Angular Dynamics: 150Nm, 75rpm, 9 & 23 Revs. Determine Inertia & Friction

[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

 

[haruspex]

ω2²-ω1² / 2 * (9*2∏) = ang accel

Fuller use of parentheses would help: (ω2²-ω1²) / (2 * 9*2∏)

= 1963.5

That's the acceleration in the first phase, right? What about the slowing down phase?

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

OK, so flesh that out. Put in symbols for the two quantities to be determined and write out the torque equations using them.

 

[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?

 

[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.

 

[Nickie]

To determine the moment of inertia, we can use the equation τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. In this case, the torque is constant at 150Nm, and the angular acceleration can be calculated using the equation ω22-ω12 / 2 * (9*2∏) = ang accel = 1963.5. Therefore, we can rearrange the equation to solve for the moment of inertia, giving us I = τ / α = 150Nm / 1963.5 rad/s2 = 0.076 kgm2. This is the moment of inertia of the rotor.

To determine the bearing friction, we can use the fact that on decelerating, the inertia torque is equal to the bearing friction. Therefore, we can rearrange the equation to solve for the bearing friction, giving us bearing friction = 150Nm - Inertia Torque = 150Nm - 0.076 kgm2 * 1963.5 rad/s2 = 2.8 Nm. This is the bearing friction of the turbine rotor.

In summary, the moment of inertia of the turbine rotor is 0.076 kgm2 and the bearing friction is 2.8 Nm. These values are important in understanding the dynamics of the turbine rotor and can be used to make adjustments and improvements to its design and operation.