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Unwinding Cylinder-Dynamics

  1. Dec 3, 2008 #1
    Unwinding Cylinder--Dynamics

    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

    v_vec = -v(y-hat) and omega_vec = - omega(z-hat)



    2. Relevant equations

    Torque = I(angular_acceleration)
    F= m(linear_acceleration)

    3. The attempt at a solution

    I'm having trouble with part B
    T is the tension of the spring
    if I goes to zero, the object will be turning faster in the -z direction.

    The tension of the spring does less work on its rotation and more work on its linear momentum.

    and so

    If the T went to 0, then linear a should be g right?
    because the tension does no work on either the angular and linear momentum and so the cylinder will just fall right?

    the answer is the second choice
    right?

    if T went to infinity, then there would not be acceleration in the in the -y direction.
    so if T went to infinity, a should be zero right?

    if T equaled mg then linear acceleration would also be zero because the forces of the weight and T cancel out

    right?
     

    Attached Files:

  2. jcsd
  3. Dec 3, 2008 #2

    turin

    User Avatar
    Homework Helper

    Re: Unwinding Cylinder--Dynamics

    Um ... What is the problem asking you to solve? As far as I can see, it is just a bunch of information, but no question.
     
  4. Dec 4, 2008 #3
    Re: Unwinding Cylinder--Dynamics

    im having trouble with part B

    i think when t=0, a = 0 because

    if I goes to zero, t no longer does work on rotation but does work on linear acceleration
    but if the tension goes to 0 then the object will continue to spin but simply fall in the -y direction due to gravity

    right?
     
  5. Dec 4, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Unwinding Cylinder--Dynamics

    I don't quite understand your reasoning (and you seem to have contradicted yourself).

    Why not apply Newton's law to translation and rotation and come up with some equations. Much easier to understand things when you put Newton to work.
     
  6. Dec 5, 2008 #5
    Re: Unwinding Cylinder--Dynamics

    ur right

    i contradicted myself

    anyway

    my logic tells me

    if t = 0, there is no tension and the object should just fall
    and so -y direction its a=g (linear)


    but the equations tells me otherwise
    that when t = 0

    a = 0

    (as seen in picture)

    that does not make sense to me though

    why wouldnt the cylinder fall? gravity is still acting upon it

    and since T=0 and I = 0

    it cant be spinning right?
     

    Attached Files:

  7. Dec 5, 2008 #6

    Doc Al

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    Staff: Mentor

    Re: Unwinding Cylinder--Dynamics

    OK. (But you have to show that T = 0.)


    The equations don't say that. (Please simplify your equation for acceleration. That could be throwing you off.)
     
  8. Dec 5, 2008 #7
    Re: Unwinding Cylinder--Dynamics

    ok

    for vertical a in the -y direction

    a_y = ( -gmr^2 ) / (I + mr^2)

    i dont see T in the equation

    so but it has to affect a_y

    but i do see I

    and if i make that equal to 0

    i get

    a_y = -g

    WOWOW

    thats amazing

    thanks!
     
  9. Dec 5, 2008 #8
    Re: Unwinding Cylinder--Dynamics

    is that right?

    now that im looking at the problem more

    if T = mg
    shouldnt a_y equal to 0?
     
  10. Dec 5, 2008 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Unwinding Cylinder--Dynamics

    Yep.

    Sure, but T ≠ mg.
     
  11. Dec 6, 2008 #10
    Re: Unwinding Cylinder--Dynamics

    im having trouble with part c

    it says its a yoyo

    and the string is wrapped around the axis

    and the radius casing >>>> radius of axis

    but i didnt that would matter because the string is wrapped around the axis

    but then the problem says

    I >>>> m(r_axis)^2

    and then i go ????

    this is what i am thinking

    using the equations

    for then

    T = (gIm) / (I + m(raxis)^2)

    and so since I = infinity

    T = mg

    and when i plug that into a_y equation

    a_y = -(T(r_axis)^2)/I <--am i using the right r?

    and I = infinity

    i get


    a = 0


    is that right??

    another question
    kinda random

    how do you quote previous response?
     

    Attached Files:

  12. Dec 6, 2008 #11

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Unwinding Cylinder--Dynamics

    Good.

    Click on the "Quote" button at the bottom right of the post you want to quote. (Just make sure that the part you want to quote is between quote tags. Just try it.)
     
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