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Acceleration of a Rolling Cylinder

  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that a=⅔g sin θ will find the acceleration of a cylinder rolling down an incline of angle θ.

    2. Relevant equations
    a=⅔g sin θ

    3. The attempt at a solution
    I don't understand how to do this without numbers. I have a feeling the principle of conservation of mechanical energy should be used somehow?
     
  2. jcsd
  3. Oct 22, 2011 #2

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    Hi Megzzy! :smile:

    You can basically do this 2 ways.

    You can set up an equation for the conservation of mechanical energy and rewrite it to find "a" (using the derivative).

    Or you can set up a diagram with the forces.
    Set up the equations of movement and find "a".


    Either way, you will need a the moment of inertia of a cylinder.
    And you will need to set up a relation between the angle the cylinder has rolled and the distance it has rolled.

    How much do you already know?
     
  4. Oct 22, 2011 #3
    thanks for the help!

    I have no values so that is why I am having problems I think. I know the moment of inertia of a cylinder is I=1/2MR^2. I'm still lost though.
     
  5. Oct 22, 2011 #4

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    That's okay.
    And yes, that is the right moment of inertia. :smile:

    Don't worry yet whether you have numbers or not.
    Start with any equations you know that might be useful.

    Which (relevant) formulas do you know?

    Like the one for kinetic energy of a linear movement?
    And the one for kinetic energy of a rotating movement?
    Energy of gravity?

    Or the one for the force of gravity?
    The equation for the resulting force on the cylinder?
    The equation for the resulting moment on the cylinder?
     
  6. Oct 22, 2011 #5
    I know KE rotational= 1/2Iw^2
    KE linear=1/2mv^2
    PE=mgh

    I'm not sure where to go from there.
     
  7. Oct 22, 2011 #6

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    This will get you on the way.

    Now suppose the cylinder rolls down a certain distance d.
    What will be the change in angle of the cylinder?
    And by what amount will its height decrease?

    Btw, what I'm aiming for are the relations between v, w, h, and a.
    But to get there, we're introducing d and the angle of the cylinder.
     
  8. Oct 22, 2011 #7
    So I am picturing the right triangle?
    d is therefore the hypotenuse and the height is opposite the angle of inclination.

    I'm sorry this process is taking so long! The question is just really strange to me.
     
  9. Oct 22, 2011 #8

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    Exactly. :)
     
  10. Oct 22, 2011 #9
    alright thank you!
    I am still unsure of where to go from there and how to connect that with v,w,h and a.
     
  11. Oct 22, 2011 #10

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    Well, you pictured a right triangle with d on the hypotenusa and h as the opposite of the angle of inclination theta.

    So which formula describes the relation between d, h, and theta?
     
  12. Oct 22, 2011 #11
    all I can think of is sin theta=height/distance
     
  13. Oct 22, 2011 #12

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    That's the one!

    Can you rewrite that in the form:
    height = ?
    And substitute it in the total energy formula?


    Btw, I'm off to bed now. :zzz:

    I'll give you another hint to work on.
    Can you find a formula that relates the angular velocity w of the cylinder to the linear velocity v of the cylinder?
     
  14. Oct 23, 2011 #13
    I'd get height=sin theta x distance.
    I don't understand how it plugs into the total energy formula though.
     
  15. Oct 23, 2011 #14

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    Let's see...

    Your total energy (which must be conserved) is:
    E = KE rotational + KE linear + PE
    [itex]E= {1 \over 2}I \omega^2 + {1 \over 2}m v^2 + mgh[/itex]

    If the ball rolls a distance d along the slope, the height h decreases by (sin theta x distance).
    So the new energy is:
    [itex]E= {1 \over 2}I \omega^2 + {1 \over 2}m v^2 + mg(h - \sin(\theta) \times d)[/itex].
    Note that w and v will have greater values in this equation to conserve energy.


    Now when the cylinder has rolled down a distance d, how many radians did it turn?
     
  16. Oct 24, 2011 #15
    radians=degrees x pi/180?
    Or am I supposed to be making an actual estimate here?
     
  17. Oct 25, 2011 #16

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    Yes, that is the relation between radians and degrees.
    For problems like this, it is easiest to work in radians.

    Suppose the cylinder rolled over an angle of 1 radian, which distance would it have rolled?
     
  18. Oct 27, 2011 #17

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    Let me try to make it clearer.

    The circumference of the cylinder is [itex]2\pi R[/itex].
    This means that an angle of [itex]2\pi[/itex] radians corresponds to a distance of [itex]2\pi R[/itex].

    So if the cylinder rolls over an angle of 1 radian, it covers a distance of [itex]d = {2\pi R \over 2\pi} = R[/itex].

    Are you still with me?
     
  19. Oct 28, 2011 #18
    Sorry for taking so long to get back to you! It's been a busy week.
    I seem to understand what you have explained so far.
     
  20. Oct 28, 2011 #19

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    Okay, so if the cylinder rolls 2 radians the distance is 2R.
    More generally the [itex]distance = \phi R[/itex]
    if [itex]\phi[/itex] is the angle over which the cylinder has rolled.

    The speed [itex]v[/itex] of the cylinder is the change in distance per unit of time.
    This is the derivative. Did you know that?

    So [itex]v = {d \over dt} distance[/itex].
    Similarly the angular velocity is the change in angle per unit of time.
    That is, [itex]\omega = {d\phi \over dt}[/itex].

    Could you apply this to [itex]d = \phi R[/itex] to find the relation between [itex]v[/itex] and [itex]\omega[/itex]?
     
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