Question

# A particle of mass $2kg$ is on a smooth horizontal table and moves in a circular path of radius $0.6m$. The height of the table from the ground is $0.8m$ If the angular speed of the particle is $12rad{s}^{âˆ’1}$. The magnitude of its angular momentum about a point on the ground right under the center of the circle is

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Solution

## Step 1. Given Data,The mass of a particle $=2kg$The radius of the circular path of motion of the body, $R$$=0.6m$Height $=0.8m$The angular speed of the particle, $\mathrm{Ï‰}=12rad{s}^{âˆ’1}$Step 2. Formula Used,The angular momentum of an object having mass ($m$) and linear velocity ($v$) is: $|\stackrel{â†’}{L}|=m|\stackrel{â†’}{v}||\stackrel{â†’}{r}|\mathrm{sin}\left(\mathrm{Î¸}\right)$ $|\stackrel{â†’}{L}|$ is the magnitude of a vector quantity of the angular momentum of the body, $|\stackrel{â†’}{r}|$ is the magnitude of the radius of the circular path of motion of the body,$|\stackrel{â†’}{v}|$is the magnitude of another vector quantity, the velocity with which the body is travelling, $\mathrm{Î¸}$ is the angle between the radius vector and the velocity vector.$\stackrel{â†’}{L}$= $\stackrel{â†’}{r}Ã—\stackrel{â†’}{p}$$\stackrel{â†’}{p}=mv$$\stackrel{â†’}{L}$= Angular Momentum$m$ = mass of the particle$\stackrel{â†’}{p}$= linear momentum$v$ is the velocity of the particleFrom Pythagoras theorem,$H=\sqrt{{P}^{2}+{B}^{2}}$$H$ is the resultant radius vector, $B$ is the base, and $P$ is the height.Step 3. Calculating the magnitude of its angular momentum,We consider $H$as the resultant radius vector and we can denote it as ${r}_{perpendicular}$The perpendicular distance between the point and mass ${r}_{perpendicular}$ is given by,${r}_{perpendicular}=|\stackrel{â†’}{r}|=\sqrt{{\left(0.6\right)}^{2}+{\left(0.8\right)}^{2}}=1m$The angle between the resultant radius vector and the linear velocity vector is $\mathrm{Î¸}={90}^{âˆ˜}$. The angular momentum is,$|\stackrel{â†’}{L}|=m|\stackrel{â†’}{v}||\stackrel{â†’}{r}|\mathrm{sin}\left(\mathrm{Î¸}\right)$$|\stackrel{â†’}{L}|=m|\mathrm{Ï‰}R||\stackrel{â†’}{r}|\mathrm{sin}\left({90}^{o}\right)$$â‡’\left|\stackrel{â†’}{L}\right|=2Ã—12Ã—0.6Ã—1Ã—1=14.4kg{m}^{2}{s}^{âˆ’1}$Thus the magnitude of angular momentum, $L=14.4kg{m}^{2}{s}^{âˆ’1}$.Therefore, the magnitude of its angular momentum about a point on the ground right under the center of the circle is$14.4kg{m}^{2}{s}^{-1}$.

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