The time dependence of the position of a particle of mass m=2
By Problem: The time dependence of the position of a particle of mass $m=2$ is given by $\vec{r}(t)=2t\,\hat\imath-3t^2\,\hat\jmath$. Its angular momentum, with respect to the origin, at time $t=2$ is (JEE Mains 2019)
- $48(\hat\imath+\hat\jmath)$
- $36\hat{k}$
- $-34(\hat{k}-\hat\imath)$
- $-48\hat{k}$
Solution: The velocity of the particle is given by \begin{align} \vec{v}(t)&=\mathrm{d}\vec{r}/\mathrm{d}t\\ &=2\,\hat\imath-6t\,\hat\jmath. \nonumber \end{align} The position and velocity of the particle at $t=2$ are \begin{align} \vec{r}&=2(2)\,\hat\imath-3(2)^2\,\hat\jmath\\ &=4\,\hat\imath-12\,\hat\jmath, \nonumber\\ \vec{v}&=2\,\hat\imath-6(2)\,\hat\jmath\\ &=2\,\hat\imath-12\,\hat\jmath. \nonumber \end{align} The angular momentum of the particle about the origin at time $t=2$ is given by \begin{align} \vec{L}&=m\vec{r}\times\vec{v}\\ &=2\, (4\,\hat\imath-12\,\hat\jmath)\times(2\,\hat\imath-12\,\hat\jmath)\\ &=-48\,\hat{k}.\nonumber \end{align}
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