Question 228 of 280

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:

A$14.4kg{m}^{2}{s}^{-1}$B$11.52kg{m}^{2}{s}^{-1}$C$20.16kg{m}^{2}{s}^{-1}$D$8.64kg{m}^{2}{s}^{-1}$

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Answer: (A) $14.4kg{m}^{2}{s}^{-1}$

$\vec{p}=mv\hat{j}=2\times v\hat{j}=2\times 0.6\times 12=7.2\times 2=14.4\hat{j}\left(v=r\omega \right)$

$\vec{r}=(0.8\hat{k}+0.6\hat{i})$$\Rightarrow \vec{L}=\vec{r}\times \vec{p}$

$=\left(0.6\hat{i}+0.8\hat{k}\right)\times 14.4\hat{j}$

$=0.6\times 14.4\hat{k}-0.8\times 14.4\hat{i}$

$=14.4\left(0.6\hat{k}-0.8\hat{i}\right)$

$\Rightarrow |\vec{L}|$$=14.4\sqrt{{\left(0.6\right)}^{2}+{\left(0.8\right)}^{2}}$

$=14.4\sqrt{0.36+0.64}$

$=14.4\times 1$

$=14.4kg{m}^{2}{s}^{-1}$