The angular momentum of a single particle mov | vedclass

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(B) The angular momentum $\vec{L}$ of a particle is given by $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.For a particle moving with

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remains same in magnitude and direction

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(B) The angular momentum $\vec{L}$ of a particle is given by $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.For a particle moving with constant speed $v$ in a circular path of radius $r$,the magnitude of angular momentum is $L = mvr \sin(90^{\circ}) = mvr$,which is constant.The direction of angular momentum is given by the right-hand rule,which is perpendicular to the plane of the circular path (along the axis of rotation).Since the speed and radius are constant and the plane of motion does not change,both the magnitude and the direction of the angular momentum vector remain constant throughout the motion.

The angular momentum of a single particle moving with constant speed along a circular path:

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