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Rotation with Newtons second law and work energy

  1. May 31, 2008 #1
    Heres the problem


    Now I have to work it out using Newtons second law and Work energy....

    For Newtons second Law (F = ma).....would I be using

    [tex] I = mk^2 [/tex]

    [tex] M = I\alpha [/tex]

    than setting up F = ma...?

    and for work energy im using kinetic and potential....so the setup without using friction would just be using kinetic and potential?....and is the potential an elastic potential since its a wheel?
  2. jcsd
  3. May 31, 2008 #2
    Use [tex] \tau= I\alpha[/tex] where [tex]\tau[/tex] is the net torque external.
    Let tension = T. Apply newton's Second Law for rotational rotation T and angular acceleration

    [tex]TR=I\alpha[/tex], where R is the radius of the pulley.

    For the second part, Total Energy = Kinetic Energy + Potential Energy + Rotational Kinetic Energy
    where Rotational Kinetic Energy = [tex]\frac_{1}{2}I\omega^2[/tex]
    Last edited: May 31, 2008
  4. May 31, 2008 #3
    Sorry for double posting; I can't seem to edit my previous post, weird.

    Rotational kinetic energy = [tex]\frac{1}{2}I\omega^2[/tex]
  5. May 31, 2008 #4
    For this equ. that you had posted


    is that relating to F = MA ?

    also the T in the equ is tension, correct?..and if so where does the [tex]\tau [/tex]come in...

    and just for clarification....my professor has been using the term "magnitude of the couple" instead of torque as you did....but the equ is the same...so im assuming there one in the same....
    Last edited: May 31, 2008
  6. May 31, 2008 #5
    Yes, T is the tension. As for magnitude of the couple, I believe it is the same thing as torque. (I've never really seen that term used before)

    In order to relate the suspended mass with Newton's 2nd law, you can draw a free diagram to see more cleary.

    We know that,
    [tex]\Sigma F_{net}=ma_{y}[/tex], m is the mass of the suspended object
    [tex]\Rightarrow mg-T=ma_{t}[/tex] Equation *, where T is the tension

    But, we know that tangential acceleration [tex]a_{t}=R\alpha[/tex], R is the radius, alpha is the angular acceleration.
    Then, [tex]\frac{mg-T}{m} = R\frac{TR}{I}[/tex]
    Then, [tex]T= \frac{mg}{1+(mR^2/I}[/tex]
    plug T back into Equation (*)

    Solve for [tex]\alpha_{t} = \frac{1}{1+I/(mR^2)}[/tex]
    Plug back R, solve for angular acceleration. Then solve for angular velocity.
    (Please note that this is for when there is no friction between the rope and the pulley!)

    Now, for the second part, using conservation of momentum. Let the distance that the mass drop = d.

    By Conservation of momentum,
    [tex]U_{f}+K_{f} = U_i + K_f = 0 +0 = 0[/tex]

    Then, [tex]U_{f}=U_{object} + U_{pulley}=m_{object}g(-d)+m_{pulley}g(-d/2)[/tex]
    [tex]K_{f} = ... = \frac{1}{2}(m_{object}+m_{pulley})v^2 [/tex]

    After some simplification, [tex]v=\sqrt{2m_{object}gd/(m_{object}+m_{pulley}}}[/tex]
    For d, plug in 5 revolutions from the problem. 5 revolutions are equal to 5*circumference of the pulley.
    Then, solve for angular velocity.

    Lastly, we know that [tex]\frac{1}{2}I_{pulley}\omega^2= \frac{1}{2}m_{pulley}v^2[/tex] by the rotational kinetic energy!

    Hope that helped.
    Last edited: May 31, 2008
  7. May 31, 2008 #6

    D H

    Staff: Mentor

    No, it is not. T (better written as the Greek letter tau, [itex]\tau[/itex]) is torque, not tension. Tension is a force. Torque has units length*force. It is a different beast.
  8. May 31, 2008 #7
    Yes, I know. I just have a habit of writing tension as T and I believe I did use tau as torque and specified T as tension.
  9. May 31, 2008 #8
    My equ looks like this for the first section ..I just need to solve for Tension....

    (also I replaced tau with M which is the magnitde of the couple, which is given in the problem)

    [tex]\frac{mg-T}{m} = r\frac{M}{I}[/tex]

    also is the mass on each side of the equ. of the suspened object?
  10. May 31, 2008 #9
    Yes, the m=mass of the suspended object (both m's), same as [tex] m_{object}[/tex] for the second method. mass of the pulley = [tex]m_{pulley}[/tex]
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