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(A) The angular momentum $L$ of a particle about a point $O$ is given by $L = m(\vec{r} 	imes \vec{v})$. For a particle of mass $m$ moving with speed

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$va(1 + \cos 2	heta)$

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(A) The angular momentum $L$ of a particle about a point $O$ is given by $L = m(\vec{r} 	imes \vec{v})$. For a particle of mass $m$ moving with speed $v$,the magnitude is $L = mvr_{\perp}$,where $r_{\perp}$ is the perpendicular distance from the origin $O$ to the line of velocity.From the geometry of the circle,the position vector $\vec{r}$ makes an angle $	heta$ with the $x$-axis. The velocity vector $\vec{v}$ is tangent to the circle at the particle's position. The angle between the position vector $\vec{r}$ and the velocity vector $\vec{v}$ is $(90^\circ + 	heta)$.The distance of the particle from the origin $O$ is $r = 2a \cos 	heta$. The angular momentum $L = mvr \sin(90^\circ + 	heta) = mvr \cos 	heta$.Substituting $r = 2a \cos 	heta$,we get $L = mv(2a \cos 	heta) \cos 	heta = 2mva \cos^2 	heta$.Using the identity $2 \cos^2 	heta = 1 + \cos 2	heta$,we get $L = mva(1 + \cos 2	heta)$.Assuming unit mass $(m=1)$,the angular momentum is $va(1 + \cos 2	heta)$.

$A$ particle is moving in a circular path of radius $a$ with a constant speed $v$ as shown in the figure. The centre of the circle is marked by $C$. The angular momentum about the origin $O$ can be written as:

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