A thin uniform rod (X) of mass M and length L is pivoted at a height ( L3 ) as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is . (g = gravitational acceleration) [figure]
- 32 gL
- 3gL
- 3 2 gL
- 1 2 gL
Answer
(2) 3gL
Solution
The rod rotates about the pivot, which is at L3 from its lower end.
Centre of mass is at L2 from the lower end distance from pivot d = L2 - L3 = L6
Initially the centre of mass is L6 above the pivot; on the table it is at the level of the pivot.
Fall of centre of mass h = L6
Moment of inertia about the pivot (parallel axis theorem):
I = ML212 + M ( L6 )2 = 3ML2 + ML236 = ML29
Energy conservation: 12I 2 = Mgh
12 ML29 2 = Mg L6
2 = 3gL
= 3gL
Answer: (2)