Homework Help: Angular dynamics

1. Apr 9, 2013

MMCS

A constant torque of 150Nm applied to a turbine rotor is sufficient to overcome the constant bearing friction and to give it a speed of 75rpm from rest after 9 revolutions. When the torque is removed, the rotor turns for a further 23 revolutions before stopping. Determine the moment of inertia of the rotor and the bearing friction

Equations

ω22-ω12 / 2 * (9*2∏) = ang accel = 1963.5

On accelerating
150Nm - Inertia Torque - Bearing friction = 0

On decelerating
Inertia Torque = bearing friction

2. Apr 9, 2013

haruspex

Fuller use of parentheses would help: (ω22-ω12) / (2 * 9*2∏)
That's the acceleration in the first phase, right? What about the slowing down phase?
OK, so flesh that out. Put in symbols for the two quantities to be determined and write out the torque equations using them.

3. Apr 25, 2013

MMCS

Ok so, decelleration

471.22/(2*(23*2∏)) = -768.3

So two formulas

accelerating
150 - bearing friction - (1963.13 * Moment of inertia) = 0

decelerating
bearing friction - (768.3 * Moment of inertia) = 0

Combined

150 - (768.3*moment in inertia)-(1963.13*moment of inertia)=0
150-2731.4*moment of inertia = 0
moment of inertia = 0.0549
This is incorrect as i have the answer to be 198kgm2

however, if i use my value of 0.0549 to find bearing friction i get

bearing friction - 768.3 * moment of inertia = 0
bearing friction - 768.3 * 0.0549 = 0
bearing friction = 42.2, which i have to be the correct answer
it seems odd that solving them simultaneously would give me one correct answer and one incorrect, have i made a mistake?

4. Apr 25, 2013

haruspex

I didn't check the details of your arithmetic before. You seem to have used 75rpm as though it's revs per second. Looks like that error was self-cancelling in calculating the bearing friction but not in calculating the MI.