Calculating inertial moment of a disk

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    Disk Inertial Moment
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Discussion Overview

The discussion revolves around calculating the moment of inertia (MI) of a disk system, particularly when additional loads are placed on it. Participants explore the effects of mass, shape, and radius on the moment of inertia, as well as implications for rotational motion under constant torque.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning
  • Experimental/applied

Main Points Raised

  • One participant questions whether the moment of inertia of a disk with an added load can be simplified to (m + 1kg)r²/2.
  • Another participant asserts that the moment of inertia of the added mass depends on its shape and position, suggesting a more complex calculation for the total moment of inertia.
  • There is a claim that changing the radius of the disk, while keeping mass constant, will affect the moment of inertia.
  • A participant references external resources for understanding the superposition of moments of inertia for irregular shapes.
  • Questions arise about the behavior of the disk under constant torque, specifically whether it can achieve a constant rotation speed.
  • Another participant explains that speed will increase until frictional losses reach an equilibrium speed.
  • Mathematical relationships between torque, moment of inertia, and angular acceleration are discussed, with calculations provided for a specific scenario involving a DC motor.
  • There is a correction regarding the acceleration calculation, with a participant clarifying the conversion from radians to degrees.
  • Participants discuss the requirements for torque and RPM in relation to motor selection, including considerations for gear ratios and power requirements.
  • Suggestions are made regarding mechanical components and control systems for achieving desired rotational speeds.

Areas of Agreement / Disagreement

Participants express differing views on the effects of mass shape and radius on moment of inertia, and there is no consensus on the simplification of the moment of inertia formula. The discussion on the behavior of the disk under torque also reveals varying interpretations of the dynamics involved.

Contextual Notes

Some assumptions about the system's ideal conditions, such as frictionless environments and constant torque, are present but not fully explored. The implications of these assumptions on the calculations and discussions remain unresolved.

Who May Find This Useful

This discussion may be of interest to those studying rotational dynamics, engineering mechanics, or anyone involved in designing systems that involve rotating bodies and torque applications.

Trainee Engineering
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Hi,

I have a disk of diameter r, and the mass of the disk is 1kg. I'm going to rotate the disk at its center. my question is:
1. let's say I put a load of m kg on top of the disk, does the moment inertia of the system is as simple as (m + 1kg)r2/2?
2. does the shape of the load put on top of disk will affect the MI of the system? or just the mass of the load that will affect the MI of the system?
3. let's say I decrease/increase the radius of the disk with negligible change in mass. will that affect MI of the system?

Thanks
 
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Trainee Engineering said:
1. let's say I put a load of m kg on top of the disk, does the moment inertia of the system is as simple as (m + 1kg)r2/2?

2. does the shape of the load put on top of disk will affect the MI of the system? or just the mass of the load that will affect the MI of the system?

The moment of inertia of mass m will depend on the shape and position of mass m.

The moment of inertia of the combination (disc and mass m) is equal to the sum of the moments of inertia of the parts:

Itotal = Id + Im

3. let's say I decrease/increase the radius of the disk with negligible change in mass. will that affect MI of the system?

Yes. The moment of inertia of the disc Id = 0.5mR2 so if you change R then Id changes and so Itotal changes.
 
another question. if the disk receives constant torque, in a ideal frictionless environment, does that mean the rotation speed will keep getting faster as time goes by? it's impossible to have a constant rotation speed with constant input of torque?
lets say the MI is 3600 kgm2, and it's connected to a 31.41 Nm DC motor. this means 4 degrees rotation takes around 4s, correct?
thanks
 
Last edited:
Trainee Engineering said:
it's impossible to have a constant rotation speed with constant input of torque?
The speed will increase until frictional losses bring it to an equilibrium ('terminal') speed. (That includes the losses in the driving motor.)
 
+1

Newtons laws apply to rotation as well as linear motion...

Linear...

F = ma
where F=Net Force, m=mass, a=acceleration

Rotation...

T = Iα
where T=Net Torque, I=Moment of Inertia, α=angular acceleration

see the similarity?
 
Trainee Engineering said:
lets say the MI is 3600 kgm2, and it's connected to a 31.41 Nm DC motor. this means 4 degrees rotation takes around 4s, correct?
thanks

From above..
T = Iα
so
α = T/I
= 31.41/3600
= 0.009 rads/s/s

eg it accelerates at 0.009 radians per second per second (=about 0.057 degrees per second per second).

You can modify the equations of motion/SUVAT to calculate how fast it will be going after time t or how many revolutions it takes to reach a certain speed etc.
 
CWatters said:
From above..
T = Iα
so
α = T/I
= 31.41/3600
= 0.009 rads/s/s

eg it accelerates at 0.009 radians per second per second (=about 0.057 degrees per second per second).

You can modify the equations of motion/SUVAT to calculate how fast it will be going after time t or how many revolutions it takes to reach a certain speed etc.

hi, just a quick observation, but 0.009 rad = 0.5 degrees? not 0.005?
secondly, is DC motor with torque output above 50 Nm considered common?
is the number considered high-powered (industrial grade), high end or expensive?
thanks
 
Oops yes should be 0.57 degrees.
 
  • #10
High torque can be achieved with gearing. What RPM do you need?
 
  • #11
CWatters said:
High torque can be achieved with gearing. What RPM do you need?

slow RPM. 1/6 deg per second is good enough. lower is also fine. basically, I just need to get ~4 degrees in 4 secs
 
  • #12
That's quite slow..

1/6th degree/s = 0.003 Rads/s = 0.03rpm

Power = Torque * angular velocity
so the power required is just
= 0.003 * 50
= 0.15W

Small DC motors are available for a few $ that could easily be provided that much power - but at perhaps 10,000 rpm.

So you actually need a small motor and a large/strong very high ratio reduction gearbox. The gearbox will cost more than the motor. Backlash might also be a big issue in such a high ratio gearbox.
 
  • #13
Use toothed belts or PolyV belts . Very high single stage reductions are possible with both .

Something to decide straight away is whether you are going to drive your turntable open loop or use a proper feed back control system .

Detailed considerations of positioning system may dictate how you arrange the mechanical components .
 
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