A particle of mass $m$ is moving with constant velocity $v$ parallel to $x$ -axis in $x - y$ plane as shown in figure. Calculate its angular momentum with respect to origin at anyinstant ' $t$ '.

mrv

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mvb

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$\frac{1}{2} m v b$

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2mvb

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Solution

The correct option is B
mvb

$\to L$ = $\to r \times \to p$

$| \to L |$ = $(r s i n θ) (m v)$ ${| \to A \times \to B |}$

= $m v b$ {perpendicular to $\to v$ and $\to r$ }

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A particle of mass m is moving with constant velocity v parallel to x-axis in x−y plane as shown in figure. Calculate its angular momentum with respect to origin at anyinstant 't'.

The correct option is B mvb →L = →r × →p |→L| = (r sin θ)(mv) {|→A×→B|} = mvb{perpendicular to →v and →r}

A particle of mass $m$ is moving with constant velocity $v$ parallel to $x$ -axis in $x - y$ plane as shown in figure. Calculate its angular momentum with respect to origin at anyinstant ' $t$ '.

The correct option is B
mvb

$\to L$ = $\to r \times \to p$

$| \to L |$ = $(r s i n θ) (m v)$ ${| \to A \times \to B |}$

= $m v b$ {perpendicular to $\to v$ and $\to r$ }