Question

Q15. A particle of mass m is moving with constant velocity v parallel to the x-axis as shown in the figure. Its angular momentum about origin O is

(1) mvb
(2) mva
(3) mv√(a²+b²)
(4) mb(a+b)​

Answer 1

Topic :-

Mechanics

Given :-

A figure which shows a particle of mass 'm' moving with constant velocity 'v' parallel to the x-axis.

To Find :-

Angular momentum of particle about origin O.

Concept :-

Angular Momentum

The moment of linear momentum of any particle about a point.

$\sf{\overrightarrow{L}=\overrightarrow{r}\times (m\overrightarrow{v})}$

$\sf {L=mvr\sin\theta}$

where

$\sf{L=Angular\:Momentum}$

$\sf{m=Mass \:of\:the\:particle}$

$\sf{v=Velocity\:of\:the\:particle}$

$\sf {r=Distance \:between\:point\:and\:particle}$

$\sf{\theta=Angle\:between\:point\:and \:particle\:from \:a\:fixed\:reference}$

Solution :-

Refer to attachment for basic idea of solution.

From Pythagoras Theorem,

$\sf{r^2=a^2+b^2}$

$\sf{r=\sqrt{a^2+b^2}}$

From trigonometry,

$\sf{\sin\theta=\dfrac{b}{r}}$

$\sf{\sin\theta=\dfrac{b}{\sqrt{a^2+b^2}}}$

$\sf{(\because r=\sqrt{a^2+b^2})}$

Now applying formula,

$\sf {L=mvr\sin\theta}$

$\sf {L=mv\sqrt{a^2+b^2}\cdot \dfrac{b}{\sqrt{a^2+b^2}}}$

$\sf{(\because r=\sqrt{a^2+b^2})}$

$\sf{\left(\because \sin\theta=\dfrac{b}{\sqrt{a^2+b^2}}\right)}$

Cancelling terms we get,

$\sf {L=mvb}$

Answer :-

Angular Momentum of the particle about origin O is mvb. Hence, option (1) is correct option.

Answer 2

Answer:

here is the answer.

L= mvb