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Homework Help: Gravitation Questions

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data

    (1) A uniform wire with mass M and length L is bent into a semicircle. Find the magnitude of the gravitational force this wire exerts on a point with mass m placed at the center of curvature of the semicircle.

    (2) Mass M is distributed uniformly along a line of length 2L. A particle with mass m is at a point that is a distance a above the center of the line on its perpendicular bisector. For the gravitational force that the line exerts on the particle, calculate the components perpendicular and parallel to the line.

    2. Relevant equations

    (1,2) [tex]\vec{F}=\frac{GMm\hat{r}}{r^2}[/tex]

    3. The attempt at a solution

    (1) Looking at the system, I notice that the radius between the wire and the point mass is always [tex]L/\pi[/tex], so, subsituting into the general equation:

    [tex]\vec{F}=\frac{\pi^2GMm\hat{L}}{L^2}[/tex]. This isn't the right answer. Then, I tried solving for the radius between the center of mass of the semi-circle and and point mass to get:


    (2) I've approached this problem from the rod's center of mass but the force depends on the variable L; then, I tried attacking it from the center of mass of each of the halves of the rod and got some ridiculously complicated form for the gravitational force and it didn't work. What bodies and masses am I looking at for this one?
  2. jcsd
  3. Nov 15, 2008 #2


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    1) Note the symmetry in the setup. The force must be directed towards the center of the wire, lets denote it A. All differentials of mass in the wire will exert gravitational force that can be split into one component of force directed towards A and one component that will cancel the component of the mirroring part of the wire. Integrate up the components of force towards A from the wire to get the result (in the integral, symmetry can once again make it a tiny bit easier, can you see it? In this case it is an easy integral either way, but learn to spot these kind of simplifications).

    Maybe one can see it even easier from the symmetry, bit that's the way I would have done it.

    2) Quite the same as 1) but with different integrals. Once again symmetry makes your life a bit easier.
  4. Nov 15, 2008 #3
    I noticed the symmetry and my submission wasn't correct. Can you exhibit both of these integrals? Ugh.
  5. Nov 15, 2008 #4


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    1) I guess the integral would be something like
    [tex]2 \int_0^{\pi/2} r d \phi \sin \phi G \frac{M}{L} \frac{m}{r^2}[/tex] where [tex]r[/tex] is the radius you had calculated, [tex]\phi[/tex] is the integrating angle. [tex]\frac{M}{L}[/tex] is the mass per length unit. At least it gives a simple expression, I have no means of checking the validity right now.
    Last edited: Nov 15, 2008
  6. Nov 15, 2008 #5
    Why does this integral represent the gravitational force exerted by the wire on the point? Isn't the radius constant at [tex]L/\pi[/tex]? I guess I'm not seeing the concept here.
  7. Nov 15, 2008 #6


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    The radius is constant, it is the arc described by [tex]\phi[/tex] I am integrating over. I add up every infinitesimal contribution to the force from every point of the arc, and I disregard all the contributions that cancel anyway because of the symmetry. I have pictured me a semicircle in the upper two quadrants and I disregard all horizontal components since they cancel.

    Maybe this is overcomplicating stuff, perhaps your problem is meant to be solved by just looking up an expression for the center of mass for the semicircle and just use that value in the force expression. What I did (was trying to at least) was really to calculate that value in the gravitation formula directly by integrating which is more general, but if one isn't comfortable with integrals, this maybe isn't the place to start and it may not be required in your course either.

    http://mathworld.wolfram.com/Semicircle.html gives the center of mass for a semicircle.
  8. Nov 15, 2008 #7
    I'm comfortable with integrals; is your integral correct (for what you're trying to do)?

    And I tried substitution [tex]4r/3\pi[/tex] in for [tex]R[/tex], where [tex]r=L/\pi[/tex], in the general gravitational formula but it didn't seem to work. What do you get? :/
  9. Nov 15, 2008 #8


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    I am not sure my integral is correct since I don't know the answer to the question, but I know the procedure I am describing is a working one. Was some time since I calculated these kind of problems though.

    [tex]4r/3\pi[/tex] isn't what you want, that is center of mass for a half disc.

    Show the answer to the question, that will save time for people trying to help.
  10. Nov 15, 2008 #9
    I don't have the answer yet. Why isn't [tex]4r/3\pi[/tex] what I want? It doesn't apply for a half-ring? :X
  11. Nov 15, 2008 #10


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    Ok, I assumed you had the right answer since you said in your first post that the answer you got was wrong.

    Read the Mathworld page again, you don't have a half disc in your problem, you have a semicircle. The centre of mass isn't the same.
  12. Nov 15, 2008 #11
    The integral worked perfectly. What about for the rod? I guess it'd be a very similar integral?

  13. Nov 15, 2008 #12


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    I would abandon the circular coordinates due to the geometry of the problem. This problem requires a bit more difficult integration and even though the answer isn't especially complex it seems like more difficult work than is intended, maybe it's meant to be done another way, but what do I know?

    Yes, the same reasoning as for the integral in 1) holds:
    Integrate up the contribution from every infinitesimal bit of rod. Symmetry makes the integral a bit nicer. Use mass per length unit in the formula. The "radius" becomes a bit more complicated in this case. The equivalent of the [tex]\sin \phi[/tex] in 1) also becomes a bit more complicated, but can be expressed through very basic trigonometry. I used a table of integrals to calculate the last step.
  14. Nov 15, 2008 #13
    I can't seem to do it. Is the integral:

  15. Nov 16, 2008 #14


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    No. The boundaries are wrong (the rod is of length 2L), you don't define which variable you are integrating over, what is your [tex]\theta[/tex] and why do you use sine on it? The expression for r has to be more elaborate, and it should be squared.

    M/2L is the mass per unit length, that is correct.

    As I said you should abandon circular coordinates and use Cartesian. I thought of the rod lying along the x-axis with its middle in the origin and the point mass m a distant a above the origin. You need to express r in this particular system (it is not constant for every bit as in 1)) and integrate up every addition from the infinitesimal bits of rod to the resultant force. What fraction of the force from every infinitesimal bit is directed vertically (this is the equivalent of the sine-function in 1))? Express it directly in x instead of a new angle. As before the horizontal components will cancel.
  16. Nov 16, 2008 #15
    [tex]\int_0^{2L}\frac{GMm}{2Lr^2}dr[/tex], where [tex]r=\sqrt{L^2+a^2}[/tex]?
  17. Nov 16, 2008 #16


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    You are integrating over r, but at the same time you set r to a fixed value [tex]\sqrt{L^2 + a^2}[/tex]. It has to be a variable if you're going to integrate over it if you want some meaning from it in this case.

    And you don't have a term that describes the vertical component of the gravitational force.

    The formula is [tex]F = G \frac{Mm}{r^2} [/tex]. The problem is that since we're not looking at a point of mass, r isn't constant and M is a distributed mass. We need to modify those.

    When we make infinitesimal pieces we need to describe the mass per piece. That is why we divide the mass by the length. The pieces are then situated at different distances from m, that is why we need to integrate over r. Introduce a coordinate x that runs from eg -L to L, every point of the line. Integrate over it so you get every addition to the force from every piece.

    Add a factor that makes it only consider the vertical parts of the force since the horizontal parts cancel anyway in the end because of symmetry. That factor will come out easily from trigonometry. Draw the picture, split the force into two components, find the expression for the vertical part. It can be expressed in terms of x. Now you have the integral.
  18. Nov 16, 2008 #17
    UGH, this is hard. I think I've figured it out (or am getting close, at least)...

    [tex]\vec{F}=\frac{GMm\hat{r}}{r^2}\rightarrow\vec{F}=\frac{GMm\hat{r}}{2Lr^2}[/tex], where [tex]\vec{r}=\sqrt{x^2+a^2}\rightarrow\vec{F}=\int_0^{2L}\frac{GMm}{2L(x^2+a^2)}dx[/tex], giving a final value:

  19. Nov 16, 2008 #18
    BUMP. Can anyone help with this integration?
  20. Nov 17, 2008 #19


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    You will have to motivate your answers more so we know why you came to them. If you just throw out an answer we can say "Wrong, try again" and nothing will be learnt.

    2) is very simliar to 1), look at the motivations for the integral in 1) so you're really sure about why it looks exactly like that. Then apply the same technique in 2).

    In your last integral you had the wrong boundaries and you mismatch vectors and scalars. You've written that [tex]\vec{r}=\sqrt{x^2+a^2}[/tex], on your LHS you have a vector and on your RHS a scalar.

    My reasoning about the horizontal parts cancelling makes the integral easier as one knows the direction of the force and can calculate just the size.
  21. Nov 17, 2008 #20
    Setting the origin at L and integrating only the first quadrant (due to symmetry of the second quadrant negating the canceled, horizontal forces due to gravity), we integrate:

    [tex]\vec{F}=\int_L^{2L}\frac{M\hat{L}}{2L}\frac{Gm\hat{r}}{r^2}dr[/tex], where [tex]r=\sqrt{x^2+a^2}[/tex] and [tex]dr=\frac{x}{\sqrt{x^2+a^2}}dx[/tex]. Substituting into the integral, we find:

    [tex]\vec{F}=2\int_L^{2L}\frac{M\hat{L}}{2L}\frac{Gm\hat{\sqrt{x^2+a^2}}}{x^2+a^2}\frac{x}{\sqrt{x^2+a^2}}dx[/tex]... I think I'm heading in the wrong direction...But...it follows to give:

  22. Nov 17, 2008 #21


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    When you use symmetry in an integral so that you don't have to integrate both parts you multiply the integral with 2 and only integrate one part since the other part will be equal to that one.

    You should put the _middle_ of the rod at origo and let the rod lie along the x-axis. Draw the picture - do you want to integrate från x=L to x=2L?

    Factor by factor in the formula:

    G - that doesn't change. It is the same for every bit of the rod.

    M - that changes to mass per unit length. Take the total mass of the rod and divide by the total length of the rod.

    m - that doesn't change, it is always the same no matter where you evaluate your function.

    r - the distance to the point mass from every piece of the rod. Simple geometry gives [tex]r = \sqrt{a^2 + x^2}[/tex]

    If one were to calculate this directly one would have to use vector notation to get the force in the x-direction and the y-direction separately. BUT - since we see that the accumulated forces in the x-direction will cancel, we skip that calculation altogether. It will yield 0 anyway.

    So, we want to express what fraction of the force in every point that is acting in the y-direction. By looking at a point, eg at x=L, drawing the vector arrow of the force from that point and splitting it into x- and y-components we can find what fraction of F that is directed in the y-direction. It can be expressed in terms of x and a using basic trigonometry.

    Now we have the factors in our integral. Multiply them and integrate the result over the rods length.

    By doing it very systematically it is very easy to look for errors and see what parts are causing trouble.

    As I said in my last post - if you understand the integral in 1), then 2) should only be more difficult because there is a bit more tricky integration involved in the last steps of the calculation.
  23. Nov 17, 2008 #22
    My integral above was exactly correct except I used the wrong trig. function, putting x in the numerator rather than a (and I forgot to carry the 2 down from the second line to the third). Thank you for your help, I understand integrals way better now.
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