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				   prove that the angular momentum is equal to twice the product of its mass and aerial velocity.


5 years ago

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										Dear charan,
The areal velocity vector is
$\frac{d \vec{A}}{d t} = \lim_{\Delta t \rightarrow 0} {\Delta \vec{A} \over \Delta t} = \lim_{\Delta t \rightarrow 0} {\vec{r}(t) \times \vec{r}(t + \Delta t) \over 2 \Delta t}$  $= \lim_{\Delta t \rightarrow 0} {\vec{r}(t) \times [ \vec{r}(t) + \vec{r}\,'(t) \Delta t ] \over 2 \Delta t}$   $= \lim_{\Delta t \rightarrow 0} {\vec{r}(t) \times \vec{r}\,'(t) \over 2} \left( {\Delta t \over \Delta t} \right)$   $= {\vec{r}(t) \times \vec{r}\,'(t) \over 2}.$
But, $\vec{r}\,'(t)$ is the velocity vector $\vec{v}(t)$of the moving particle, so that
$\frac{d \vec{A}}{d t} = {\vec{r} \times \vec{v} \over 2}.$
The areal velocity vector can be placed at the moving point B. As the  particle moves along its path in space, it sweeps out a cone-shaped  surface. The areal velocity vector is perpendicular to this surface,  and, in general, varies in both magnitude and direction. In planar  problems, such as the orbit of a planet about the sun, the direction of  the areal velocity vector is perpendicular to the orbital plane. Kepler's 2nd law is a statement of the constancy of the rate at which the position  vector of a planet sweeps out area, with the sun taken as origin. The  path of the planet is an ellipse, with the sun at one focus.
The angular momentum of the particle is

$\vec{L} = \vec{r} \times m \vec{v},$
and hence
$\vec{L} = 2 m \frac{d \vec{A}}{d t}$. The direction of the angular momentum vector L is always the  same as that of the areal velocity vector. Angular momentum is conserved  if and only if the areal velocity is a constant vector.

All the best.
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Sagar Singh
B.Tech, IIT Delhi


5 years ago

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