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

d L → d t - d r → d t × p → = 0

d L → d t - d r → d t × p → = 0

d L → d t + r → × d p → d t = 0

d L → d t + d r → d t × p → = 0

d L → d t - r → × d p → d t = 0

d L → d t - d r → d t × p → = 0

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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 $\vec{r}, \vec{p} a n d \vec{L}$ ?

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$\frac{d \vec{L}}{d t} + \vec{r} \times \frac{d \vec{p}}{d t} = 0$

$\frac{d \vec{L}}{d t} + \frac{d \vec{r}}{d t} \times \vec{p} = 0$

$\frac{d \vec{L}}{d t} - \vec{r} \times \frac{d \vec{p}}{d t} = 0$

$\frac{d \vec{L}}{d t} - \frac{d \vec{r}}{d t} \times \vec{p} = 0$

answer is D.

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As $\vec{L} = \vec{r} \times \vec{p}$

Differentiating both sides with respect to time, we get

$\frac{d \vec{L}}{d t} = \frac{d}{d t} (\vec{r} \times \vec{p}) = \frac{d \vec{r}}{d t} \times \vec{p} + \vec{r} \times \frac{d \vec{p}}{d t}$

$(∵ \frac{d \vec{r}}{d t} \times \vec{p} = 0)$

$= \vec{r} \times \frac{d \vec{p}}{d t} \frac{d \vec{L}}{d t} - \vec{r} \times \frac{d \vec{p}}{d t} = 0$

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