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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