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Moment of Inertia and Torque problem

  1. Aug 2, 2011 #1
    1. The problem statement, all variables and given/known data
    A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +44 N·m is applied to the wheel for 23 s, giving the wheel an angular velocity of +520 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.
    b) Find the frictional torque, which is assumed to be constant.

    2. Relevant equations
    Torque = Intertia x angular acceleration
    Inertia = (mr^2)/4
    This is the inertia for a solid disk
    angular acceleration = angular velocity/time


    3. The attempt at a solution
    So heres my attempt I started by changing the 520 rev/min to...
    520 rev/min * 1 min /60 sec = 8.67 rev/sec
    8.67 rev/sec * 2 pi rads/1 rev = 17.33 pi rads /sec

    After that I wanted the angular velocity so I did...
    17.33 pi rads/sec / 23 seconds = 2.368 rads / sec^2

    With the angular velocity I found the angular acceleration to be...
    a = 2.368/23 sec = .1029 rads / sec^2

    I know I'll have to use that to solve for the Torque (I think)

    What I'm really stuck on is getting a) The moment of intertia of the wheel.
    I = (mr^2)/4
    I don't know where to get the r from, and maybe that equation is wrong overall which is another thing I am confused about.

    If I can get that I believe I would just use the angular acceleration I got and the Intertia to get the Torque.

    Any help towards the right direction would be greatly appreciated, thanks :)
     
  2. jcsd
  3. Aug 2, 2011 #2
    You have torque and angular acceleration; use that to calculate the moment of inertia of the wheel.
     
  4. Aug 2, 2011 #3
    Uhm If I use the acceleration I got .1029 rads/sec^2 into the equation T = Inertia x angular acceleration

    That would be Intertia = Torque/angular acceleration
    Intertia = 44 / .1029
    Intertia = 427.6 kg m^2

    That didn't get me the right answer and it doesnt look like the right one. Is that what you
    ment.
     
  5. Aug 2, 2011 #4

    SteamKing

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    Your calculation of angular velocity and angular acceleration are incorrect. (Do you really think that angular velocity has units of rad/s^2?)
     
  6. Aug 2, 2011 #5
    Ah true...wouldn't the angular velocity and acceleration be this instead then

    520 rev/min * 1min/60 sec = 8.67 rev/sec
    8.67 rev /sec *2pi rads / rev = 54.454 rads /sec (This would be my actual angular velocity)

    So then my angular acceleration would be
    a = w/t
    a = 54.454 (rads/sec) / 23 seconds = 2.367577 rads/sec^2

    This looks right to me, but even if I take this and plug it into Torque = Inertia x angular acceleration the answer for a is wrong.
     
  7. Aug 3, 2011 #6
    Ah I finally got the problem, I had to also take into consideration the acceleration of when the wheel was slowing down as well. Add the two accelerations and then use that for
    the Intertia.

    b) was just using the found intertia to solve for torque

    Thanks a bunch to those that helped!
     
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