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Rotational joy!

  1. Dec 7, 2004 #1

    A disk of mass 15 kg and radius 25 cm is rotating initially at 20 rad/s. A constant force of 40 N is applied tangentially to it's edge, positively accelerating the disk.

    a) At t= 5 seconds, calculate the-

    1) rotational velocity of the disk
    2) the instantaneous power supplied to the disk
    3) the angular momentum of the disk
    4) the work done on the disk
    5) the angular displacement of the disk

    b) calculate the same quantities as above if instead the force applied to the disk is given as F(t)= 6t (N).

    so from what i read the disk is already rotating counter-clockwise at 20 rad/s when the force is applied... right? how do i go about finding the new rotational velocity?

    thanks a lot to anyone who helps out. :wink:
  2. jcsd
  3. Dec 7, 2004 #2
    Remember that torque = I*(angular acceleration) (analogous to F=ma, where moment of inertia I replaces inertia m). Also remember that torque is the cross product of force and the radial vector, so all you have to do is solve for angular acceleration to get started on your rotational kinematics.
    PS. You may not have studied the cross product in detail and may just know the scalar relationship torque = |F||r|sin(theta), which is just |F||r| when the force acts perpendicularly to the radius (tangent to the rotation).
  4. Dec 8, 2004 #3
    These formulae might help you

    [tex]\Tau = I\alpha = Fr[/tex]

    Remember the kinematics equations
    [tex]v = v_0 + at[/tex]
    [tex]x = x_0 + (1/2)at^2[/tex]
    [tex]v^2 = (v_0)^2 + 2ax [/tex]

    They are all analogous to rotational motion.
    Just replace [tex]v[/tex] and [tex] v_0[/tex] by
    [tex]\omega [/tex] and [tex]\omega_0[/tex] and
    [tex]x [/tex] and [tex] x_0 [/tex] by [tex]\theta [/tex] and [tex]\theta_0[/tex]
    and [tex] a [/tex] by [tex]\alpha[/tex].
    Last edited: Dec 8, 2004
  5. Dec 8, 2004 #4
    cool. the reason it was taking me so long was that i failed to register that the force was constant (stupid me!). anyways, thanks a lot for the help hypermorphism and prasanna. :wink:

    one last question: for the second part that says "calculate the same quantities as above if instead the force applied to the disk is given as F(t)= 6t (N).", you just do the same calculations over again for a F(5)= 30 N, right?
  6. Dec 8, 2004 #5


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    Nope,in this case,everything changes,since the force that accelerates the disk in not constant,but time dependent,thankfully,linearly.
    The eq.u used will not work in this case.Try to find another ones.Integrate the relation I times alpha is equal to the force to find that this time the angular velocity will not be linear anymore,by quadratic wrt to time.

  7. Dec 8, 2004 #6
    isn't I times alpha equal to the torque? anyways, should i integrate 6t= I times alpha? sorry if i'm a liittle slow catching on to this cos we've never done this in class.
  8. Dec 8, 2004 #7


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    Homework Helper

    No,you're not "slow-catching",i made a blunder.The torque is force times distance as it should,and that should be put equal to I times "alpha".It's 5 am in Belgium,maybe i should quit.Obviusly concentrating is impossible.
    Sorry again,and hopefully smb else will help,as i'm about to get me a nap.
  9. Dec 8, 2004 #8
    lol, no prob... sweet dreams. :tongue:

    anyone else??
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