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.

Explanation:

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

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