1.

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 ?

Answer»

`(momega^(2)L^(2)sintheta)/(6)`
`(momega^(2)l^(2)sin2theta)/(6)`
`(momega^(2)l^(2)sin^(2)theta)/(6)`
`(momega^(2)l^(2)cos^(2)theta)/(6)`

Solution :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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