Rotational Motion Question 171
Question: 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
Options:
A) $14.4kgm^{2}{{s}^{-1}}$
B) $8.64kgm^{2}{{s}^{-1}}$
C) $20.16kgm^{2}{{s}^{-1}}$
D) $11.52kgm^{2}{{s}^{-1}}$
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Angular momentum,
$L _{0}=mvr\sin 90^{o}$
$=2\times 0.6\times 12\times 1\times 1$
$
$AsV=r\omega ,Sin90^{o}=1$
$
$So,L _{0}=14.4kgm^{2}/s$
