Question

A particle of mass $m$ is moving with a constant velocity along a line parallel to the positive direction of $X$ -axis. The magnitude of its angular momentum with respect to the origin

• A. is zero
• B. goes on increasing as $x$ increases
• C. goes on decreasing as $x$ increases
• D. remains constant for all positions of the particle

Answer 1

Answer: (D)

According to the question,
Given, mass of the particle $=m$

$\therefore$ Linear momentum of the particle, $p=mV$
or $=mV\hat{i}$
Position of particle at time $t$, $r=x\hat{i}+y\hat{J}$
$=Vt\hat{i}+y\hat{J}\left(\because x=Vt\right)$
$\therefore$ Angular momentum of the particle about $O$
$L=r\times p=\left[Vt\hat{i}+y\hat{J}\right]\times mV\hat{i}$
$L=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ vt& y& 0\\ v& 0& 0\end{array}\right|$
$L=\hat{i}\left(0-0\right)-\hat{j}\left(0-0\right)+\hat{k}\left(0-vy\right)$
$=-Vy\hat{k}$ (constant)
$(\because$ Particle is moving in +ve $x$-direction )
Hence, angular momentvun of particle w.r.t the origin remains constant for all positions of the particle