Question

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)

• A. mg/L
• B. $(mg/3L)^{\frac{1}{2}}$
• C. g/L
• D. $(3g/L)^{\frac{1}{2}}$

Answer 1

Answer: (D)

$(3g/L)^{\frac{1}{2}}$

Answer 2

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