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 $12rad{s}^{-1}$. 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$$=0.6m$Height $=0.8m$The angular speed of the particle, $\omega =12rad{s}^{-1}$Step 2. Formula Used,The angular momentum of an object having mass ($m$) and linear velocity ($v$) is: $|\stackrel{\to }{L}|=m|\stackrel{\to }{v}||\stackrel{\to }{r}|\mathrm{sin}\left(\theta \right)$ $|\stackrel{\to }{L}|$ is the magnitude of a vector quantity of the angular momentum of the body, $|\stackrel{\to }{r}|$ is the magnitude of the radius of the circular path of motion of the body,$|\stackrel{\to }{v}|$is the magnitude of another vector quantity, the velocity with which the body is travelling, $\theta$ is the angle between the radius vector and the velocity vector.$\stackrel{\to }{L}$= $\stackrel{\to }{r}×\stackrel{\to }{p}$$\stackrel{\to }{p}=mv$$\stackrel{\to }{L}$= Angular Momentum$m$ = mass of the particle$\stackrel{\to }{p}$= linear momentum$v$ is the velocity of the particleFrom Pythagoras theorem,$H=\sqrt{{P}^{2}+{B}^{2}}$$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,${r}_{perpendicular}=|\stackrel{\to }{r}|=\sqrt{{\left(0.6\right)}^{2}+{\left(0.8\right)}^{2}}=1m$The angle between the resultant radius vector and the linear velocity vector is $\theta ={90}^{\circ }$. The angular momentum is,$|\stackrel{\to }{L}|=m|\stackrel{\to }{v}||\stackrel{\to }{r}|\mathrm{sin}\left(\theta \right)$$|\stackrel{\to }{L}|=m|\omega R||\stackrel{\to }{r}|\mathrm{sin}\left({90}^{o}\right)$$⇒\left|\stackrel{\to }{L}\right|=2×12×0.6×1×1=14.4kg{m}^{2}{s}^{-1}$Thus the magnitude of angular momentum, $L=14.4kg{m}^{2}{s}^{-1}$.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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