Question 312: Centre of Mass and Rotational Motion

Question

If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the center of the Sun, its areal velocity is :

Accepted Answer

dA = $${1 \over 2}$$ r
2
d$$	heta $$
$$ 	herefore $$   $${{dA} \over {dt}} = {1 \over 2}{r^2}{{d	heta } \over {dt}}$$
$$ \Rightarrow $$   $${{dA} \over {dt}} = {1 \over 2}{r^2}\omega $$    . . . . . (1)
We know,
angular momentum,
L = $$mvr$$
=  $$m\left( {\omega r} ight)r$$
=  mr
2
$$\omega $$
$$ 	herefore $$   $$\omega $$ =  $${L \over {m{r^2}}}$$     . . . . . (2)
Put value of $$\omega $$ in equation(1),
$${{dA} \over {dt}}$$ = $${1 \over 2}{r^2}$$ ($${L \over {m{r^2}}}$$)
=  $${L \over {2m}}$$

Answer

$${L \over m}$$

Answer

$${4L \over m}$$

Answer

$${L \over 2m}$$

Answer

$${2L \over m}$$

Centre of Mass and Rotational Motion

If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the center of the Sun, its areal velocity is :

Choose the correct answer: