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Angular momentum question - Strange silly mistake

  1. Feb 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Q] A particle of mass m is moving in a circle of radius r. The centripetal acceleration([itex]a_c = Kt^2[/itex]), where K is a positive contant and t is time. The magnitude of the time rate of change of angular momentum of the particle about the centre of the circle is:

    [itex]\textrm{B]}~\sqrt{m^2 Kr^3}[/itex]

    2. Relevant equations

    all concerned with rotational mechanics.

    3. The attempt at a solution

    The [itex]\textrm{Key}~\textrm{Idea}[/itex] here is that despite the variable centripetal force, the radius of rotation remains constant. Which means, the net torque on the particle needs to vary with time. At any time, 't' the centripetal force is related to the angular velocity as:

    \textrm{F}_\textrm{c} = \textrm{Kt}^2 = \textrm{mr}\omega^2


    \omega = \sqrt{\frac{\textrm{Kt}^2}{\textrm{mr}}} = \textrm{t}\sqrt{\frac{\textrm{k}}{\textrm{mr}}}

    Now, the moment of inertia of the particle is given by: [itex]\textrm{I} = \textrm{mr}^2[/itex]

    Hence, the angular momentum at time 't' is given by:

    \textrm{l} = \textrm{I}\omega = \textrm{mr}^2 \textrm{t}\sqrt{\frac{\textrm{k}}{\textrm{mr}}} = \textrm{t}\sqrt{mKr^3}

    To find the rate of change w.r.t time, we have:

    \frac{\textrm{dl}}{\textrm{dt}} = \sqrt{mKr^3}

    But it ain't one of the options.. i know i'm making a silly mistake somewhere.. just can't catch it..
  2. jcsd
  3. Feb 12, 2008 #2
    alpha=a_c / r = (Kt^2) /r


    put alpha from first eqn. into second one and then integrate.
  4. Feb 12, 2008 #3
    I don't think that's right. [itex]\alpha = \frac{a}{r}[/itex], where for a particle in rotation, [itex]\alpha[/itex] is the angular acceleration, 'r' is the radius of rotation and 'a' is the linear acceleration (i.e. the tangential acceleration) of the paritcle and not the centripetal acceleration.
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