Question

A thin uniform rod AB of mass M and length L is hinged at one end A to the horizontal floor. Initially it stands vertically. It is allowed to fall freely on the floor in the vertical plane. The angular velocity of the rod when its end B strikes the floor is ____ 

(A) √(g/L) 

(B) √(2g / L) 

(C) √(3g / L) 

(D) 2√(g/L)

Answer 1

Correct option: (C) √(3g / L)

Explanation:

Loss in P.E = gain in Rotational K.E. centre of Mass of rod is at (L/2).

hence loss in P.E = Mg(L / 2)

Gain in R.E = (1/2)Iω² = (1/2)(ML² / 3)ω² 

hence

(MgL / 2) = (1/6)ML²ω² 

(g/2) = (Lω² / 6)

ω² = √(3g / L) hence ω = √(3g / L)

Comments

How come h = L/2 in mgh formula???