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Belt around pulley's

  1. Apr 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A belt passes over two pulleys O1 (radius R1= 0.4 m) and O2 (radius R2 = 0.6 m) as shown, and points A and B are on the rim of the pulleys. The normal acceleration at A is 0.9 m/s2 and the angular acceleration of pulley O1 is 3.6 rad/s2. Find (a) the belt velocity υb, (b) the belt acceleration ab, and (c) the acceleration of B.


    2. Relevant equations
    a n= v^2/R
    angular acceleration and velocity equations

    3. The attempt at a solution
    I got a by using a n=v^2/R... so .9 = v^2/.4, v = .6 which is correct. I do not know how to go about finding the other values at all.

    answers: (a) 0.6 m/s (b) 1.44 m/s2 (c) 1.56 m/s2
     
  2. jcsd
  3. Apr 21, 2009 #2
    Draw free body diagrams for your two pulleys, label all the forces acting on them, including torques, and use newtons second law to derive equations that can help you solve for your unknowns. You may have to look up the moment of inertia of a pulley (a cylinder I suppose?)
     
  4. Nov 15, 2010 #3
    How's VTech? haha, I just finished it, and I'll post now:

    so we have:

    r1=.4 m
    r2=.6 m
    Ana= .9 m/s^2 (this is normal acceleration for point A)
    alpha1= 3.6 rad/s^2 (this is angular acceleration for O1)

    In order to find Belt Velocity we...

    Ana=(wa^2)(r1) => wa=sqrt((Ana)/r1)) = 1.5 => v=wr=(1.5)(.4)=.6m/s

    In order to find Belt Acceleration we...

    at (tangential acceleration) = (alpha1)(r1) = 1.44 m/s^2

    In order to find acceleration of B:

    we know that at above =1.44m/s^2

    we can solve for Anb=(v^2)/(r2)=(.6^2)/(.6)=.6
    so TOTAL acceleration of aT

    aT=sqrt((at^2)+(Anb^2))=1.56 m/s^2

    Sorry about the sloppy notation :-/
     
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