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Question involving flywheel.

  1. Mar 30, 2004 #1
    I missed a week of school so I'm a bit behind, i do have a book so i need just a bit of guidance. Any extra help would be appreciated.

    Question Part 1
    The flywheel of a steam engine runs with a constant angular speed of 115 rev/min. When steam is shut off, the friction of the bearings and the air (assumed constant) brings the wheel to rest in 2.20 hours. How many rotations does the wheel make before coming to rest?

    Part 2
    Consider a particle that is located at a distance of 37.0 centimeters from the axis of rotation when the flywheel is turning at 0.390 rev/min. What is the magnitude of the net linear acceleration of this particle?
     
  2. jcsd
  3. Mar 30, 2004 #2
    Angular motion formulas have a direct analog to kinematics formulas. The formula relating velocity and acceleration in kinematics,

    v = at

    simply becomes

    [tex]\Delta \omega = \alpha \Delta t[/tex]

    where omega is the angular velocity and alpha is the angular acceleration. From that formula, you can calculate the acceleration. Then we have the equation

    [tex]\Delta(v^2) = 2a\Delta x[/tex]

    becomes

    [tex]\Delta(\omega^2) = 2\alpha \Delta \theta[/tex]

    from which you can calculate theta.

    For the second part, centripetal acceleration is related to linear velocity by

    [tex]a = \frac{v^2}{r}[/tex]

    and linear velocity is related to angular velocity by

    [tex]v= \omega r[/tex]

    Combine the two and you're done.

    cookiemonster
     
  4. Mar 30, 2004 #3
    Got the first part, and now workin to understand the second. I'll let you know. Thanks.
     
  5. Mar 30, 2004 #4
    For Part two, this is what i've come up with and I'm stuck on.
    w = 115 rev/min
    r = .37 m

    Does v^2 in the first equation equal (wr)^2?

    I missed all the work on Omega and such, i'm having alot of trouble with some work.
     
  6. Mar 31, 2004 #5
    Yes. v^2 = (wr)^2.

    cookiemonster
     
  7. Mar 31, 2004 #6
    Therefore..
    (wr)^2/.37 should give me my answer?
     
  8. Mar 31, 2004 #7
    Yes.

    cookiemonster
     
  9. Mar 31, 2004 #8
    I'm getting 4893.25.. m/s^2 and that's not right.

    Should i be converting 115 rev/min to something?
     
  10. Mar 31, 2004 #9
    radians/min?
     
  11. Mar 31, 2004 #10
    I'll look up how to do that. or does anyone know off hand?
     
  12. Mar 31, 2004 #11

    Doc Al

    User Avatar

    Staff: Mentor

    One revolution = 2π radians;
    One minute = 60 seconds.
     
  13. Mar 31, 2004 #12
    Didn't you say that it was going .390 rev/min?

    (.390rev/min)(1min/60sec)(2pi rad/1rev) = .0408 rad/s

    cookiemonster
     
  14. Mar 31, 2004 #13
    Consider a particle that is located at a distance of 37.0 centimeters from the axis of rotation when the flywheel is turning at 0.390 rev/min. What is the magnitude of the net linear acceleration of this particle?

    Ya you're right Cookie.. I was lookin at the first part.
    (wr)^2/r
    (.0408*.37)^2/.37

    a =.0006159 m/s^2.. Does that look right?
     
  15. Mar 31, 2004 #14
    .39 rev/min isn't very fast.

    cookiemonster
     
  16. Mar 31, 2004 #15
    so my answer is on target?
     
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