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

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Homework Help Overview

The discussion revolves around a problem in angular dynamics, specifically focusing on determining the moment of inertia and bearing friction of a turbine rotor subjected to a constant torque. The rotor accelerates from rest to a speed of 75 rpm over 9 revolutions and continues to rotate for an additional 23 revolutions after the torque is removed.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equations governing angular acceleration and torque, questioning the application of values and the use of parentheses in calculations. There is an exploration of the relationship between torque, moment of inertia, and bearing friction during both acceleration and deceleration phases.

Discussion Status

The discussion is ongoing, with participants providing various equations and calculations. Some guidance has been offered regarding the arithmetic and assumptions made in the calculations, particularly concerning the interpretation of rotational speed. There is an acknowledgment of discrepancies in the results obtained for moment of inertia and bearing friction, prompting further inquiry into potential errors.

Contextual Notes

Participants are working under the constraints of a homework problem, which may impose specific requirements for the approach taken. There is a noted confusion regarding the interpretation of the rotational speed and its impact on the calculations.

MMCS
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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
 
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MMCS said:
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.
 
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?
 
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.
 

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