Conceptual Problems 6.1 Figure P6.1 shows a flywheel with moment of inertia J 0.

Question

Conceptual Problems 6.1 Figure P6.1 shows a flywheel with moment of inertia J 0.5kg-m2 that is initially rotating at an angular velocity 60 40 rad/s. The flywheel is subjected to friction, which is modeled by linear viscous friction torque bé, with friction coefficient b 0.06 N-m-s/rad. Use Simulink to obtain the dynamic response and plot the angular position (t) (in rad) and angular velocity 0(1) (in rad/s). In addition, use the simulation to integrate the rate of energy dissipation and plot dissipated energy (in J) vs. time. Show that the total dissipated energy as computed by the simulation is equal to the system's initial energy. Flywheel, J Viscous friction, b Figure P6.1 Repeat Problern 6.1 with the addition of dry (Coulomb) friction torque. Try sgn()), where Td y = 0.1 N-m. Linear viscous friction torque (b = 0.06 N-m-s/rd) also acts on the flywheel. 6.2

Explanation / Answer

6.1

given moment of inertia of the flywheel J = 0.5 kgm^2

initial angular velocity thetao'= wo = 40 rad/s

viscous torque = b*theta' = bw

b = 0.06 N m s / rad

now, at time t, let the angular velocity be w

then

torque = J*dw/dt

-bw = J*dw/dt

-b*dt/J = dw/w

integrating

-bt/J = ln(w) + k

now, at t = 0, w = wo

hence

ln(wo) = -k

so, -bt/J = ln(w) - ln(wo)

so w = wo*e^(-bt/J)

so the angular velocity of the wheel as a function of time is given by

w = 40*e^(-0.06t/0.5)

w = 40e^(-0.12t)

now power being dissipated at this time

P = torque*w = -bw^2

energy dissipated in time dt

dE = -bw^2dt = -0.06*(40*e^(-0.12t))^2dt

dE = -0.06*(40*e^(-0.12t))^2dt = -96e^(-0.24t)dt

total energy dissipated = integrate dE from t = 0 to t = infinity

E = 96[e^(-0.24t)]/0.24 = 400[0 - 1] = -400 J

initial energy of disc = 0.5Jwo^2 = 0.5*0.5*40^2 = 400 J

hence we can see that energy dissipated is equal to the initial energy of the system

6.2

if we add a constant dissipative torque of Tdry = 0.1 Nm

then

-(b(w + 0.1/b)) = jdw/dt

-b*dt/J = dw/(w + 0.1/b)

integrating

-bt/J = ln(w + 0.1/b) + k

at t= 0, w = wo

k = -ln(wo + 0.1/b)

so, -bt/J = ln((w + 0.1/b)/(wo + 0.1/b))

w + 0.1/b = (wo + 0.1/b)e^(-bt/J)

w = (wo + 0.1/b)e^(-bt/J) - 0.1/b

w = (40 + 0.1/0.06)e^(-0.06t/0.5) - 0.1/0.06

w = 41.667e^(-0.12t) - 1.6667

dE = -(bw + 0.1)wdt = (-bw^2 + 0.1w)dt

dE = (-0.06(41.667e^(-0.12t) - 1.6667)^2 + 0.1(41.667e^(-0.12t) - 1.6667))dt

integrating

E = -400 J

hence

energy dissipated = initial energy

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