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B Is Center of Mass a vector or scalar quantity?

  1. Dec 21, 2016 #1
    I am a little bit confused on Center of Mass and Center of Gravity about what are they vectors or scalars . As I may think they are neither because they are simply two points . Am I saying right or wrong .
  2. jcsd
  3. Dec 21, 2016 #2
    They're points, hence they're vectors. Points are defined by vectors, but of course since vectors represent more 'differences' between points, you need to define a reference frame first. If for example you have a cube of side ##l## and uniform density, and you place the origin in its bottom left corner the center of mass is going to be at ##[l/2,l/2,l/2]## which is a vector.
  4. Dec 21, 2016 #3
    I did not quite get it can you give more explanation please .
  5. Dec 21, 2016 #4
    I am talking about are they vector quantities.
  6. Dec 21, 2016 #5

    Namely, they are vector quantities.
  7. Dec 21, 2016 #6
    How are they vector quantities take any frame of your choice and plz prove it to me .
  8. Dec 21, 2016 #7
    I gave you an example above, with the cube. There's nothing to prove really. Points in three-dimensional space require three numbers to be specified, therefore are vectors. The center of mass is a point. Therefore, the center of mass is specified by a vector.
  9. Dec 21, 2016 #8
    Yes, the center of mass is indeed a vector quantity

    The center of mass is itself a position, thus it requires an exact position which is given by Coordinates x,y,z (if working in Cartesian coordinates).

    The position vector R of the center of mass is given by:

    R =\frac{1}{M} \sum_{\alpha = 1} ^N m_\alpha r_\alpha

    where m##_\alpha## is the mass of an individual particle, and r##_\alpha## as the position of the individual particle.

    M is the total mass of the system.

    Note that the each r##_\alpha## is a position of each particle thus r = r(x,y,z)

    If r is a function of the coordinates of (x,y,z), then the final position vector, R must be a function of x,y,z as well thus R = R(x,y,z).

    The mass is a scalar function and r is a vector, a scalar times a vector is a vector.

    Hope this helps!
  10. Dec 21, 2016 #9
    A point in 3 dimensional space is a vector ??
  11. Dec 21, 2016 #10
    The individual coordinates of the position vector, R, of the center of mass can be broken down into its 3d components (X,Y,Z) if each particle has a position (x,y,z)

    X = \frac{1}{M} \sum_{\alpha = 1}^N m_\alpha x_\alpha \\
    Y = \frac{1}{M} \sum_{\alpha = 1}^N m_\alpha y_\alpha \\
    Z = \frac{1}{M} \sum_{\alpha = 1}^N m_\alpha x_\alpha
    <Moderator's note: LaTeX fixed>
    Last edited by a moderator: Dec 21, 2016
  12. Dec 21, 2016 #11


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  13. Dec 21, 2016 #12
    A point is any arbitrary position (0,4,6), a vector is the point's displacement relative to another point. a point would just be the displacement from the origin. If you move the origin and the point of your choice equally, the magnitude of the vector does not change. The direction is simply the components (x,y,z) of the point in space.
  14. Dec 21, 2016 #13
    Remember in 2D Kinematics (projectile motion) where you broke the velocity of the object to both the x and y components. Both the components had a magnitude and direction (either the x-direction or the y-direction). The position of the object at any given time had a x component and y component. (It had height above the ground and a distance from where it started).

    The center of mass is a position and position (like velocity and acceleration) have a magnitude and direction thus it is a vector.
  15. Dec 21, 2016 #14
    These things are a little bit difficult to understand but as far as i get, the conclusion for this is that center of mass is position in 3D space so it will have both magnitude and direction .
  16. Dec 21, 2016 #15
    Pretty much. Depending on what your origin of the reference system is, "magnitude" will be the distance from that origin, and "direction", well, the direction in which you have to go to reach it.

    It's perhaps simpler to visualise in 2D space. If you have a treasure map and it's got an X marking the spot, and you landed in Dead Man's Bay, I can tell you "well, starting from Dead Man's Bay, to reach the treasure you have to walk 3 km to the North and 4 km to the East". And that's a vector with magnitude (5 km total) and direction (something like North-East-East).
  17. Dec 21, 2016 #16
    As I understand it the original question can be rephrased as:

    Is a point in three dimensional space a vector or scalar?

    As such I don't think the question makes much sense. If the question was more specific eg asking about the location of the point then it could be defined in terms of vectors.
  18. Dec 21, 2016 #17


    Staff: Mentor

    I would say they are neither. I would say that they are members of an affine space


    Affine spaces are definitely closer to vector spaces, but with some subtle differences.
  19. Dec 21, 2016 #18
    I think there is some confusion in this thread. In order to specify position we have to compare to something, so we pick an arbitrary coordinate system with an arbitrary origin. Position is then a vector from the origin to the object. However this vector is not an intrinsic physical property of the object. The origin isn't anything real. The position vector is not an intrinsic property of the physical system. This is apparent when you pick a new origin. All the position vectors change even though the physical situation has not. This is distinguished from vector quantities like displacement, force, velocity, or electric field which intrinsically have direction. What's more the associated vector direction and magnitude does not depend on the choice of coordinate system. Those are intrinsically vector quantities. They can't be contemplated without direction. So when someone asks if center of mass is a vector quantity, I think the answer has to be no. It has no intrinsic directional physical property that is invariant under coordinate transform.
  20. Dec 21, 2016 #19
    True that. I had the impression the question was more at a beginner level and referring to practical matters such as "how should I treat the center of mass in calculations", to which the answer is "like a vector" because anyway any calculation is usually performed in a given system of reference and therefore the position vector of the center of mass will be all that features. But yeah, it's more of a habit to consider them as if they were one and the same while they actually are two slightly different things.
  21. Dec 21, 2016 #20


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    Sometime I think the most elementary issue is taken way too far.

    Let's to back to the foundational mathematics that is relevant here. Let's do this one step at a time, shall we?

    @PHYSICS5502 : Look at this diagram below, which should not be anything new to you.

    Do you understand that the vector r points to a location in space with respect to the coordinate axis defined in the diagram?

  22. Dec 21, 2016 #21
    A point in 3D space itself is not a vector, but it's displacement from the origin is.

    The displacement is the magnitude and the component (x,y,z) is the direction. Vectors are displacement from point a to point b.

    In physics, we often are more concern with the displacement rather than the path of motion it took to get from point a to point b, If I started at position x = 0 walked to x = 5 and then walked back to x = 0 (my original position), then my displacement vector is zero. Likewise if I start at x = 0 and end at x = 3, my vector is 3$/hat{x}$

    You will see in a lot of physics classes, that a lot of physical changes are only concern about the starting point vs the ending point. If you start at a bottom of the hill, and walk up, you are doing work to convert kinetic energy into potential energy, but as you go back down to your starting point, work is being done on you to change potential energy back into kinetic energy. Thus you end in the same amount of kinetic energy and by the work-energy theorem, no work was ever done. The point is although it may seem like you did work, you started at the same point and ended at the same point, so your displacement is zero, thus no work is done. Does this make sense? It doesn't matter if you walked straight up the hill or at an angle around the side of the hill, your starting height and ending height is all that matters, physics doesn't really care how you got up the hill or how you got down. (in terms of measuring kinetic and potential energy).
  23. Dec 21, 2016 #22
    I agree with Zapper Z that this discussion has gone too far afield given the level of the question. From the Wikipedia article on center of mass, we have for a collection of n particles of mass mi, total mass M, and position vectors ri:


    The center of mass, R, is a vector.
  24. Dec 21, 2016 #23
    How can you define the direction of displacement, velocity etc. without an external coordinate system? These quantities don't have an intrinsic direction.
    But then how can you describe position without direction; only by using a scalar quantity?
    I'm sorry, but what you said doesn't make sense to me. In my view, position ( and therefore centre of mass) is a vector.
  25. Dec 22, 2016 #24
    I am surprised you don't see the difference. They HAVE direction whether you define a coordinate system or not. The electric field points from this charge toward that charge. A test charge placed in the field will move in that direction. The car is moving in that direction. In a time interval it will move from here to there. No choice of coordinates will change which direction the force points. Turning your head doesn't prevent the ball from falling down.
  26. Dec 22, 2016 #25
    I agree! But that's what makes math and physics so exciting is the discussions that come out of simple questions. Some of the greatest scientific breakthroughs came from simple questions. It's in our nature to try to go in depth as much as possible and to explore the validity of what we believe is true. :wink::smile:
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