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Acceleration of a Pulley

  1. Dec 7, 2007 #1
    1. The problem statement, all variables and given/known data

    A string is wrapped around a uniform solid cylinder of radius R, cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass, m.

    Find the magnitude "alpha" of the angular acceleration of the cylinder as the block descends in terms of r, radius, and g, acceleration due to gravity.


    2. Relevant equations

    F=ma
    F=mg-T
    I x alpha= -rT
    I=1/2mr^2
    alpha= -a/r



    3. The attempt at a solution

    I've been trying to solve this system of equations to find alpha in terms of g and r, but everytime I do so, I end up with g and alpha cancelling out.

    I keep ending up with -alpha x r = mg + 1/2r x alpha

    I think perhaps I'm subsituting something incorrectly? Somebody want to lead me in the correct direction please??
     
  2. jcsd
  3. Dec 7, 2007 #2

    Shooting Star

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    Does the block descend with accn g? If not, where does the energy go? Use the total energy to make the block accelerate and the cylinder rotate.
     
  4. Dec 7, 2007 #3
    I'm assuming it descends with accn g, because it doesn't tell me otherwise... and I'm not sure what you mean by total energy?
     
  5. Dec 7, 2007 #4

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    Possibly that's why those are canceling out.

    The block has a energy of mgh at a height of h. As it descends, this energy goes into the sum of KE+PE of the block + the rotational KE of the cylinder. So, the block can't descend with accn g, but at a lesser rate.

    Let the height of the block be y below the axis of the cylinder at time t, and w be the angular speed of the clndr at t. You can set up eqns relating (dy/dt)^2 and w^2 through energy consvn. Then differentiate to find alpha, a etc. Right now only this comes to my mind as the easiest.
     
  6. Dec 7, 2007 #5
    AHH, when I entered an answer that implied it was equal to g, it gave me feedback that it had to be a lesser rate. Okay, I've never solved these equations through differentiation but I suppose I can try??
     
  7. Dec 7, 2007 #6

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    Of course you can try. We are all here.

    Initially, just equate the energies of both the block and clndr at time t to the initial energy. That won't require differentiation. But geometry. After that, we'll see.
     
  8. Dec 7, 2007 #7
    Alright, it told me to substitute for certain values in equations, and then to solve the system of equations by eliminating T, tension. I'm not sure where the energies come into play here...

    I've literally spent hours the past 3 days on this problem, I'm sick of not understanding this danged thing.
     
  9. Dec 7, 2007 #8
    This is what I've gotten...

    F=ma
    F=mg-T
    -T=1/2m r alpha
    a= -alpha r

    I then substitute to get F=m(-alpha r)=mg-T

    -alpha r= g-T/m

    but if I substitute in for T I will cancel alpha which is what I'm looking for!
     
  10. Dec 8, 2007 #9

    rl.bhat

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    Since cylinder is moving freely, the block is accelerating with g. Hence the force mg rotates the wheel. The torque is given by mgR. And this must be equal to I*alpha.
    So mgR = 1/2*m*R^2*alpha. Hence alpha = 2g/R
     
  11. Dec 8, 2007 #10

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    Ummm...the block is not falling with accn g, because it has to impart energy to itself, as well as increase the rotational KE of the cylinder.

    Suppose T is the tension in the string. The torque on the cylinder is T*r.

    For the falling block, force = m*d^y/dt^2 = mg –T. – (1)

    For the rotating cylinder, torque = T*r = I*A => T = IA/r -- (2), where I’m writing A for alpha. So, A=dw/dt, where w is the ang speed of the cylinder.

    Also, if v is the linear speed of the circumference of the cylinder, then
    dy/dt = v = rw => d^2y/dt^2 = r*dw/dt = rA.

    Then, from (1) and (2),

    mrA = mg – IA/r = mg – (mr^2/2)A/r (putting I = mr^2/2)
    => (mr + mr/2)A = mg
    => A = 2g/3r.
     
  12. Dec 8, 2007 #11

    rl.bhat

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    Yes. You are right.
     
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