MHT CET202313 May 2023Evening ShiftPhysicsMotion in Two DimensionsActual
A thin uniform rod A B of mass ' m ' and length ' l ' is hinged at one end A to the ground level. Initially the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its end B strikes the ground is ( g = acceleration due to gravity)
Options
- Ag l
- Bmg l
- C3 ~g l
- Dmg 3 l
Correct answer
C. 3 ~g l
Step-by-step solution
We know, Loss in P.E = Gain in rotational K.E. When the centre of mass of the rod falls through a distance L 2 , Loss in P.E =m g L 2 ...(i) and, Gain in Rotational K.E = 1 2 I ^2= 1 2 [ ML ^2 3 ] ^2 ...(ii) Equating (i) and (ii) aligned & MgL 2 = 1 2 ML ^2 3 ^2 & ^2= 3 ~g ~L & = 3 ~g ~L = 3 ~g l ( here L =l) aligned ( here L =l)