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Homework Help: Moment of Inertia and flywheel

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Part One:

    A flywheel in the form of a heavy circular disk of diameter 0.725 m and mass 127 kg is mounted on a frictionless bearing. A motor connected to the flywheel accelerates it from rest to 1340 rev/min. What is the moment of inertia of the flywheel? Answer in units of kg times meters squared.

    Part Two:
    How much work is done on it during this acceleration?

    Part Three:
    After 1340 rev/min is achieved, the motor is disengaged. A friction brake is used to slow the rotational rate to 973 rev/min. What is the magnitude of the energy dissipated as heat from the friction brake? Answer in units of J.

    2. Relevant equations

    Honestly, I have no idea where to even begin! I know that torque = rFsin(theta) = I(angular acceleration), but this does not seem to generate a reasonable answer. I'm in an algebra-based physics class, so there must be SOME algebraic equation that can help me out? Any tips in the right direction are extremely helpful, thanks!
  2. jcsd
  3. Oct 20, 2009 #2


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    Part One
    Your textbook must have a list of moments of inertia about axes usually through the center of mass. Look up "disk" or "cylinder" and plug in.

    Parts Two and Three
    Use the work-energy theorem for rotational motion.
  4. Oct 20, 2009 #3
    I for a disk = 1/2(M.R^2)
    R: radius of disk
    Put the values and you'll get the answer.

    work done= (torque).(angular displacement)

    Energy dissipated will be equal to Work done by the brake.

    Buddy, if you know only this much about rotation, you have to study a LOT MORE.
    Knowing only the algebraic equations won't help either.
    Just take up any book/study material and learn the concepts of rotation.

    And yes.. "I" isn't angular acceleration as you have stated.
    Last edited by a moderator: Oct 20, 2009
  5. Oct 20, 2009 #4
    I found I = (1/4)(m)(R squared), so would this be I = (1/4)(127 kg) (0.725 m /2)^2 ? I am wary of using this equation, as it does not in any way utilize 1340 rev/min.

    Do you mean W=ΔK=(1/2)Iω(final)^2−(1/2)Iω(initial)^2? I understand how to use this for Part Two, but Part Three seems to require heat in the equation. Would this just be the energy difference between 1340 rev/min and 973 rev/min?

    I am afraid I am very new at these units (rev/min) as well, and am wondering if I will need to convert them to meters, maybe?
  6. Oct 20, 2009 #5
    :( Wow, okay, I even managed to find the wrong equation for Part One.

    I was writing the equation for torque...which is "I" times the angular acceleration.

    Could you maybe clarify why energy dissipated is equal to the work done by the brake? I see now that I have really missed entire concepts in this unit. :( Oh no.
  7. Oct 20, 2009 #6


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    The correct expression for a disk is I = (1/2)mR2. You don't need the 1340 rev/min for this part. The disk does not need to be rotating to have a moment of inertia much like a block does not need to be moving to have a mass.
    Yes, you need to convert but to radians/second for all calculations. How many radians are there in one revolution? Multiply that number by "revolutions/second" to get "radians/second". Don't forget to convert revolutions/minute to revolutions/second first.
  8. Oct 20, 2009 #7
    All the answers given by kuruman are correct..
    But what about your concepts??
    I guess you are just learning the algebraic results...which we are NOT supposed to do until we have thoroughly understood all the concepts. Maybe you should get yourself a book or something to help you understand the basic concepts first.
    And you also need to revise the Work-Energy Conservation.

    By the way, what's your age??
    And why don't you Admins and Helpers here recommend this guy to get himself some material to understand the concepts.

    I am from India and dont know what course is followed in other nations.
    So, guys please help each other.
    Last edited by a moderator: Oct 20, 2009
  9. Oct 20, 2009 #8
    I am sorry to have analyzed your statement wrong.

    And please don't get disheartened.
    It won't take long to get your concepts cleared.
    All you need is some good study material and some passion...that's all.
    A lot of people here can recommend some basics' material.

    Best of Luck !!
    Last edited by a moderator: Oct 20, 2009
  10. Oct 20, 2009 #9
    I completed the first part correctly, but I can't seem to understand how to convert the units properly. :(

    At first, I converted rev/min to rad/s and got 44.67pi rad/s. After plugging this into the work equation, however, I got the wrong answer.

    Since the answer is supposed to use units of joules, I thought I'd convert 1340 rev/min to m/s and got 0.50867621 m/s, but after plugging that into the work equation, I still got the wrong answer!

    Does anyone know what I am doing incorrectly? I cannot find any problem similar to this one in the textbook, unfortunately.
  11. Oct 21, 2009 #10


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    You cannot convert 1340 rev/min to something in m/s. The first has dimensions of Time to the negative one power and the second has dimensions of Length times time to the negative one power. You don't seem to understand what radians are. Read about them at


    then do dimensional analysis on (1/2)Iω2.
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