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Bashyboy
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Homework Statement
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 possesses 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
Homework Equations
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.