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Moment of Inertia question

  1. Apr 16, 2007 #1
    1. The problem statement, all variables and given/known data

    A 1200kg car is being unloaded by a winch. At the moment shown the gearbox shaft of the winch breaks and the car falls from rest. During the car's fall, there is no slipping between the (massless) rope, the pulley, and the winch drum. The moment of inertia of the winch drum is 320kgm^2 and that of the pulley is 4kgm^2. The radius of the winch drum is 0.8m and that of the pulley is 0.3m.

    Find the speed of the car as it hits the water?

    Note: The winch pulley is on one side of the system with the smaller pulley in the middle. The car is then on the other side of the system hanging. The car is also 5m above the water.

    2. Relevant equations

    I will states these in the form of linear motion because I don't know how to make omega and tao...etc

    vf = v1 +at
    Torque = I * Alpha
    Acceleration Tangential = radius * Alpha
    L (momentum) = I * omega

    3. The attempt at a solution

    I have tried all kinds of attempts at the solution. I tired conservation of energy but I don't think that is the way. I tried conservation of momentum and I'm not sure how to set it up properly. I figure that once you get the tangential acceleration you can easily solve for tangential velocity.

    I don't know how to relate two pulleys together either. The equations ask for a radius and I know two radiuses. The book doesn't help much either.

    Thanks in advance. Any help is great.
  2. jcsd
  3. Apr 16, 2007 #2
    It looks like there are forces involved so Newtons second.

    Since there is no slippage, three items will accelerate: the car, the first pulley, and the winch drum. Tensions tie it all together. Also since there is no slippage, the tangential speed over the two rotating parts will be the same.

    Can you do an FBD? That might be the best place to start.
  4. Apr 16, 2007 #3
    I'm still not getting much of anywhere. I solved for T= m(g-a) when the positive y direction is pointing towards the ground. Plugging Acceleration tangential into into a I get T=m(g-r*alpha). Now I'm guessing that Tension has to be related to the inertial mass of the pulley and the winch. Do I sum up Torque with I*alpha of the first pulley and I*alpha of the winch?

    I worked backwards a little with the answer and couldn't find any numbers that were noticible. The answer is 8.23m/s and acceleration tangential is 6.77329m/s^2.

  5. Apr 16, 2007 #4


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    Conservation of energy.

    energy before = mgh ( car above water)
    energy after = 1/2*m*v^2 + 1/2 *I1*O1^2 + 1/2*I2*O2^2
    where O1 and O2 are the angular velocites of the rotating elements and I1 and I2 their MoI's. If you relate O to v through the radius, you have one equation in v.

    It is completely elementary.
  6. Apr 16, 2007 #5
    I'll give it a shot. Maybe I was thinking WAY to hard for this one.... I wasn't there the day the teacher did most of this chapter.
  7. Apr 16, 2007 #6
    You can try Mentz's approach, but should know how to do problem with Newtons laws. In general, best practice is to start with one end which you did.

    ma=T1-mg Then go around system using new tensions.

    Note for pulleys such as these T*R=Torque=I*alpha; multiplying again by R,
    T*R^2=I*a where a is the tangential or linear acceleration.

    so T1-T2=a*I/R^2

    Finally T2*R'=I'*alpha', where R and I are for the drum. T2=aI'/R'^2

    T1-T2=a*I/R^2 (for pulley)
    simplify and collect terms,
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