# B Calculating inertial moment of a disk

1. Feb 12, 2017

### Trainee Engineering

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

Thanks

2. Feb 12, 2017

### CWatters

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

Yes. The moment of inertia of the disc Id = 0.5mR2 so if you change R then Id changes and so Itotal changes.

3. Feb 12, 2017

### CWatters

4. Feb 13, 2017

### Trainee Engineering

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: Feb 13, 2017
5. Feb 13, 2017

### sophiecentaur

The speed will increase until frictional losses bring it to an equilibrium ('terminal') speed. (That includes the losses in the driving motor.)

6. Feb 13, 2017

### CWatters

+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?

7. Feb 13, 2017

### CWatters

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

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.

8. Feb 13, 2017

### Trainee Engineering

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

9. Feb 14, 2017

### CWatters

Oops yes should be 0.57 degrees.

10. Feb 14, 2017

### CWatters

High torque can be achieved with gearing. What RPM do you need?

11. Feb 14, 2017

### Trainee Engineering

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. Feb 14, 2017

### CWatters

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. Feb 14, 2017

### Nidum

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 .