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24 April, 03:33

If the only force acting on the rock is its weight, what is the magnitude of the rate of change of its angular momentum at this instant?

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  1. 24 April, 05:57
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    The rate of chage of angular momentum = the net torque on the object.

    The net torque is the weight "mg" times the perpendicular distance from the line of action of the weight to the origin, x

    dL/dt = mgx = mgrCos (36.8)

    Or it is the component of the weight perpendicular to the position vector, mgCos (36.8), times the magnitude of the position vector, r;

    dL/dt = {mgCos (36.8}]r

    Obviously the same eq.

    Everything is given, so just plug in the values. The particle velocity is not needed to find rate of change of L.
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