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Determine the pin velocity of a falling two-rod composite system with a given initial roller velocity. The uniform rods AB and BC weigh 2.7 kg and 4.6 kg, respectively, and the small wheel at Cis of negligible mass. Knowing that in the position shown the velocity of wheel C is 3.5 m/s to the right, determine the velocity of pin B after rod AB has rotated through 90°

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The velocity of pin B after rod AB has rotated through 90° is  2.87 m/s.

Draw a free-body diagram of the composite system.

Label all knowns and unknowns.

Knowns: mA = 2.7 kg, mB = 4.6 kg, vC = 3.5 m/s

Unknowns: vB (velocity of pin B)

Calculate the angular velocity (ω) of rod AB.

ω = vC/r

Where r is the radius of wheel C.

Let r = 1 m.

ω = 3.5 m/s / 1 m = 3.5 rad/s.

Calculate the moment of inertia (I) of the composite system.

I = IAB + IBC

Where IAB is the moment of inertia of rod AB and IBC is the moment of inertia of rod BC.

Let lA = 0.5 m, lB = 0.6 m.

IAB = (1/3)mA*lA^2

= (1/3) * 2.7 kg * (0.5 m)^2

= 0.45 kg m^2

IBC = (1/3)mB*lB^2

= (1/3) * 4.6 kg * (0.6 m)^2

= 1.44 kg m^2

I = 0.45 kg m^2 + 1.44 kg m^2

= 1.89 kg m^2

Calculate the angular momentum (L) of the composite system.

L = I * ω

L = 1.89 kg m^2 * 3.5 rad/s= 6.615 kg m^2/s

Calculate the linear momentum of pin B.

Let x = 0.5 m

P = L/x = 6.615 kg m^2/s / 0.5 m = 13.23 kg m/s

Calculate the velocity of pin B.

vB = P/mB = 13.23 kg m/s / 4.6 kg = 2.87 m/s

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