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Atwood problem

  1. Apr 27, 2007 #1
    Hi all -

    Here's the problem I am having trouble with -

    Basic Atwood machine setup -

    Two masses are hanging from a pulley

    m pulley = 2.0kg

    frictional torque from pulley = .50 Nm

    radius of pulley = .06m

    mass1 = 4.0kg

    mass2 = 2.0kg

    The system is at rest with mass1 exactly 1m above ground. The problem is asking for the amount of time it will take for mass1 to drop the 1m to ground contact.

    I think that T1 and T2 should be relative to the acceleration, but I am getting lost when I try to put the whole thing together. Here's what i have:

    m1g - T1 = m1a

    T2 - m2g = m2a

    Torque net = angular accel. * moment of inertia of pulley (1/2MR^2)

    Torque net should = T1r - T2r - .50Nm (torque from pulley) ??

    I know I need to find accel(y) of the system in order to use a normal knematic equation to solve, so I know that I need (w*r) to finish the problem

    It seems like the tensions need to disappear through some type of substitution, but I am not sure how to make them go away. Also, I am unsure as to what role the frictional torque of the pulley plays in this system.

    ANY help or advice would be much appreciated - Also, I can attach a picture later if the description of the problem seems confusing.


  2. jcsd
  3. Apr 28, 2007 #2
    This is a tough problem. You have 4 forces to account for in addition to two tensions. Personally I have never solved one of this type, but no one is jumping in to help, you have made an effort, and at the risk of further confusing you...

    First the two tensions are not equal. You have masses on each end, which exert pull according to mg. The accelerations of the rope need to be constant. So the a's on the masses can be considered equal and opposite. The two flies in the ointment are the moment of inertia of the pulley, and the frictional torque. The flywheel effect of the pulley can be simplified in linear terms as the Moment of inertia divided by R^2. Finally we are given a frictional torque. This would cause a difference in tension beteen the two sides. The key I believe is converting frictional torque to "displacement" torque by multiplying by the radius of the wheel to once again get tangential acceleration.

    In other words the key is to get this reduced to a a problem where a's and alphas are not mixed, nor masses and I's. Of course after getting a, it becomes a kinematics expression, but I'm thinking thats no sweat. See if this leads anywhere.

    An addendum, that just occurred to me--sort of like a train of three blocks where the middle block is rotational and sliding thru sand while the other two are on ice.
    Last edited: Apr 28, 2007
  4. Apr 28, 2007 #3
    Doc -

    Thanks for your insight - I'm glad someone else thinks this is a difficult problem, as I was beginning to feel like i just wasn't catching on. I am going to take another look at this problem this afternoon when I finish some other work, and I will post my progress. One question - When you refer to the tangential acceleration of the pulley, that is just the product of the frictional torque and the radius of the pulley - How can this be used in an equation? Also, I am thinking off the top of my head that because a(y)= ang accel *r, I can try using that substitution in a force-component style equation. Does that make sense?

    More later......

    Thanks again,

  5. Apr 28, 2007 #4
    Thats where I was headed,

    the masses are easy M1(a)=M1*g-T1 M2(a)=T2-M2*(g)

    For the pulley, I believe, total torque=
    (T1-T2)*R-frictional torque=I*alpha, just need to convert this to the same a used in the above, and solve (I think)
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