L}\mathrm{{=}}{0} [/itex]

    What I miss in that equation is the variable x, which is the distance from the suspension point to A, but you wrote that the tension force in the thread is not included in the moment calculation to determine the c.o.m. of the angle. So how can I add x as a variable to the equation?
     
  13. Dec 2, 2014 #12

    SteamKing

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    The distance x is equal to the horizontal distance of the c.o.m. of angle ABC measured from point A.

    Σ T ≠ 0 about point A in this case. It would be if we needed to find the tension in the thread holding the angle.
     
  14. Dec 2, 2014 #13
    Ok, got it! The answer is [itex]x=0.75L[/itex].
    Now the second part of the question is what is going to be the angle between AB and the vertical straight line (the line that is actually parallel to the thread), should the thread will be suspended at the end point A instead of D.

    Now for that I need to understand how to use x that I have found in the new setup. There has to be a relation to that x, but I cannot figure out a way to see it.
    What I can see, though, is that the angle I need to find is in a right triangle, whose hypotenuse is L, but the new distance (say y) between A and the straight vertical line is unknown. If I can express it as a function of the length L, then I can calculate [itex]\theta[/itex], as
    [itex]\sin\mathit{\theta}\mathrm{{=}}\frac{y}{L}[/itex]
    and [itex]\theta[/itex] is the required angle.

    What do you suggest?
     
  15. Dec 2, 2014 #14

    SteamKing

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    You know from calculating 'x' where the c.o.m. of the angle must be in relation to point B. You need to find the vertical location of the c.o.m. measured from point C. That imaginary point in the corner of the angle is the c.o.m. for the entire angle. If the angle is suspended from point A, then the c.o.m. of the angle and point A must lie on the same line. You can use this relationship to determine the angle theta which rod AB makes with the vertical.
     
  16. Dec 2, 2014 #15
    Here's the equation I tried to build and work with - unsuccessfully. What do you think is wrong with it?
     

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  17. Dec 2, 2014 #16

    SteamKing

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    You haven't used the proper moment arm for rod BC when taking moments about point B. Check your diagram. According to the moment equation you wrote, the distance of the c.o.m. for rod AB from the vertical line thru point B is the same distance as the c.o.m. for rod BC from this same line. That's not what your sketch shows.
     
  18. Dec 2, 2014 #17
    You're absolutely right. But now that I've re-sketched it and wrote the equation again it gets even more complexed.
     

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  19. Dec 2, 2014 #18

    SteamKing

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    I haven't checked all your algebra, but you stopped your calculations before you got an answer.

    Using your last equation:

    cos θ - 3 sin θ = 0

    you should be able to use some trig definitions and solve for the angle θ. Here, let algebra take its course.
     
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