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 L3 from its lower end, while its centre of mass is L2 from the lower end, so the distance from the pivot is d = L2 - L3 = L6.
Initially the centre of mass is L6 above the pivot and on the table it is at the level of the pivot, so it falls through h = L6.
By the parallel axis theorem, the moment of inertia about the pivot is
I = ML212 + M ( L6 )2 = 3ML2 + ML236 = ML29
Energy conservation, 12I 2 = Mgh, gives
12 ML29 2 = Mg L6
2 = 3gL
= 3gL
Answer: (2)