A particle moves in a circular path with decr | vedclass

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(D) $1$. In circular motion,the angular momentum is given by $\vec{L} = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.$2$. For a particle moving

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The direction of angular momentum remains constant.

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(D) $1$. In circular motion,the angular momentum is given by $\vec{L} = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.$2$. For a particle moving in a circular path,the position vector $\vec{r}$ and the velocity vector $\vec{v}$ lie in the plane of the circle.$3$. The angular momentum vector $\vec{L}$ is perpendicular to the plane of the circle (along the axis of rotation) according to the right-hand rule.$4$. Even if the speed $v$ is decreasing,the direction of the axis of rotation remains fixed perpendicular to the plane of the circular path.$5$. Therefore,the direction of the angular momentum vector remains constant,although its magnitude changes as the speed changes.

$A$ particle moves in a circular path with decreasing speed. Determine the correct statement.

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