Problem 2 Bar AB is supported with a pin joint (frictionless) at A. The bar has

Question

Problem 2 Bar AB is supported with a pin joint (frictionless) at A. The bar has mass m= 5kg. A constant moment of M=20Nm is applied on the bar. It is know that the center of mass of the bar is at point G and that it has a radius of gyration about point G of kG=0.206m. Using the force mass acceleration approach, a) Determine the expression for the angular acceleration as a function of angle   . b) If the bar is initially at rest when   0, determine the angular velocity when ?/2 c) Determine the expression for the magnitude of the joint reactions at joint A as a function of angle  

Explanation / Answer

a)Moment of inertia about centre of mass, Icm = mk^2

Icm = 5 x 0.206^2 = 0.212 kg m^2

I about pivot point = Icm + md^2 = 0.212 + (5 x (0.6 - 0.2)^2)

I = 1.012 kg m^2

Applying net torque = I x alpha

20 + (0.4 x 5 x 9.8 x cos@) = 1.012 (alpha)

alpha =19.76 + 19.37cos@ ...............Ans

b) alpha = w dw/d@

w dw = (19.76 + 19.37cos@)d@

w^2 = 19.76@ + 19.37sin@

w is from 0 to w and @ is from 0 to pi/2.

w^2 - 0 = 19.76(pi/2) + 19.37 * 1

w = 7.1 rad/s

c) at angle @ in horizontal,

Fx = m w^2r cos@

Fx = m ( 19.76@ + 19.37sin@ ) (0.6 - 0.2)cos@

Fx = 39.52@cos@ + 38.74sin@cos@

Fy = mg + mw^2rsin@

      = (5 x 9.8) + (5)(( 19.76@ + 19.37sin@ ) (0.6 - 0.2)sin@

    = 49 + 39.52@cos@ + 38.74sin@cos@

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