The center of gravity drops by L( 1-cosθ)/6, and the moment of inertia of the rod relative to the rotating shaft is ML 2/9.
Mgl (1-cos θ)/6 = j ω 2/2, and the answer can be calculated as a.
2. Choose D, the farther the mass distribution is from the axis of rotation, the greater the moment of inertia, and vice versa. Of the three ABC companies, B is obviously the smallest.
Compared with D, I calculated it by integral, and the ratio of their moment of inertia is (a+3b)sinθ/(4b), where θ is the included angle between the B side and the axis in Figure D (it is not difficult to calculate, but it is very complicated, so it is difficult for me to type it into the computer), so it can be concluded that the moment of inertia in Figure B is the smallest.
If you use qualitative analysis, you can also imagine b>& gta, basically the moment of inertia ratio between them is 4: 3.