Rotational MotionHard
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.14.4 kg m2s-1
B.8.64 kg m2s-1
C.20.16 kg m2s-1
D.11.52 kg m2s-1
Solution
L0 = mvr sin 90°
= m(0.6ω)r
= 2 × 0.6 × 12 × 1
= 14.4 kgm2/s
= m(0.6ω)r
= 2 × 0.6 × 12 × 1
= 14.4 kgm2/s
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