JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 \(11.52\)
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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