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Rotational Mechanics - Pulley and Bucket

  1. Jan 19, 2007 #1
    1. The problem statement, all variables and given/known data

    "A cylindrical 4.91 pulley with a radius of 0.338 m is used to lower a 3.80 kg bucket into a well. The bucket starts from rest and falls for 4.06 s. The acceleration of gravity is 9.81 m/s². What is the linear acceleration of the falling bucket? Answer in units of m/s²."

    \\ m_{p}=4.91 kg\\

    \\ r=0.338 m\\

    \\ m_{p}=3.80 kg\\

    \\ v_{i}=0 m/s\\

    \\ \Delta t=4.06 s\\

    \\ a_{t}=?

    2. Relevant equations

    \\ a_{t} = r\alpha\\

    \\ \Tau\=\I\alpha = Fd sin \theta\\

    \\ I_{cylinder}\=\frac{1}{2} r^2\\

    3. The attempt at a solution
    \\ \tau = F_{g}r = I\alpha\\
    \\ m_{b}gr = \frac{1}{2}mr^2\alpha\\
    \\ \alpha = \frac{2m_{b}g}{m_{p}r}\\
    \\ a_{t} = \frac{2m_{b}g}{m_{p}\\

    Now, I thought that was a perfectly logical way to get the answer, but when I submitted it, the answer was wrong. Can anyone tell me what I'm doing wrong here?

    Edit: And I seem to be having difficulties putting breaks into my TeX, heh.
    Last edited: Jan 19, 2007
  2. jcsd
  3. Jan 19, 2007 #2
    torque is equal to the rate of change of angular momentum. That is-- T=d/dt(Lp+Lb)
    I think that should work out.
  4. Jan 19, 2007 #3
    your approach just doesnt take into account the mass of the bucket.
  5. Jan 19, 2007 #4
    What do you mean by this? I used the mass of the bucket to calculate [tex]F_{g}[/tex], which I used to find the torque that I could divide by [tex]I[/tex] to find [tex]\alpha[/tex].
  6. Jan 19, 2007 #5
    Well that makes sense if Fg just acted on the pulley. If it acts over the whole system, there is some difference.
    does that make sense? now you can distribute the d/dt to get your answer.
  7. Jan 19, 2007 #6
    I am going to try to put Ja4Coltrane's math into more conceptual terms.

    Note: Remember those mass/pulley systems. This problem is just a rotational analogy.

    1st: You are correct that the only force is the wieght (mg) of the bucket. However F=ma and there are two masses here. Now F is a net force so it is F=Fg - torque

    The two masses are the rotational mass (I) of the cylinder (which you included) but also includes the mass of the bucket (which you forgot).

    2nd: Treat both masses as one system and set Fg= both masses x(acceleration).

    3. use some algebra and turn angular acceleration into (translational) acceleration. I have a feeling that some factors of r might cancel out.

    I hope this is helpful although the mathematical explanation is suffecient for all calculus lovers.
    Last edited: Jan 19, 2007
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