Question

51. A thin rod AB of length / is kept vertical on
horizontal floor such that end A is in contact with
floor. If the rod is allowed to fall without slipping at
end A, then its angular velocity just before hitting
the ground is
1,root3gl
3)roptgl
(2) 2rootgl
(4)root2gl​

Answer 1

Answer:

(4) root2gl or √2gl

Explanation:

Let v be the velocity when it hits the ground.

Then

Loss in P.E= Gain in rotational K.E

mg(l/2)= 1/2Iω²=1/2ml²/3 (v/l)²

v= √2gl

Answer 2

Answer:

Option A

v = sqrt(3gL)

Explanation:

L = length of meterstick = 1 m

h = height of center of mass of meterstick = L/2 = 1/2 = 0.5 m

m = mass of meterstick

w = angular speed just before the stick hits the floor

I = moment of inertia of meterstick = mL²/3

Using conservation of energy

potential energy = rotational kinetic energy

mgh = (0.5) I w²

m g (L/2) = (0.5) ( mL²/3) (v/L)²

mgL = mv²/3

v = sqrt(3gL)