A particle is confined to rotate in a circula | vedclass

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(B) The angular momentum is given by $\vec{L} = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$. Since the particle is moving in a circular path,t

Vedclass · Accepted Answer

Only the direction of angular momentum $\vec{L}$ is conserved.

Vedclass · Answer

(B) The angular momentum is given by $\vec{L} = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$. Since the particle is moving in a circular path,the position vector $\vec{r}$ and velocity vector $\vec{v}$ are always perpendicular. The magnitude of angular momentum is $L = mvr$. Since the linear speed $v$ is decreasing,the magnitude of angular momentum $L$ is not conserved.However,for a particle moving in a circular path in a plane,the direction of the angular momentum vector $\vec{L}$ is perpendicular to the plane of motion (determined by the right-hand rule). As long as the particle remains in the same circular path,the direction of $\vec{L}$ remains constant.Since the particle is confined to a circular path,it cannot spiral towards the centre. Because the speed is changing,there is a tangential acceleration $a_t$ in addition to the centripetal acceleration $a_c$. Therefore,the net acceleration is not directed towards the centre. Thus,only the direction of angular momentum is conserved.

$A$ particle is confined to rotate in a circular path with decreasing linear speed. Which of the following is correct?

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