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Torque, spool and string

  1. Jan 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A spool of mass m and moment of inertia I (presumably with respect to the CM) is free to roll without slipping on a table. It has an inner radius r, and an outer radius R. If you pull on the string (which is wrapped around the inner radius) with tension T at an angle [tex]\Theta[/tex] with respect to the horizontal, what is the acceleration of the spool? Which way does it move?

    3. The attempt at a solution

    I got an answer for the first part, but I feel a little unsure about it. Am I on the right track?

    [tex]\tau_{CM}=f-rT\cos{\Theta}=I\alpha,\hspace{.1 in}(1)[/tex]

    [tex]m\ddot{x}=T\cos{\Theta}-f.\hspace{.1 in}(2)[/tex]

    where f is the friction force between the spool and the ground.

    The "non-slip" condition gives

    [tex]\ddot{x}=R\alpha.\hspace{.1 in}(3)[/tex]

    A little algebra on (2) yields

    [tex]f=T\cos{\Theta}-m\ddot{x}.\hspace{.1 in}(4)[/tex]

    Substituting (3), (4) into (1) gives

    [tex]\tau_{CM}=T\cos{\Theta}-m\ddot{x}-rT\cos{\Theta}=I\dfrac{\ddot{x}}{R}[/tex]
    [tex]\Rightarrow T\cos{\Theta}(1-r)=\ddot{x}\left(\dfrac{I}{R}+m\right)[/tex]
    [tex]\Rightarrow \ddot{x}=T\cos{\Theta}\dfrac{1-r}{m+\dfrac{I}{R}}.[/tex]

    [Edit] In the diagram in my book, the string is wrapped counter-clockwise around the inside of the spool so it comes out from the underside on the right and is pulled towards the right ([tex]+\hat{x}[/tex]).
     
    Last edited: Jan 11, 2010
  2. jcsd
  3. Jan 11, 2010 #2
    What sort of frictional force would you account for if the spool is rolling without slipping?
     
  4. Jan 11, 2010 #3
    There is a static frictional force between the spool and the ground that keeps the spool from slipping. Without this force the spool would not roll smoothly along the ground. This is the basis of the non-slip condition.
     
    Last edited: Jan 11, 2010
  5. Jan 11, 2010 #4
    I see...though you wrote the net torque as a difference between the applied torque and the frictional force in the first line (the difference does not look like it gives another torque value).
     
  6. Jan 11, 2010 #5
    You are correct. That should have read

    [tex]\tau_{CM}=Rf-rT\cos{\Theta}=I\alpha[/tex].

    Thank you.

    Other than that mistake and the obvious changes it will make in the rest of my algebra, does the work look correct logically?
     
  7. Jan 11, 2010 #6
    Other than that, your work looks good (though you might want to wait for other opinions).
     
  8. Jan 11, 2010 #7
    Thank you, Gear300. The final answer I obtain after your correction is now

    [tex]\ddot{x}=T\cos{\Theta}\dfrac{R-r}{mR+\dfrac{I}{R}}[/tex]

    which at least has the dimensions of an acceleration now.
     
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