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Rotational/Linear motion & friction

  1. Mar 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A 2.4 kg block rests on a 30° slope and is attached by a string of negligible mass to a solid drum of mass 0.80 kg and radius 5.0 cm, as shown in Fig. 10.29. When released, the block accelerates down the slope at 1.2 m/s2. What is the coefficient of friction between block and slope?

    2. Relevant equations

    Ff=mu*N

    a=alpha*r

    Torque=I*alpha


    3. The attempt at a solution

    Ok the first thing I did was draw a free-body diagram of the block and the drum. I have Fnet=mg-Tension-mu*Fnormal then and set Tension=torque and got T=1/2*Mdrum*a and put that back into the original equation and solved for mu, but I didn't get the right answer. Is this even the right way to do this?
     
  2. jcsd
  3. Mar 11, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Nick_L! Welcome to PF! :smile:
    I don't get it … what's the drum doing? :confused:

    is it rotating on a fixed horizontal axis, with the string unwinding, or on a fixed vertical axis, or is it sliding down the hill on its base?
     
  4. Mar 11, 2009 #3
    Here is the picture associated with the problem

    12-47.gif
     
  5. Mar 11, 2009 #4

    tiny-tim

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    ok … rotating on a fixed horizontal axis, presumably with a frictionless axle, and with the string unwinding …

    so use the work-energy theorem … work done = energy lost, using the mass of the object and the moment of inertia of the drum :wink:
     
  6. Mar 11, 2009 #5
    Ok so I would have Ffriction*dcos(30)=1/2*massbox*v2+1/2I*omega2? If that's right it makes sense, but how would I find displacement or velocity when I only have the acceleration?
     
  7. Mar 11, 2009 #6

    tiny-tim

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    (you left out gravity)

    sorry, I missed the acceleration :redface:

    in that case, just use F = ma and torque = Iα and slug it out! :smile:
     
  8. Mar 11, 2009 #7
    Alright, thanks for the help :biggrin:
     
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