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Question: Consider a particle of mass m having linear momentum at position relative to the origin O. Let \(\ov...

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

A

dL→dt+r→×dp→dt=0\frac{d\overrightarrow{L}}{dt} + \overrightarrow{r} \times \frac{d\overrightarrow{p}}{dt} = 0

B

dL→dt+dp→dt×p→=0\frac{d\overrightarrow{L}}{dt} + \frac{d\overrightarrow{p}}{dt} \times \overrightarrow{p} = 0

C

dL→dt+dr→dt×p→=0\frac{d\overrightarrow{L}}{dt} + \frac{d\overrightarrow{r}}{dt} \times \overrightarrow{p} = 0

D

dL→dt−r→×dp→dt=0\frac{d\overrightarrow{L}}{dt} - \overrightarrow{r} \times \frac{d\overrightarrow{p}}{dt} = 0

Answer

dL→dt−r→×dp→dt=0\frac{d\overrightarrow{L}}{dt} - \overrightarrow{r} \times \frac{d\overrightarrow{p}}{dt} = 0

Explanation

Solution

As L→=r→×p→\overset{\rightarrow}{L} = \overset{\rightarrow}{r} \times \overset{\rightarrow}{p}

Differentiate both sides with respect to time, we get

dL→dt=ddt(r→×p→)\frac{d\overset{\rightarrow}{L}}{dt} = \frac{d}{dt}\left( \overset{\rightarrow}{r} \times \overset{\rightarrow}{p} \right)

=dr→dt×p→+r→×dp→dt= \frac{d\overrightarrow{r}}{dt} \times \overrightarrow{p} + \overrightarrow{r} \times \frac{d\overset{\rightarrow}{p}}{dt}

=r→×dp→dt(∴dr→dt×p→=0)= \overrightarrow{r} \times \frac{d\overset{\rightarrow}{p}}{dt}\left( \therefore\frac{d\overset{\rightarrow}{r}}{dt} \times \overset{\rightarrow}{p} = 0 \right)

dL→dt−r→×dp→dt=0\frac{d\overset{\rightarrow}{L}}{dt} - \overset{\rightarrow}{r} \times \frac{d\overset{\rightarrow}{p}}{dt} = 0