A small mass m is attached to a massless stri | vedclass

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Question

(C) The angular momentum of a particle about a point is given by $\vec L = \vec r 	imes \vec p = m(\vec r 	imes \vec v)$.For $\vec L_o$:The position

Vedclass · Accepted Answer

$\vec L_o$ remains constant while $\vec L_p$ varies with time.

Vedclass · Answer

(C) The angular momentum of a particle about a point is given by $\vec L = \vec r 	imes \vec p = m(\vec r 	imes \vec v)$.For $\vec L_o$:The position vector $\vec r$ is in the $x-y$ plane and the velocity $\vec v$ is tangential to the circular path. The cross product $\vec r 	imes \vec v$ points along the $z$-axis. Since the magnitude $|\vec L_o| = mvr \sin 90^{\circ} = m(R\omega)R = mR^2\omega$ is constant and the direction is fixed along the $z$-axis,$\vec L_o$ remains constant.For $\vec L_p$:The position vector $\vec r'$ from $P$ to the mass $m$ changes direction as the mass moves in a circle. The angular momentum $\vec L_p = \vec r' 	imes \vec p$ is perpendicular to both $\vec r'$ and $\vec v$. As the mass moves,the vector $\vec L_p$ traces a cone around the $z$-axis. Thus,while the magnitude $|\vec L_p| = mvr' \sin 	heta$ remains constant,the direction of $\vec L_p$ changes continuously with time.

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