20.1 Angular mometum (check your understanding)

Here, moment arms of the particles are $y_1$ and $y_1$ . Let the mass of each particle be “m” and speed of each particle is “v”. Then, the angular momentum of the individual particles about the origin of coordinate system are :

$$\begin{array}{l}{\ell }_1=m{y}_1\mathbf{j}\mathbf{x}{v}_1\mathbf{i}=-m{y}_{1}v\mathbf{k}\\ {\ell }_2=m{y}_2\mathbf{j}\mathbf{x}(-v_2)\mathbf{i}=m{y}_{2}v\mathbf{k}\end{array}$$

Since angular momentum are along the same direction, we can find the net angular momentum by arithmetic sum as :

$$\begin{array}{l}\ell ={\ell }_1+{\ell }_2=mv(y_2-y_1)\mathbf{k}\end{array}$$

We see that the magnitude of net angular momentum is directly proportional to the perpendicular distance between the parallel lines. Further, the perpendicular distance is fixed for a given pair of parallel paths. Hence, net angular momentum is constant for a given pair.

Hence, options (a) and (c) are correct.

This question can be evaluated using cross product in component form as :

ℓ = | i j k | | 0 1 1 || 2 -1 1 |

$$\begin{array}{l}\Rightarrow \mathbf{\ell }=(1x1-1x-1)\mathbf{i}+(1x2-0)\mathbf{j}+(0-2x1)\mathbf{k}\\ \Rightarrow \mathbf{\ell }=2\mathbf{i}+2\mathbf{j}-2\mathbf{k}\end{array}$$

Hence, option (b) is correct.

The velocity of the projectile at the highest point is equal to the horizontal component of velocity as vertical component of velocity is zero. Further, there is no acceleration in the horizontal direction. Therefore, horizontal component of velocity is same as the horizontal component of the velocity at projection. It means that velocity of projection at the highest point is :

Projectile motion

$$\begin{array}{l}{v}_x=v\cos \theta \end{array}$$

Here, the moment arm i.e. the perpendicular drawn from the point of projection on the extended line of velocity is equal to the height attained by the projectile. Thus,

$$\begin{array}{l}{r}_{\perp }=H=\frac{v^2{\sin }^2\theta }{2g}\end{array}$$

The angular momentum as required is :

$$\begin{array}{l}\Rightarrow \ell =m{r}_{\perp }v=\frac{m{v}^2{\sin }^2\theta }{2g}xv\cos \theta \\ \Rightarrow \ell =\frac{m{v}^3{\sin }^2\theta \cos \theta }{g}\end{array}$$

Hence, option (a) is correct.