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Q. A thin uniform rod ''AB'' of mass m and length L is hinged at one end A to the level floor. Initially it stands vertically and is allowed to fall freely to the floor in the vertical plane. The angular velocity the rod, when its end B strikes the floor is (g is acceleration to gravity)

Solution:

The velocity with which the point 'B' strikes the ground is $V = \sqrt{3rg}$
but $\nu = r \omega$
$\therefore $ $r^2 \, \omega^2 = 3rg \, \Rightarrow \, \omega = \sqrt{\frac{3g}{r}} = \sqrt{\frac{3g}{L}}$