A Particle Of Mass 2 Kg Is On A Smooth Horizontal Table And Moves In A Circular Path Of Radius 0.6m. The Height Of The Table From The Ground Is 0.8m If The Angular Speed Of The Particle Is 12rads −1, The Magnitude Of Its Angular Momentum About A Point On The Ground Right Under The Centre Of The Circle Is:

Sol

Given:

Mass = 2 kg

$$w=12rad^{-1}$$

Distance (perpendicular) between point and mass, =

$$r_{\lambda } = \sqrt{(0.6)^{2} + (0.8)^{2}} = 1m$$

Angular momentum about the point on the ground under the centre of table = mvr

= $$m (wr)\eta$$

= 2 * 12 * 0.6 * 1

= 24 * 0.6 kg $$m^{2} s^{-1}$$

= 14.4 kg $$m^{2} s^{-1}$$

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