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A rod of mass `m` and length `l` is rotating about a fixed point in the ceiling with an angular velocity `omega` as shown in the figure. The rod maintains a constant angle `theta` with the vertical. What is the rate of change of angular momentum of the rod ? A. `(momega^(2)l^(2)sintheta)/(6)`B. `(momega^(2)l^(2)sin2theta)/(6)`C. `(momega^(2)l^(2)sin^(2)theta)/(6)`D. `(momega^(2)l^(2)cos^(2)theta)/(6)` |
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Answer» The vertical component of the angular momentum remains constant while its horizontal component keeps changing the direction. `dL=(momegalsin(2theta))/(6)xxomegadtrArr(dL)/(dt)=(momega^(2)l^(2))/(6)sin(20)` |
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