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 ........ \(kg\, m^2\,s^{-1}\)

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Answer & Solution

Correct Answer

(A) \(14.4\)

Step-by-step Solution

Detailed explanation

Angular momentum,

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

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