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
remains constant for all positions of the particle
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Solution and Explanation
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
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Top Questions on System of Particles & Rotational Motion
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