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21 June, 12:40

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.

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  1. 21 June, 14:43
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    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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