Question

The thin bar, hinged as shown is released from rest. There is no friction in the hinge.

L1=0.31m and L2=0.75m. g=9.8m/s ².

Hint: The mass is not required. Carry the mass of the bar through as a variable and it should cancel.
What is the angular acceleration of the bar at the instant shown?

Answer 1

Here is the Diagram for the question:

Answer 2

i would work the torque around the fulcrum, and relate that to angular acceleration via \(\tau = I \alpha \)

assuming constant density, the centre of mass of rod from fulcrum is \(\bar x = \dfrac{L_1 + L_2}{2} - L_1 = \dfrac{L_2-L_1}{2}\)

\(\tau = mg \bar x\)

then parallel axis theorem for I: \(\large I_{fulcrum} = I_{CoM} + m \bar x^2\)

and then put that all together, with some messy algebra.....