Each of the two uniform slender bars OA and BC has a mass of 8 kg. The bars are

Question

Each of the two uniform slender bars OA and BC has a mass of 8 kg. The bars are welded at A to form a T-shaped member and are rotating freely about a horizontal axis through O. If the pendulum is released from rest in the shown position where OA is horizontal, determine when OA is 45 ? from the horizontal, the angular velocity of the frame, the angular acceleration of the frame and the reaction force at the pivot O.

Each of the two uniform slender bars OA and BC has a mass of 8 kg. The bars are welded at A to form a T-shaped member and are rotating freely about a horizontal axis through O. If the pendulum is released from rest in the shown position where OA is horizontal, determine when OA is 45 ? from the horizontal, the angular velocity of the frame, the angular acceleration of the frame and the reaction force at the pivot O.

Explanation / Answer

MOI of OA about O is (ml^2)/3.

MOI of BC about A is (ml^2)/12.

By parallel axis theorem, MOI of BC about O is (ml^12)/12 + ml^2 = 13/12*ml^2

Total inertia about O, I = (ml^2)/3 + (13/12)*ml^2 = (17/12)*ml^2 = (17/12)*8*0.5^2 = 2.83 kg-m^2

Distance of CG of assembly from O, r = [(0.5/2)*8 + 0.5*8]/(8+8) = 0.375 m

After rotation, Torque T = [(2m)*g]*rCos = (2*8)*9.81*0.375 = 58.86*Cos Nm

T = I

Angular acceleration = '' = (58.86*Cos )/2.83 = 20.8*Cos

At = 45 deg, = 20.8*Cos 45 = 14.707 rad/s^2

= d/dt = (d/d)(d/dt) = (d/d)

(d/d) = 20.8*Cos

d = 20.8*Cos d

Integrating both the sides, ^2/2 = 20.8*Sin + c

Initial condition, at = 0, we have = 0. This gives c = 0.

Therefore, ^2/2 = 20.8*Sin

At = 45 deg, we get = (2*20.8*Sin 45) = 5.424 rad/s

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