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:

Step-by-step solution

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 &#16