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Homework Help: Rotational Mechanics - Pulley and Bucket

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

    Problem:
    "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."

    Givens:
    [tex]
    \\ m_{p}=4.91 kg\\

    \\ r=0.338 m\\

    \\ m_{p}=3.80 kg\\

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

    \\ \Delta t=4.06 s\\
    [/tex]

    Unknown:
    [tex]
    \\ a_{t}=?
    [/tex]

    2. Relevant equations

    [tex]
    \\ a_{t} = r\alpha\\

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

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

    [/tex]
    3. The attempt at a solution
    [tex]
    \\ \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}\\
    [/tex]

    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.
    T=dL/dt=d(Lp+Lbucket)/dt=d(Iw+mvr)/dt
    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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