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 center of the circle is:

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 center of the circle is:, 14 .4 kg m 2 s - 1, 11.52 kg m 2 s - 1, 20.16 kg m 2 s - 1, 8.64 kg m 2 s - 1

MARKS App · Accepted Answer

p → = m v j ^ = 2 × v j ^ = 2 × 0 .6 × 12 = 7 .2 × 2 = 14 .4 j ^ v = r ω r → = ( 0 .8 k ^ + 0 .6 i ^ ) ⇒ L → = r → × p → = 0 .6 i ^ + 0 .8 k ^ × 14 .4 j ^ = 0 .6 × 14 .4 k ^ - 0 .8 × 14 .4 i ^ = 14.4 0.6 k ^ - 0.8 i ^ ⇒ | L → | = 14.4 0.6 2 + 0.8 2 = 14.4 0.36 + 0.64 = 14.4 × 1 = 14.4 kg m 2 s - 1

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