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6.System of Particles and Rotational Motion
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A particle of mass $2\, kg$ is on a smooth horizontal table and moves in a circular path of radius $0.6\, m$. The height of the table from the ground is $0.8\, m$. If the angular speed of the particle is $12\, rad\, s^{-1}$, the magnitude of its angular momentum about a point on the ground right under the centre of the circle is ........ $kg\, m^2\,s^{-1}$
A
$14.4$
B
$8.64$
C
$20.16$
D
$11.52$
(JEE MAIN-2015)
Solution

Angular momentum,
$\begin{array}{l}
{L_0} = mvr\sin {90^ \circ }\\
= 2 \times 0.6 \times 12 \times 1 \times 1\\
\left[ {AS\,\,V = r\omega ,\,\sin \,{{90}^ \circ } = 1} \right]\\
So,\,{L_0} = 14.4\,kg{m^2}/s
\end{array}$
Standard 11
Physics
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