The position vector of a particle of mass 2 kg is given as a function of time by ~r = (6 m) ˆı + (3 m/s) t ˆ. Determine the magnitude of the angular momentum of the particle with respect to the origin at time 3 s. Answer in units of kg m2 / s.

Given that mass of particle is 2kg

M=2kg

Given a position vector as a function of time

r = (6•i+3t•j) m

Angular momentum at t=3sec

Angular momentum is given as the cross product of position vector and momentum

I = r*p

Where r is position vector

And p is momentum

Momentum is given as Mass*Velocity

p=mv

The velocity can he determine from r by differentiating r with respect to t

r = (6•i+3t•j) m

v=r'

v=dr/dt = (0i+3j) m/s

v=3•j m/s

Therefore, p=mv=2*3j=6j

p=6•j kgm/s

Now, applying this to angular momentum

I=r*p

I = (6•i+3t•j) * 6•j

Note i*i=j*j=0, i*j=k, j*i=-k

I=6•i*6•j + 3t•j * 6•j

I=36• (i*j) + 18t• (j*j)

I=36•k + 18t• (0)

I=36•k kgm²/s

The angular momentum is

36•k kgm²/s