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Moment of Inertia and Torque

  1. Oct 21, 2008 #1
    A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +40 N·m is applied to the wheel for 24 s, giving the wheel an angular velocity of +610 rev/min. The external torque is then removed, and the wheel comes to rest 120 s later. (Include the sign in your answers.)

    (a) Find the moment of inertia of the wheel.
    kg·m2
    (b) Find the frictional torque, which is assumed to be constant.
    N·m

    This whole topic of angular/moment of inertia and torque taking its toll on me. Please bear with me, the more I read in textbook the more confused I get.

    Relevant Equations:
    Inertia = sum of mi * ri^2
    Inertia for continuous objects = integral of r^2 dm

    I would use the equation for inertia for continuous objects correct? since the textbook used the other one when given points.
    The problem doesn't give us any masses/radius -- How would I start this?

    For part B:
    relevant equations:
    Torque = r x F
    Tnet = I alpha

    What should I use for the Moment of Inertia for the wheel? It doesn't state it,
    some exercise from the book says consider a wheel of a bicycle to be a hoop I = MR^2

    Im not sure if this is correct,
    I did for translation equation
    Fnet = Ma
    Ext F + friction = Ma

    For rotation
    Tnet = I alpha
    RF - Rf = 1/2 M R^2 (a/R)
    = F-f = 1/2 Ma

    I added the two equations
    2F = 3/2 Ma
    a = 4F/3M

    From equation 1
    F+f = Ma
    f = F/3

    Is this even remotely correct?

    Please help, thanks so much.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 21, 2008 #2
    Hey there,

    For part a) I don't think you need any more information. In this question you wouldn't find the moment of inertia by summing or integrating.

    A word of warning: You haven't been the given the angular velocity, so to start with I would change your rev/min to rad/sec.

    See this site: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

    You need to use the rotational equivalent of v=v-0+a*t, along with Newtons second law rotational equivalent.

    Hope this gets you going for the first part
     
  4. Oct 21, 2008 #3
    It says it gave the wheel an angular velocity of +610 rev/min
    to change this to rad/sec.
    I got 63.87905062 approx 64 rad/sec.

    63.87905062 rad/sec / 24s = 2.661627109 approx 2.7 rad/s^2 is the acceleration
    Tnet = I alpha
    so Tnet/alpha = I
    40 Nm /2.661627109 rad/s^2 = 15.02840110 approx 15 is the Inertia?

    Is this correct?
     
    Last edited: Oct 22, 2008
  5. Oct 22, 2008 #4
    Could someone check if what I did for part A/B is correct?
     
  6. Oct 23, 2008 #5
    Yes your answer is the same as the one I got. For part b I'd use the same equations except this time you know the moment of inertia and initial and final angular velocities but want to find the frictional torque. You can use the same equations since they say that the frictional torque is constant.

    Sorry for the late reply.
     
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