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} $
$L=\left|\begin{matrix}\hat{i}&\hat{j}&\hat{k}\\ vt&y&0\\ v&0&0\end{matrix}\right|$
$L=\hat{i}\left(0-0\right)-\hat{j} \left(0-0\right)+\hat{k} \left(0-vy\right)$
$=-Vy\hat{k}$ (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