Answer :
Answer:
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
The magnitude of the angular momentum of the particle with respect to the origin is 36[tex]\hat k[/tex] kgm²/s and it is independent of time.
Angular momentum:
The mass of the particle is m = 2kg
The position vector of the particle, [tex]\vec r=[/tex] (6m)[tex]\hat i[/tex] + (3m/s)t[tex]\hat j[/tex]
The velocity of the particle will be:
[tex]\vec v=\frac{d\vec r}{dt}\\\\\vec v=\frac{d}{dt}[6\hat i+3t\hat j]\\\\[/tex]
v = 3[tex]\hat j[/tex] m/s, constant velocity in y-direction
So, the momentum of the particle is:
p = mv
p = 2×3[tex]\hat j[/tex] kgm/s
p = 6[tex]\hat j[/tex] kgm/s
Now the angular momentum is defined as:
L = r × p
L = [6[tex]\hat i[/tex] + 3t[tex]\hat j[/tex]]×6[tex]\hat j[/tex]
L = 36[tex]\hat k[/tex] kgm²/s
So, the angular momentum is independent of time.
Learn more about angular momentum:
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