Physics question

Question

Please help me with this physics question.  Please show steps so I can better understand.  Particles 1 and 2, each of mass m, fixed to the ends of a rigid massless rod of length L1+L2 with L1=20cm and L2=80cm.  The rod is held horizontally on the fulcrum and then released.  What are the magnitudes of the initial accelerations of (a) particle 1 (b) particle 2?

Answer 1

The rod will rotate clockwise because the torque is greater on the right side. Calling this rotation positive, the angular acceleration "α" is related to the net torque as;
T(net) = I(α)

Initially the rod is horizontal and the net torque due to the vertical weights is
T(net) = mgL₁ - mgL₂

The moment of inertia of the system is;
I = mL₁² + mL₂²

So the angular acceleration, which is the same for both masses is;
(α) = T(net)/I

= g(L₁ - L₂)/(L₁² + L₂²)

The Linear accelerations are then; a = radius*α
a₁ = L₁(α) = gL₁(L₁-L₂)/(L₁² + L₂²) , down
a₂ = L₂(α) = gL₂(L₁-L₂)/(L₁² + L₂²) , up

use g=980 cm/s² to get the acceleration's in cm/s². Or use g=9.8 m/s² and change lengths to meters to get the acceleration's in m/s².