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Rotational Motion and linear accerlation

  1. Dec 18, 2009 #1
    1. The problem statement, all variables and given/known data

    A solid cylinder of weight 50 lb and radius 3.0 inches has a light thin tape wound around it. The tap passes over a light smooth fixed pulley to a 10 lb body hanging vertically in the air. If the plane on which the cylinder moves is inclined 30 degrees to the horizontal, find (a) the linear acceleration of the cylinder down the inclined plane, and (b) the tension of the tabpe, assuming no slippage.

    2. Relevant equations

    [tex]\sum _\tau = I_c\alpha[/tex]

    [tex]I_c = \frac{M_c R_0^2}{2}[/tex]

    3. The attempt at a solution

    I'm having an issue seeing the relationship linear acceleration and Newton's 2nd law for rotational motion.

    [STRIKE]Am I correct in thinking about substituting [tex]\alpha = \omega^2 R[/tex] into the 2nd law, then applying appropriate forces? [/STRIKE]

    Edit: See next post
    Last edited: Dec 18, 2009
  2. jcsd
  3. Dec 18, 2009 #2
    Ok... scratch what I said earlier.

    [tex]\alpha = \frac{\tau}{I_c} [/tex]

    [tex]\alpha = \frac{a_t}{R} [/tex]


    [tex]\frac{\tau}{I_c} = \frac{a_t}{R}[/tex]

    which means [tex]\tau R = a_t I_c[/tex] (*)

    A force diagram tells me

    [tex]\tau = m_Ag sin\vartheta - m_Bg[/tex]

    The cylinder rolls down the incline [tex]\vartheta[/tex] under the influence of gravity and the weight generated by the blocks mass detracts from that value via the tape.

    So, plugging [tex]I_c[/tex] for a solid cylinder and the derived value for [tex]\tau[/tex] into * we get:

    [tex]m_Ag sin\vartheta - m_Bg = \frac{m_AR^2a_t}{2} [/tex]

    Meaning that [tex]a_t[/tex] (or the linear accleration) would be:

    [tex]a_t = 2 (\frac{gsin\vartheta}{R} - \frac{m_Bg}{m_aR^2}) [/tex]

    Granted.. I didn't take the units into consideration.. but I am hoping for a more conceptual and general understanding of the problem, units aside.

    Am I closer to being on the right track?
    Last edited: Dec 18, 2009
  4. Dec 18, 2009 #3

    Doc Al

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    Careful. What you are calling a_t cannot simply be the acceleration of the cylinder down the incline, since the tape is also accelerating up the incline. (Figure out the connection between α and the two linear accelerations.)

    Careful. Since the cylinder and the block have different accelerations, don't try to apply Newton's 2nd law in one step. Instead, write a separate force equation for each, then combine them later.
  5. Dec 18, 2009 #4
    The force of the tension in the tape should be subtracted from the force of gravity compelling the cylinder to roll down the incline. So [tex]\tau = m_Ag sin\vartheta - T[/tex] where T is tension. Correct?

    I'm calculating [tex]T = m_Bg[/tex] as the m_B is free hanging and pulling on the tape.

    Wouldn't the linear acceleration of the cylinder and the block (or body as I described in the problem statement) be the same as they are a system (linked by the tape)?
  6. Dec 18, 2009 #5
    I've attached a picture of the set up.... just for reference.

    Thanks for the assistance.

    Attached Files:

  7. Dec 18, 2009 #6

    Doc Al

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    It's true that the net force on the cylinder is mAgsinθ - T. (I don't know why you are labeling that net force 'tau', since 'tau' usually stands for torque.) Now apply Newton's 2nd law to the cylinder.

    No. If the tension equaled the weight of the hanging mass, the mass would not accelerate.

    No. The cylinder unwinds the tape. (Their accelerations are connected, just not equal.)
  8. Dec 18, 2009 #7

    Doc Al

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    Interesting. I imagined the orientation of the cylinder and tape reversed. (With the tape being flat against the incline.)
  9. Dec 18, 2009 #8
    Well... I have had trouble with Torque vs Force.... So I'm guessing [tex]\alpha = \frac{\tau}{I_c}[/tex] isn't valid? If it is.. how to we calculate tau?

    Concerning Tension...

    Would [tex]T_{Net} = m_Agsin\vartheta - m_Bg[/tex] ?
  10. Dec 18, 2009 #9
    Does that change your suggestions?
  11. Dec 18, 2009 #10

    Doc Al

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    What's the definition of torque?

    What do you mean by 'net' tension? By setting up equations, you'll solve for the tension.

    Is that a diagram you created or was it provided with the problem? Where the cylinder touches the incline, is there friction?
  12. Dec 18, 2009 #11
    Well mathematically [tex]\tau = r \times F[/tex] ... with r being radius and F being force

    I suppose that means that torque is force applied radially. So wouldn't that mean that the torque of the tape around the cylinder affect the tension in the tape? Also, does the Force in the equation for torque equal the force of the weight of the block pulling through the tape?

    Idea: Edit:see better idea

    [tex]\tau = Rm_Agsin\vartheta[/tex]


    [tex]R^2m_Agsin\vartheta = \frac{m_AR^2a_t}{2}[/tex]


    [tex]gsin\vartheta = \frac{a_t}{2}[/tex]

    [tex]a_t = \frac{gsin\vartheta}{2}[/tex] right? closer at least?

    Better idea:

    [tex]\tau = r \times F[/tex]

    [tex]\tau = r m_Agsin\vartheta -m_Bg[/tex]

    actually.. this is right back where I started........... :confused:
    I'm denoting net tension as the tension in the tape as deduced by the force from the cylinder and the force from the block.

    The pulley has negligible mass and is fixed... so wouldn't the tension in either piece of the tape be the same?

    The diagram is an mspaint translation of the one provided with the problem. Friction is neglected.
    Last edited: Dec 18, 2009
  13. Dec 18, 2009 #12

    Doc Al

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    OK, so what forces act on the cylinder? What torque do those forces exert about its center of mass.

    Only forces with a tangential component exert a torque.
    Not sure what that means.
    No. The force that the tape pulls does not equal the weight of the block.

    Rather than try to figure out the tension in your head, set up equations for each mass then solve for the tension.

  14. Dec 18, 2009 #13
    So wouldn't that mean that the torque of the tape around the cylinder affect the tension in the tape?
    Would the torque generated by the cylinder apply a force to the tape?
  15. Dec 18, 2009 #14
    I'm not sure I follow what you mean by set up equations for each mass . My best guess leads me back to the same world I've been trapped in for nearly 3 hours....

    [tex]\tau = R \times F = Rm_Agsin\vartheta + T[/tex]

    [tex]F_B = ma = m_Bg -T[/tex]

    .:. [tex]Rm_Agsin\vartheta = -m_Bg[/tex]

    There's no unknown in this equation. What is it that I am missing? Should I be solving for a system?



    [tex]\tau = R \times F = RTsin\vartheta[/tex]

    However.... I'm still seeing [tex]F_B = ma = m_Bg -T[/tex] which says [tex]T = -m_Bg[/tex] which you've told me isn't the case.............
    Last edited: Dec 18, 2009
  16. Dec 18, 2009 #15

    Doc Al

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    Identify the forces on each mass (the block and the cylinder). Apply Newton's 2nd law to each. You'll get two equations for translational forces/accelerations and a third for the rotational torque/acceleration for the cylinder. Combine those three equations (plus a constraint equation relating the rotational and translational accelerations) and you can solve for the two accelerations and the tension.

    Do it step by step.
  17. Dec 18, 2009 #16
    Ok... so

    Forces for cylinder:


    [tex]m_Agsin\vartheta - T = m_Aa[/tex] =>
    [tex]T = m_Agsin\vartheta -m_Aa[/tex]


    [tex]\tau = R \times F = R m_Aasin\vartheta = I_c\alpha[/tex] =>
    [tex]\alpha = \frac{R m_Aasing\vartheta}{I_c}[/tex]

    Forces for block:

    [tex]T- m_Bg = m_Ba[/tex] =>
    [tex]T = m_Ba + m_Bg[/tex]


    Solving system for T => [tex]m_Agsin\vartheta -m_Aa = m_Ba + m_Bg[/tex]

    So [tex]a = \frac{m_Agsin\vartheta - m_Bg}{m_A + m_B}[/tex]

    BUT... linear acceleration from angular:

    [tex]a = R\alpha[/tex]

    I don't believe that I'm using the right value for Force in the torque equation.... perhaps other problems....

    Am I even using the right base equations.... it's starting to seem like maybe that's a no.

    I do, however, appriciate the ever so gentle guidance (in all sincerity).... I'll have done every possible algebraic operation to these equations before it's over....

    I look forward to seeing what angle I should work next.
  18. Dec 19, 2009 #17

    Doc Al

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    OK, but label that acceleration a_A, since it does not equal the acceleration of the block. You've chosen the acceleration such that it will have a positive value if the cylinder goes down the incline; nothing wrong with that.

    No. Of the forces acting on the cylinder, which one exerts a torque about its center?

    Again, give that acceleration a label such as a_B. Here you chose a positive value of 'a' as up. (I always let 'a' stand for the positive magnitude of the acceleration and I always write my equations assuming the most likely direction for the object's acceleration. Here I would assume that the block accelerates down.)

    The big problems are:
    (1) Not recognizing that the two linear accelerations are different.
    (2) Using the wrong force in your torque equation.
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