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EXAMPLE: ROLLING CYLINDER WITH COUPLE

c) To determine how far the wheel moves, the translational and angular acceleration equation is used and then integrated twice:

ẍ = − .4(3.2) = − 1.28(m)/(s²) ẋ = − 1.28t(m)/(s) x = − (1.28)/(2)t² x = − 5.76m

Since the origin is at location where the motion starts and the wheel is initially at rest, there are no constants of integration (they are zero).

d) The velocity of the wheel is most easily determined using the principles of kinetic energy and work.

W = ΔKE = ⎲⎳(FΔx) = (1)/(2)mV²_c + (1)/(2)I_zω²

Recall that the angular velocity can be related to the translational velocity ω = (V_c)/(r)]. Substituting the values:

− 40(2ΔX_c) + 60(ΔX_c) = (1)/(2)(10)V²_c + (1)/(2)(.3)²(10)(V²_c)/((.4)²) − 80X_c + 60X_c = (1)/(2)(10)V²_c + (1)/(2)(.3)²(10)(V²_c)/((.4)²) 20(3) = 5V²_c + (.45)/(.4²)V²_c (6.0)/(7.81) = V²_c V_c = 2.77(m)/(s)