1.

A uniform rod of length 4m and mass 2sqrt(2)kg revolves with constant angular velocity omega what a vertical axis through a smooth joint A at one extermly of the rod so that it describes a cone of semi vertical angle 53^(@) as shown in the figure. Choose the correct statement (s). (g=10m//s^(2))

Answer»

The value of angular velocity `OMEGA` is `2.5rad//s`
The angle between HINGE reaction at point `A` and vertical axis `AC` is `45^(@)`
The magnitude of hinge reaction at point `A` is `40N`
The angular MOMENTUM of the ROD is constant.

Solution :`F=2xx(mu_(0)I_(1))/(2piR)xxI_(2)l=(2xx2xx10^(-7)xx50xx20xx0.50)/(0.25)=8xx10^(-4)N`


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