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Newton's 2nd law - cylinder

  1. Jul 14, 2008 #1
    1. The problem statement, all variables and given/known data
    http://ready2goxtr.googlepages.com/problem1055.jpg

    A cylinder having a mass of 2.0kg can rotate about its central axis through point O. Forces are applied as show: F1 = 6.0N, F2 = 4.0N, F3 = 2.0 N, and F4 = 5.0N. Also, r = 5.0cm and R = 12 cm. Find the (a) magnitude and (b) direction of the angular acceleration of the cylinder.




    2. Relevant equations

    Tnet = I [tex]\alpha[/tex]
    T = rFsin(theta)
    I = 1/2MR^2



    3. The attempt at a solution

    So far I figured i should find the Net torque. So..

    T1 = F1R = 6 * .12 = .72 N * m
    T2 = F1R = 4 * .12 = .48 N * m
    T3 = F1Rsin(theta) =??
    T4 = 0

    How would I find T3?
     
  2. jcsd
  3. Jul 14, 2008 #2

    rock.freak667

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    Is T3 just not 2.0*0.05 since the angle is 90?
     
  4. Jul 14, 2008 #3
    Its 90 with r not R
     
  5. Jul 14, 2008 #4

    rock.freak667

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    Why would you need to find it with R?

    Torque=Frsine(theta)...r is the distance of the force from the centre of rotation.
     
  6. Jul 14, 2008 #5
    If that was the case. couldnt I say T4 has a 90 degree angle too?
     
  7. Jul 14, 2008 #6

    rock.freak667

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    T4 has an angle with the length of R.But since you are taking torques about the centre,O, F4 produces no torque since the distance between the force and centre of rotation is 0.
     
  8. Jul 14, 2008 #7
    Okay
    So then i think I can do the Tnet => -F1 + F2 + F3 = Tnet Tnet = -.14 N * m

    Tnet = Ialpha

    Tnet/I = Alpha?
     
  9. Jul 14, 2008 #8

    rock.freak667

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    Yes, that should be correct.
     
  10. Jul 14, 2008 #9
    hey. =) Im pretty good with paint huh+ D=
     
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