Question

a particle starts from the point (0,8) metre and moves with uniform velocity of v= 3i^m/s. What is the angular momentum of the particle after 5s about origin (mass of particle is 1 kg)

Answer 1

Particle starts from (0m , 8m) from rest with uniform velocity of 3 m/s in the x-direction so after 5 seconds the distance a particle will traverse along the x-axis is 3*5= 15m

so new position of the particle is ​(15m , 8m).

So the distance of this new position to the origin is  

\square-root(15- 0)2 + (8 - 0)2

= 17

Angular velocity about origin is given by

W = v/r

= 3/17 rad/sec

Answer 2

Goven

Initial posion vector 0i + 8j

Velocity acts only along the x axis i.e

V=3i

Hence the angular momentum remains constant and

L=R(perpendicular)*(cross) mv

= 8j * 3i

= -24k. (i*j=k , j*i= -k )

Hope it helps