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Second Law of Motion

Consider a pa...

Question

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

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is D $\frac{d \to L}{d t} - \to r \times \frac{\to d p}{d t} = 0$
$\begin{matrix} \to L = \to r \times \to F \Rightarrow \frac{d \to L}{d t} = \to r \times \frac{d \to P}{d t} \Rightarrow \frac{d \to L}{d t} - \to r \times \frac{d \to P}{d t} = 0 \end{matrix}$
Hence,
option $(D)$ is correct answer.

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