91Ó°ÊÓ

Three masses and their velocity vectors are given.

Angular momentum is equal to the moment of linear momentum. The angular momentum is the cross product of the radius vector with the linear momentum.

Use the concept of angular momentum to find the net angular momentum about each point given.

Formulae are as follows:

$\stackrel{→}{L}=\stackrel{→}{r}×m\stackrel{→}{v}$

Here,$\stackrel{→}{L}$is angular momentum,$\stackrel{→}{r}$is the position vector,$\stackrel{→}{v}$is velocity, and m is mass.

Angular momentum about any point is,

$\stackrel{→}{L}=\stackrel{→}{r}×m\stackrel{→}{v}$

We can write magnitude as,

$\left|\stackrel{→}{L}\right|=\left|\stackrel{→}{r}\right|\left|m\stackrel{→}{v}\right|sin\mathrm{θ}$

Here,$\stackrel{→}{L}$is angular momentum,$\stackrel{→}{r}$is the position vector,$\stackrel{→}{v}$is velocity vector,$\mathrm{θ}$is an angle between$\stackrel{→}{r}$and$\stackrel{→}{v}$. We can write r for$\left|\stackrel{→}{r}\right|sin\mathrm{θ}$.

So, it can be written as

$L=rmv$

.

The angular momentum is positive for counter-clockwise direction and negative for the clockwise direction.

Now, calculate net angular momentum about point a:

Let$l$is the side of the square.

$L_a=\left(mv\right)0+\left(mv\right)l+\left(mv\right)l$

$L_a=2mvl$

Now, calculate net angular momentum about point b:

$L_b=\left(mv\right)0+\left(mv\right)0+\left(mv\right)l$

$L_b=mvl$

Now, calculate net angular momentum about point c:

$L_c=(−mv)l+\left(mv\right)0+\left(mv\right)0$

$L_c=−mvl$

Now, calculate net angular momentum about point d :

$L_d=\left(mv\right)l+(−mv)l+\left(mv\right)0$

$L_d=0$

Now, calculate net angular momentum about point e:

$L_e=\frac{(−mv)l}{2}+\frac{\left(mv\right)l}{2}+\frac{\left(mv\right)l}{2}$

$L_e=\frac{mvl}{2}$

Thus, ranking of magnitude of angular momentum is:

$L_a>(L_b=L_c)>L_e>L_d$

Therefore, the magnitude of net angular momentum can be ranked using the angular momentum concept.