A particle of mass 2, kg is on a smooth horiz | vedclass

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(B) The angular momentum $\vec{L}$ of a particle about a point $O$ is given by $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.Here,the

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$8.64$

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(B) The angular momentum $\vec{L}$ of a particle about a point $O$ is given by $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.Here,the particle moves in a horizontal circle of radius $r = 0.6\, m$ at a height $h = 0.8\, m$ above the ground.Let the point on the ground directly below the center of the circle be $O'$. The position vector $\vec{r}$ of the particle relative to $O'$ has a horizontal component $r = 0.6\, m$ and a vertical component $h = 0.8\, m$.The velocity of the particle is $\vec{v}$,which is tangential to the circular path,so its magnitude is $v = r\omega = 0.6 	imes 12 = 7.2\, m/s$.The angular momentum about $O'$ is $L = |\vec{r} 	imes m\vec{v}| = mvr_{\perp}$,where $r_{\perp}$ is the perpendicular distance from the axis of rotation passing through $O'$ to the velocity vector. Since the velocity is horizontal and the point $O'$ is on the vertical axis,the perpendicular distance from the line of velocity to the point $O'$ is the radius $r = 0.6\, m$.Thus,$L = mvr = (2\, kg) 	imes (7.2\, m/s) 	imes (0.6\, m) = 8.64\, kg\, m^2\, s^{-1}$.

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