Question

Statement -1 —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 ends B strikes the floor √(3g / L) 

Statement -2 — The angular momentum of the rod about the hinge remains constant throughout its fall to the floor. 

(A) Statement - 1 is correct (true), Statement - 2 is true and Statement- 2 is correct explanation for Statement - 1 

(B) Statement -1 is true, statement -2 is true but statement-2 is not the correct explanation four statement -1. 

(C) Statement - 1 is true, statement-2 is false 

(D) Statement-2 is false, statement -2 is true

Answer 1

Correct option: (D) Statement-2 is false, statement -2 is true

Explanation:

Loss in PE = gain in KE (rotational)

As the centre of mass of the rod falls through the distance (L/2),

Loss in PE = mg (L/2)         (1)

gain in PE = (1/2) Iω² = (1/2)(mL² / 3)ω²    (2)

(1) = (2)

hence

[(mgl) / 2] = [(mL²ω²) / 6]

ω = √(3g / L)

And angular momentum of rod will change throughout its falls to the floor.