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Homework Help: Angular Momentum and Kepler's Second Law

  1. Jun 19, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m moves along a straight line with constant velocity v in the x direction, a distance b from the x axis. (a) Does the particle possess any angular momentum about the origin? (b) Explain why the amount of its angular momentum should change or should stay constant. (c) Show that Kepler's second law is satisfied by showing that the two shaded triangles in the figure have the same area when t_D - t_C = t_B - t_A

    2. Relevant equations

    3. The attempt at a solution

    Here are my answers to part (a) and (b):

    (a) Examining one of the forms of angular momentum, [itex]\vec{L} = \vec{r} \times m \vec{v}[/itex], we can easily see that neither the position vector function, nor the velocity, is the zero vector at any time, meaning that angular momentum is some nonzero value. For [itex]\vec{r}[/itex], we see that the y-distance (projection of the [itex]\vec{r}[/itex] on the [itex]\hat{j}[/itex] will remain constant; however, because the object with a velocity in the x-direction, the x-component of the position vector function is varying as some function of time. So, the position vector function will have the form [itex]\vec{r}(t) = f(t) \hat{i} + b \hat{j} [/itex] Now, the derivative of [itex]\vec{r}(t)[/itex] should equal [itex]\vec{v}_0[/itex].

    [itex]\vec{r}'(t) = f'(t) \hat{i}= \vec{v}_0[/itex] Since the velocity does not vary with time (it is constant) [itex]f'(t) = a[/itex]

    That the three quantities in the expression for angular momentum are never zero implies that angular momentum will never be zero.

    (b) Since [itex]\vec{r}[/itex] varies with its x-component, which varies with time, and since [itex]\vec{L}[/itex] depends upon [itex]\vec{r}[/itex], then [itex]\vec{L}[/itex] must vary.

    Does this appear correct? Even if it is correct, I would be interested in knowing other ways of answering this particular question.

    For part (c), I am not sure of how to employ kepler's law. I feel as the object travels farther and farther, it will take longer to sweep out certain angles.
  2. jcsd
  3. Jun 19, 2013 #2
    To answer (a) and (b), you just need to find out what L is exactly. For (b), it might also b useful to think whether conservation of angular momentum applies here.

    For (c), you don't need to employ Kepler's law, you need to do what it says: show that the areas are equal.
  4. Jun 19, 2013 #3
    Voko, has the way I answered the question not in fact answer it? Also, what do you mean by, "...find out what L is exactly;" haven't I described what L is sufficiently?
  5. Jun 19, 2013 #4
    I might accept the answer to (a), but it is very lengthy and leaves the impression of hand waving (I mean no offence). The answer to (b) is not good at all, a vector product of varying entities can be constant.
  6. Jun 19, 2013 #5
    What Voko means is calculate L explicitly. In fact it would do to just calculate the modulus of L With the info you have you can easily calculate this: v is constant in direction and magnitud, you can obtain r vía the pythagoras theorem. All you need is now is the sine of the angle these two vectors form. Once you have done this, (a) and (b) are solved and all that's left to do is explain the result ( that's the important part: anyone can string out a load of equations, but knowing how to tell someone what they mean makes the difference between simple calculations and PHYSICS).
  7. Jun 19, 2013 #6
    As per your suggestion, Joey21, I believe I have properly solved part (b). I found the modulus of angular momentum, [itex]\vec{L} = mv(r \sin \theta)[/itex]. I then drew a few triangles and took note of the fact that, even though that the angle, [itex]\theta[/itex] and the magnitude of the position vector function, [itex]r[/itex], were varying, the quantity [itex]r \sin \theta[/[/itex] as a whole was not varying; for this is the distance between the object and the x-axis, which is known to be "b". So, angular momentum is in fact constant, and it is [itex]L = mvb [/itex]. Does this seem correct?
  8. Jun 19, 2013 #7
    Exactly. It all cancels out. The form of L's magnitude is in agreement with that angular momentum with respet to a point depends on the particle's perpendicular position with respect to the point. Also, the fact that it is constant is in agreement with the fact that the particle moves with constant velocity: the sum of forces acting on it is 0, and therefore the torque acting on it, which turns out to be the time derivative of L, is 0. So the angular momentum is constant.
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