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Angular velocity, acceleration. and moment of inertia

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A grinding wheel is a uniform cylinder with a radius of 8.70 cm and a mass of 0.400 kg.
    a)calculate the moment of inertia about the center.
    b)Calculate the applied torque needed to accelerate it from rest to 1950 rpm in 6.00 s if it is known to slow down from 1250 rpm to rest in 57.5 s.

    3. The attempt at a solution
    first i changed the given radius to .087 m..

    a) wasnt that hard, I=mR^2, (.4)(.087)^2=.00151

    b) since i was given the ωi=0, and ωf=1950 rpm→32.5 rps, and t=6s, i used the formula ωf=ωi+αt, and got α=5.42. i used my α to solve for a with the equation a=Rα (because F=ma) and got .47, then using F=ma, F=(.4)(.47)=.19. Since τ=RF, τ=(.19)(.087)=.02, which is incorrect.

    Im guessing that my calculated I has something to do with the answer because of the way my professor asks the question, in parts where you use your answer in part a to solve for part b which is needed to solve for part c, etc, but i dont understand how my answer is wrong. however realistically i can understand how a τ of .02 wouldnt cause a cylinder to rotate from 0 to 1950 rpm in 6 sec.
     
  2. jcsd
  3. Mar 26, 2012 #2

    gneill

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    Staff: Mentor

    Your acceleration should be expressed in unit of rad/s2 in order to be consistent with other calculations. You should ALWAYS attach units to your values.

    You don't need to convert back and forth from torque to force via the radius. You can work directly with torques, keeping everything "angular" :smile:

    Why do you think they gave you the information about the wheel slowing to a halt over a given time? What would bring something in motion to a halt over time?
     
  4. Mar 26, 2012 #3
    i dont understand what you mean by "You can work directly with torques, keeping everything "angular"" in order to get torque i need the force, in order to get the force i need acceleration, so the only way to take what was given and connect it to acceleration is through the equation a=rα and α is calculated from the equation i used with ωf and ωi
     
  5. Mar 26, 2012 #4

    gneill

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    The entire problem is dealing with angular quantities; there are no linear references and none are required. You can do everything with torque, angular acceleration, and time. All of the "usual" linear kinematic equations have their angular counterparts. Thus ##\tau = I \alpha## is the angular equivalent to F = ma, for example.
     
  6. Mar 26, 2012 #5
    ok so then, τ=(.00151)(5.42)=.0082 kg*m^2*rev/sec??? still sounds like to small of a number, and i am unsure of the units
     
  7. Mar 26, 2012 #6

    gneill

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    I think that one problem is that you're mixing units inapropriately. Do your calculations in radians per second for angular velocities, and radians per second2 for angular accelerations.

    5.42 is your value for revolutions per second2. Convert that to radians per second2.

    Also, you haven't yet dealt with the information concerning the slowing of the wheel from some initial angular velocity to rest. What information does this give you?
     
  8. Mar 26, 2012 #7
    well α=34.05 rad/sec^2, but i have no clue what the information the slowing of the wheel is giving, i want to say drag, but no i highly doubt it
     
  9. Mar 26, 2012 #8

    gneill

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    Staff: Mentor

    Another word for drag is friction...

    So there's a frictional torque acting.
     
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