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Determining torque

  1. Oct 15, 2013 #1
    1. The problem statement, all variables and given/known data

    An electric trimmer blade has rotational inertia 1.55×10−4kg⋅m2 .

    What torque is needed to accelerate this blade from rest to 605rpm in 1.05s ?

    How much work was done to accelerate the blade?

    2. Relevant equations

    Torque=Ixα
    rpm=>Rad/s
    α=Δω/Δy
    3. The attempt at a solution

    Ok so i converted RPM to Rev/s first. which is 63 rad/s now substituting it in for α and multiply it by the Inertia will get me the torque of 9.35x10^-3N-m

    Angular Velocity i think is 66.15rad/s. Because ω=α*T
    For the work, it is just the change in kinetic energy right? K=(.5)(I)ω^2 i keep getting a wrong answer
     
  2. jcsd
  3. Oct 15, 2013 #2

    SteamKing

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    Homework Helper

    RPM and rad/s are measures of angular velocity, and angular velocity is not the same as angular acceleration.

    You have an initial angular velocity of 0 and a final angular velocity of 605 RPM (converted to rad/s) which is to be achieved in 1.05 s. What angular acceleration (units of rad/s^2) is required to do this? Once you find the ang. acceleration, then you'll be able to calculate the torque required to do this.

    You should have studied the basic equations of angular motion (and they are analogous to the equations of linear motion).
     
  4. Oct 15, 2013 #3
    α is angular acceleration. you used the angular velocity, 605rpm (63rad/s) to find the torque. Your KE equation is correct but you used the wrong value for angular velocity. The angular velocity was given.
     
  5. Oct 16, 2013 #4
    Yeaa, i noticed that. Wasn't thinking straight.

    I would think that i would use one of my kinematic equations ω=ωi+αt to find α which in this case is just ω=60
    (.5)(1.55x10^-4)(60^2)=.279

    IS there anything i am forgetting?
     
    Last edited: Oct 16, 2013
  6. Oct 16, 2013 #5
    Scratch that, that one is for constant acceleration. This isn't constant
     
  7. Oct 16, 2013 #6
  8. Oct 16, 2013 #7
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