# Required torque to accelerate a disk of a certain mass

1. Oct 6, 2008

### Sum1

1. The problem statement, all variables and given/known data

(Actually, this is not a homework, but a project I'm trying to accomplish
but I have almost forgotten physics...)

There is a disk weighting 100kg with a radius of 0.6 meters. What would be the torque needed to accelerate the disk from 0 rpm to 6 rpm in 10 seconds?

Variables:

m=100 kg
r=0.6 meters
rpmA=0 rpm
rpmB=6 rpm
t=10 seconds

2. Relevant equations

If I'm not wrong:

The mass moment of inertia of a disk with a uniform mass is considered to be:
(m*r^2)/2

where pi = 3.14159...

3. The attempt at a solution

So, trying to solve the problem:

moment of inertia of the disk
I = (m*r^2)/2
= (100*0.6^2)/2
= 18

Average angular velocity in radians per second
v = (rpmB-rpmA) * (pi/30)
= 0.628319

a = Angular velocity in radians per sec /t sec
= v/t
= 0.0628319

Torque
T = Angular acceleration * Moment of Inertia
= a*I
= 1.13097

Is this correct?
Is the above result of Torque in Kg-m?
What exactly is the unit of the above inertia result? Is it Kg-m^2 ?

Any help will be appreciated!

Last edited: Oct 6, 2008
2. Oct 6, 2008

### cepheid

Staff Emeritus
Your method looks correct. I haven't checked your math. Note that rpmB-rpmA is not the average angular velocity. It is the change in angular velocity. In any case, that's what you want -- the change in angular velocity. That is because, for uniform acceleration, a = (change in v)/time.

Torque has units of newton-metres.

1 N = 1 kg*m/s^2
1 Nm = 1 kg*m^2/s^2

Moment of inertia has units of kg*m^2...and multiplying by the angular acceleration to get torque gives you the missing 1/s^2.

3. Oct 6, 2008

### Sum1

Yes, it is the change in angular velocity not the average -that was stupid!

But I'm still confused about the output torque unit of the above calculation. On this site http://nmbtc.com/Calculators/converter.swf" it has Kg-m and N-m torque units. Should I consider the above result as N-m? If so, that would mean that it requires about 10 times less torque than if the output unit is Kg-m!

Last edited by a moderator: Apr 23, 2017
4. Oct 6, 2008

### Sum1

Nevermind, just found an example which used Kg and meters and the output torque unit was Newton meters! So it is N-m!