A particle of mass 2   kg is on a smooth horizontal table and moves in a circular path of radius 0 . 6…

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:

  1. 14.4 kg m2 s-1
  2. 11.52 kg m2 s-1
  3. 20.16 kg m2 s-1
  4. 8.64 kg m2 s-1

Solution



p→=mv j^=2×v j^=2×0.6×12=7.2×2=14.4j^     v=rω

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^

=14.40.6 k^-0.8 i^

⇒|L→|=14.4 0.62+0.82

=14.40.36+0.64

=14.4×1

=14.4 kg m2 s-1

Asked in: JEE Main 2015 (11 Apr Online)

Practice more Rotational Motion questions on Aicharya