- #1
jimgram
- 95
- 1
This is a real-world problem - NOT AN EXAM OR HOMEWORK (although it may sound as if it was produced by a deranged physics professor).
A three-flywheel system has two identical variable inertia flywheels, each connected to the inputs of a differential. The common output of the differential is connected to the 3rd flywheel. The initial conditions are:
1. Angular velocity of variable-inertia flywheel A, wa = 200 rad/sec
2. Angular velocity of variable-inertia flywheel B, wb = -200 rad/sec
3. Angular velocity of fixed-inertia flywheel C is from the standard equation for differentials:
wc = (wa+wb)/2, so initially, wc = 0
4. The inertia range for both flyweels A & B is: Ia, Ib = 0.4 m2*kg to 0.6 m2*kg
5. The initial setting for the inertia of flywheel A & B is: Ia, Ib = 0.5 m2*kg
6. The inertia of flywheel C is fixed: Ic = 5 m2*kg
The momentum for each flywheel is: L = I*w, so: La = 200*m2*kg/sec; Lb = -200*m2*kg/sec; Lc = 0
Ignoring friction of any sort, if the inertia of flywheel A is decreased from 0.5 m2*kg to 0.4 m2*kg, at the same time the inertia of flywheel B is increased from 0.5 m2*kg to 0.6 m2*kg, and this change occurs linearly over a period of 10 seconds, find:
1. The ending angular velocities for all three flywheels
2. The momentums of each flywheel
Owing to the work done in changing the inertia’s of wheels A & B, kinetic energy is not conserved in the flywheels.
The initial momentum can’t be zero, so I assume that the initial momentum is 400*m2*kg/s (the velocity relationship is wc = (wa-wb)/2). Therefore, the ending momentum must be 400*m2*kg/s divided into the three flywheels.
I am into my 2nd year working on this problem in one way or another. It certainly has a solution because it's actually being done. But I'm at a loss as how to model it. Anyone out there needing your brain cells stimulated? (Please don't come back with a simple solution - my ego couldn't stand it).
A three-flywheel system has two identical variable inertia flywheels, each connected to the inputs of a differential. The common output of the differential is connected to the 3rd flywheel. The initial conditions are:
1. Angular velocity of variable-inertia flywheel A, wa = 200 rad/sec
2. Angular velocity of variable-inertia flywheel B, wb = -200 rad/sec
3. Angular velocity of fixed-inertia flywheel C is from the standard equation for differentials:
wc = (wa+wb)/2, so initially, wc = 0
4. The inertia range for both flyweels A & B is: Ia, Ib = 0.4 m2*kg to 0.6 m2*kg
5. The initial setting for the inertia of flywheel A & B is: Ia, Ib = 0.5 m2*kg
6. The inertia of flywheel C is fixed: Ic = 5 m2*kg
The momentum for each flywheel is: L = I*w, so: La = 200*m2*kg/sec; Lb = -200*m2*kg/sec; Lc = 0
Ignoring friction of any sort, if the inertia of flywheel A is decreased from 0.5 m2*kg to 0.4 m2*kg, at the same time the inertia of flywheel B is increased from 0.5 m2*kg to 0.6 m2*kg, and this change occurs linearly over a period of 10 seconds, find:
1. The ending angular velocities for all three flywheels
2. The momentums of each flywheel
Owing to the work done in changing the inertia’s of wheels A & B, kinetic energy is not conserved in the flywheels.
The initial momentum can’t be zero, so I assume that the initial momentum is 400*m2*kg/s (the velocity relationship is wc = (wa-wb)/2). Therefore, the ending momentum must be 400*m2*kg/s divided into the three flywheels.
I am into my 2nd year working on this problem in one way or another. It certainly has a solution because it's actually being done. But I'm at a loss as how to model it. Anyone out there needing your brain cells stimulated? (Please don't come back with a simple solution - my ego couldn't stand it).