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The net force and centre of gravity of a one dimensional object

  1. Jun 25, 2013 #1
    I have been trying to derive a formula for the forces acting on and the centre of gravity of objects in a one dimensional space. I've started in one dimension to make things easier and intend to move on to two dimensions soon. I tried to do it without referring to any source and I got a solution which I want to check with experienced people.

    A one dimensional stick whose mass is uniformly distributed has its centre of mass at position x on a number line. It has a mass of m and a length l. There is a point mass at x = 0 of mass M.

    I used to gravitational equation inside an integral.

    [itex]x+\frac{l}{2}[/itex]​
    [itex]F_{G} = \frac{\int Fdr}{l} = mMG\int\frac{1}{r^{2}}dr[/itex]
    [itex]x-\frac{l}{2}[/itex]​

    (Sorry about the bad placement of the limits because I don't know how to put them in the editor).

    I then used this equation to find out the centre of gravity.

    [itex]\frac{\int FRdr}{\int Fdr}=\frac{\int\frac{1}{r}dr}{\int\frac{1}{r^{2}}dr}[/itex] with the limits of both integrals being the same as the limits for the equation for the force.

    I substituted the values into the integrals and got the final equations to be these:
    [itex]F_{G}=\frac{4mMG|x|}{x(l^{2}-4x^{2})}[/itex]
    (The [itex]\frac{|x|}{x}[/itex] is there to account for the direction of the force).

    [itex]x_{G}=\frac{ln(\frac{2x+l}{2x-l})(4x^{2}-l^{2})}{4l}[/itex]

    Are my reasoning and equations correct? It took me hours to get to this. (Actually, it took me two years because I didn't know the problem needed integrals and I didn't know what integrals were but once I realised that it might need integrals it took me hours).

    Edit: I have checked them with some actual values and they do seem correct. As the stick goes further away from the point mass the centre of gravity and centre of mass of the stick converge and the force on the stick gets closer to the value of the force on a particle of the same mass at the same position. But I might still be wrong.
     
    Last edited: Jun 25, 2013
  2. jcsd
  3. Jun 25, 2013 #2

    D H

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    No, your equations are not correct.

    The force integral is almost correct. It's best to use a proper sign convention indicating that force pulls the object toward the origin. You should be using [itex]F_G = -\frac{GmM}{l}\int_{x-l/2}^{x+l/2} \frac{dr}{r^2}[/itex].

    The center of gravity calculation is incorrect. This is simply the radial distance at which the force on a point mass of mass m is the same as that on your rod. There is no natural logarithm in the solution.
     
  4. Jun 26, 2013 #3
    Thanks. I understand the force now.

    To summarise for the centre of gravity. If, instead of a rod, there existed another point mass of the same mass of the rod, the position of this point mass so that it has the same force exerted on it as on the rod is the centre of gravity of the rod. Right?
     
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