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Angular Speed of a flywheel

  1. Nov 30, 2005 #1

    A flywheel with a diameter of 1.2 m is rotating at an angular speed of 200 revs/min

    What is the angular speed of the flywheel in radians per second?

    Would I just do this?

    200 revs/min = 3.33 revs/sec

    3.33 revs/sec = 6.66(pi)rads - - - answer?
  2. jcsd
  3. Nov 30, 2005 #2


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    Yes, that's right.
  4. Nov 30, 2005 #3
    ok..and is angular speed measure in revs/sec or rad/sec

    Also..another part asks

    What constant angular acceleration (in rev/min^2) will increase the wheels angular speed to 100 rev/min in 60 seconds?

    I used:

    [tex] \alpha = \frac{\Delta \omega}{ \Delta t}[/tex]

    1000 revs/min = [2000(pi)revs/min - 400(pi)revs/min] / 1(minute)

    [tex] \alpha = 1600 \frac{revs}{min^2}[/tex]
  5. Nov 30, 2005 #4
    and it also asks how many rotations did this take could I just assume since its a constant acceleration that i took 1600 revs?
  6. Nov 30, 2005 #5


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    Angular speed is usually given in rads/sec, but, depending upon the application, it can also be measured in rpm or cps, for example.

    alpha = 1600 revs/min² is correct.
    But you have changed the initial angular velocity from 200 rev/min to 2000 rev/min.
    Is this a typo, or a different situation with a different speed.

    To find the number of rotations. You are being asked to find the (angular) distance travelled.
    Do you remember this,

    v² - u² = 2as ??

    What do you think is its equivalent in circular motion ?

    Use that.
  7. Nov 30, 2005 #6
    its suppose to be increased to 1000 rpm

    isnt it [tex] \omega_f - \omega_i[/tex]
  8. Nov 30, 2005 #7
    What do the variables stand for...i.e - u, a, s, v?
  9. Nov 30, 2005 #8


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    u = initial (linear) speed
    v = final (linear) speed
    a = (linear) acceleration
    s = (linear) distance travelled

    Can you now transpose that (linear) eqn into its circular/rotational equivalent ?
  10. Nov 30, 2005 #9

    [tex] \frac{ \Delta \omega}{2 \alpha} = d[/tex]
  11. Nov 30, 2005 #10


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    Almost there, but not quite :(

    [tex]\mbox{linear: } v^2 - u^2 = 2as[/tex]
    [tex]\mbox{circular: } \omega _f^2 - \omega _i^2 = 2\alpha\theta[/tex]

    [tex] \omega _f^2[/tex] is the final angular velocity
    [tex] \omega _i^2[/tex] is the initial angular velocity
    [tex]\alpha[/tex] is the angular acceleration
    [tex]\theta[/tex] is the rotational displacement
    Last edited: Nov 30, 2005
  12. Nov 30, 2005 #11
    Ok so (2000^2 - 400^2)/2(1600) = [tex] \theta[/tex]

    = 1200

    circumferance of wheel = 2(pi)r = 3.77m

    1200m/3.77m = 318revs?
  13. Nov 30, 2005 #12


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    You are working in revs/min and revs/min² for velocity and acceleration so your displacement (theta) will be in revs.

    i.e. theta = 1200 revs (rotations)
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