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Finding an expression for linear acceleration of a hanging mass

  1. Dec 10, 2011 #1
    A rope of negligible mass is wrapped around a uniform solid cylinder, of radius R, that can rotate freely about its axis, which is horizontal and fixed. A block hangs vertically from the rope and causes the cylinder to spin. The blocks mass, M_B is 4 times the mass, M, of the cylinder. Find an expression for the magnitude of the linear acceleration of the hanging block in terms of R and/or g. (Rotational inertia of a cylinder is I=1/2*MR^2)

    I know the motion of the block is Fnet=M_B*a or M_B*g - T (tension) = M_B*a
    (btw I'm taking up to be negative y direction and down positive y direction)

    I also know that Torque= I * alpha. and also Torque = R T(tension)sin(theta) where theta =1

    so TR = 1/2*MR^2*alpha (from Torque=I*alpha)

    solving for T i have T=1/2*M*R*alpha

    Now if i plug this equation for Tension into F_net = M_B*a i get:

    M_B*g - 1/2*M*R*alpha = M_B*a **(also i know that alpha = a/r)

    solving for a; i found two answers. by computing the algebra differently for the two answers and im not sure which is correct...if either of them even are correct.

    a = g(1+(2M_B/M)) AND a = g/(1+(M/2M_B))

    Any insight is appreciated! thanks!
     
  2. jcsd
  3. Dec 10, 2011 #2

    Simon Bridge

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    After you substitute [itex]\alpha=a/R[/itex], you have:

    [tex] gM_B - \frac{1}{2}Ma = M_B a [/tex]

    How do you get two different solutions for a from that?

    Also note, [itex]M_B=4M[/itex]
     
  4. Dec 11, 2011 #3
    Im not sure what i was doing before...but after recalculating using your imput, i got

    a=2g

    is this what you calculated as well?
     
  5. Dec 11, 2011 #4

    Simon Bridge

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    Rules for homework mean I can give pointers on working but not on answers.
    Ergo - don't be afraid to post your working and reasoning: show me how you got that.

    (It looks like you need to work on your algebra.)
     
  6. Dec 13, 2011 #5
    its not homework, its just a practice problem. and i dont have my work with me at the moment but i figured it out. my algebra was off. thanks for your imput!
     
  7. Dec 13, 2011 #6

    Simon Bridge

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    Of course I'm also too lazy ;) I once wrote in an exam: "the proof of this expression is left as an exercise for the examiner."
     
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