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Required torque to accelerate a disk of a certain mass

  1. Oct 6, 2008 #1
    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?


    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:
    where m=mass, r= radius

    Converting rpm to radians/sec:
    rad/s = rpm * (pi/30)
    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

    Angular acceleration in rad/s^2
    a = Angular velocity in radians per sec /t sec
    = v/t
    = 0.0628319

    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. jcsd
  3. Oct 6, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
  4. Oct 6, 2008 #3
    cepheid, thanx for the reply!

    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!
  5. Oct 6, 2008 #4
    Nevermind, just found an example which used Kg and meters and the output torque unit was Newton meters! So it is N-m! :smile:
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