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Unwinding Cylinder-Dynamics

  1. Sep 27, 2008 #1
    Unwinding Cylinder--Dynamics

    ok check it out, ive been doing my homework all day and im on the second to last part of the last problem. its a long problem and i've solved the first like 5 parts, now im on part F and im stuck and trying not to use the hint tool.

    1. The problem statement, all variables and given/known data


    A cylinder with a moment of inertia I (about its axis of symmetry), mass m, and radius r has a massless string wrapped around it which is tied to the ceiling (Intro 1 figure) .

    At time t=0 the cylinder is released from rest at height h above the ground. Use g for the magnitude of the acceleration of gravity. Assume that the string does not slip on the cylinder. Let v_vec represent the instantaneous velocity of the center of mass of the cylinder, and let omega_vec represent the instantaneous angular velocity of the cylinder about its center of mass. Note that there are no horizontal forces present, so for this problem \vec{v} = -v \hat{y} and \vec{\omega}= - \omega \hat{z}.
    intro1
    MRB_ke_2_1_002.jpg

    Part A
    The string constrains the rotational and translational motion of the falling cylinder, given that it doesn't slip. What is the relationship between the magnitude of the angular velocity omega and that of the velocity v of the center of mass of the cylinder?
    Express omega in terms of v and r.
    omega =\frac{v}{r}
    Correct

    Part B
    Let's look at some limiting cases as a way to build your intuition and also to check your answers. If you can't answer these questions now, work through other parts of this problem first and then consider these special cases using your final analytic answer.

    In the limit that the moment of inertia I \to 0 while the mass m remains finite, what magnitudes would you expect for the tension T in the vertical section of string and the downward acceleration a of the center of mass?
    Note: This is a hypothetical cylinder with all its mass concentrated along its axis. The rest of the cylinder (i.e. the bulk) is massless.
    Choose the option that best describes the limiting values of T and a under the conditions given.

    T=0 and a=0
    T=0 and a=g
    T=mg and a=0
    T=\infty and a=g
    T=0 and a=\infty
    T=mg and a=g

    Correct

    The string provides the torque that transfers gavitational potential energy into rotational kinetic energy. In the limit I \to 0 the kinetic energy of rotation K_{\rm rot} \to 0 as well, so in this case the string has no work to do.
    Part C
    Imagine that the string is wound around the center axle of a yo-yo; the axle radius is r_axle, but the yo-yo casing has a radius r_{\rm casing} \gg r_{\rm axle} and moment of inertia I\gg m r_{\rm axle}^2. In the limit the moment of inertia of the yo-yo I \to \infty and the mass m of the yo-yo remains finite, what magnitudes would you expect for the tension T in the vertical section of string and the downward acceleration a of the center of mass?
    Choose the option that best describes the limiting values of T and a under the conditions given.

    T=0 and a=0
    T=\infty and a=0
    T=mg and a=0
    T=\infty and a=g
    T=0 and a=\infty
    T=\infty and a=\infty

    Correct

    As I \to \infty the yo-yo becomes increasingly unresponsive (has a smaller angular acceleration) to any applied torque; it will seem like it cannot begin rotating. If it can't rotate, the string can't unwind, so the center of mass can't fall, and the net force on the yo-yo must be zero.
    Part D
    Using Newton's 2nd law, complete the equation of motion in the vertical direction y_unit that describes the translational motion of the cylinder.
    Express your answer in terms of the tension T in the vertical section of string, m, and g; a positive answer indicates an upward acceleration.
    m a_y = \sum (F_y) T-mg
    Correct

    Part E
    Using the equation of rotational motion and the definition of torque \vec{\tau} = \vec{r} \times \vec{F}, complete the equation of rotational motion of the cylinder about its center of mass.
    Your answer should include the tension T in the vertical section of string and the radius r. A positive answer indicates a counterclockwise torque about the center of mass (in the z_unit direction).
    I \vec{\alpha} = \sum(\vec{\tau}) -Tr
    Correct

    Part F
    In other parts of this problem expressions have been found for the vertical acceleration of the cylinder a_y and the angular acceleration alpha of the cylinder in the z_unit direction; both expressions include an unknown variable, namely, the tension T in the vertical section of string The string constrains the rotational and vertical motions, providing a third equation relating a_y and alpha. Solve these three equations to find the vertical acceleration, a_y, of the center of mass of the cylinder.
    Express a_y in terms of g, m, r, and I; a positive answer indicates upward acceleration.


    2. Relevant equations
    m a_y = \sum (F_y) T-mg
    I \vec{\alpha} = \sum(\vec{\tau}) -Tr
    omega =\frac{v}{r}


    3. The attempt at a solution

    so i figured, in order to relate a_y and alpha i could just solve for T in both equations and then set them equal and then solve for a_y, BUT i guess its not that simple, because after solving for it i got, -\frac{I{\alpha}}{mr}-g but the program said it doesnt rely on the variable Ialpha, so i tried to take the alpha out but that didnt work either. now im just sitting wondering how i can find the 3rd equation relating the two so i can solve them and get a_y.

    any help would be appreciated.

    PS: i'm sorry i don't really know how to work the formatting when it comes to things like alpha and omega and stuff... so yeah. =x
     
  2. jcsd
  3. Sep 28, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Unwinding Cylinder--Dynamics

    The angular acceleration [itex]\alpha[/itex] relates to linear acceleration [itex]a[/itex] in the same way that [itex]\omega[/itex] relates to [itex]v[/itex].

    (Read about using Latex here: Introducing Latex & Latex command window)
     
  4. Sep 28, 2008 #3
    Re: Unwinding Cylinder--Dynamics

    ok thanks, but what exactly does that mean. does that mean like they adhere to the same laws of motions and have the same equations of motion? like a= v/t? except [tex]\alpha[/tex]=[tex]\omega[/tex]/t?

    should i just change [tex]\alpha[/tex] to a_y in the equation I[tex]\alpha[/tex] ??? then i would get Ia = -Tr ?

    but then what would i do?
     
  5. Sep 28, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Unwinding Cylinder--Dynamics

    Start with what you already know, the condition for the string not to slip:

    [tex]v = \omega r[/tex]

    Now take the derivative to find a similar relationship between acceleration and [itex]\alpha[/itex].
     
  6. Sep 28, 2008 #5
    Re: Unwinding Cylinder--Dynamics

    so something like...

    a = [tex]\alpha[/tex]r ??
     
  7. Sep 28, 2008 #6
    Re: Unwinding Cylinder--Dynamics

    ok i got it thanks a lot doc al.

    :D

    a/r = alpha is what i used.
     
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