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(D) The angular momentum $\vec{L}$ is defined as $\vec{L} = \vec{r} 	imes \vec{p}$.Differentiating both sides with respect to time $t$,we get:$\frac{

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$\frac{d\vec{L}}{dt} - \vec{r} 	imes \frac{d\vec{p}}{dt} = 0$

Vedclass · Answer

(D) The angular momentum $\vec{L}$ is defined as $\vec{L} = \vec{r} 	imes \vec{p}$.Differentiating both sides with respect to time $t$,we get:$\frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} 	imes \vec{p})$Using the product rule for cross products:$\frac{d\vec{L}}{dt} = \left(\frac{d\vec{r}}{dt} 	imes \vec{p}ight) + \left(\vec{r} 	imes \frac{d\vec{p}}{dt}ight)$Since velocity $\vec{v} = \frac{d\vec{r}}{dt}$ and linear momentum $\vec{p} = m\vec{v}$,the term $\frac{d\vec{r}}{dt} 	imes \vec{p}$ becomes $\vec{v} 	imes (m\vec{v}) = 0$ because the cross product of parallel vectors is zero.Therefore,$\frac{d\vec{L}}{dt} = 0 + \vec{r} 	imes \frac{d\vec{p}}{dt}$.Rearranging the terms,we get $\frac{d\vec{L}}{dt} - \vec{r} 	imes \frac{d\vec{p}}{dt} = 0$.

Consider a particle of mass $m$ having linear momentum $\vec{p}$ at position $\vec{r}$ relative to the origin $O$. Let $\vec{L}$ be the angular momentum of the particle with respect to the origin. Which of the following equations correctly relate $\vec{r}, \vec{p}$ and $\vec{L}$?

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