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Incline moment of inertia.

  1. May 1, 2010 #1
    1. The problem statement, all variables and given/known data
    A block with mass m = 5.00 kg slides down a surface inclined 36.9 to the horizontal. The coefficient of kinetic friction is 0.25. A string attached to the block is wrapped around a flywheel has mass 25.0 kg and moment of inertia 0.500 kgm2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.200 m from that axis.
    a) What is the acceleration of the block down the plane? my answer:1.123m/s^2
    b) What is the tension in the string? my answer:14N

    2. Relevant equations



    3. The attempt at a solution
    hey, i did post this before sorry. the reason i put it up again is that i really need to know if i am correct or not and i need to know asap.... thanks
     
  2. jcsd
  3. May 1, 2010 #2

    kuruman

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    Show how you got the numbers, then we will tell you if your approach is correct and where you went wrong, if you did go wrong.
     
  4. May 1, 2010 #3
    the 1st thing i did was: ΣF_x=F_g *sinθ-F_fric-T=ma〗
    then after that i did did this T= 1/2 M*a and subbed that into the original equation to solved for a.

    after i had a i subbed it in T= 1/2 M*a to find T. Please get back to me.. thanks.
     
  5. May 1, 2010 #4

    kuruman

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    It looks right, but why is T = (1/2)Ma ? Where did that come from? Is that always true?
     
  6. May 1, 2010 #5
    That came from τ=r*T=Iα= 1/2 M*r^2*α i think that it would always be true but im not sure, did you check the actual numbers or just the process??????
     
  7. May 1, 2010 #6

    kuruman

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    It is true only as long as you have a cylinder so that you can write its moment of inertia as (1/2)Mr2. I checked the process only and it looks fine. I trust you know how to punch numbers in your calculator.
     
  8. May 1, 2010 #7
    yeah i hope so, all pulleys are cylinders arnt they?? i have another question about a buch longer cylinder (coke can shape). it would be true for that wouldn't it??
     
  9. May 1, 2010 #8
    Yes, the moment of inertia formula for a disk and a cylinder are the same as long as the axis of rotation is through the center.
     
  10. May 1, 2010 #9
    ok, good thanks for your help.
     
  11. May 1, 2010 #10
    sorry 1 more thing, ive asked a few people this. why am i given a radius if it is not needed, is there another way of solving these that does use the radius??
     
  12. May 1, 2010 #11
    If they ask you for the torque you need to know the radius. In this case you just used the definition of torque without having to calculate torque itself. Problems sometimes give extra information to force you to think about what it useful info and what is not useful.
     
  13. May 1, 2010 #12
    ok i was just asking because i have 3 questions all that give a radius and all that i solved without it. Although, i do understand what you are saying about extra information.
     
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