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Suspended object moment of inertia

  1. Apr 30, 2010 #1
    1. The problem statement, all variables and given/known data

    A 15.0 kg bucket of water is suspended by a very light rope wrapped around a solid cylinder 0.300 m in diameter with a mass of 12.0 kg. The cylinder pivots on a frictionless axle through its centre. The bucket is released from rest at the top of a well and falls 10.0 m to the water.
    a) What is the tension in the rope while the bucket is falling?
    b) With what speed does the bucket strike the water?
    c) What is the time of the fall?
    d) While the bucket is falling, what is the force exerted on the cylinder by the axle?

    2. Relevant equations

    T-Mg=ma
    I=mr^2

    3. The attempt at a solution
    For a) i have solved as far as T-147.15=12a
    how do i find a??
     
  2. jcsd
  3. Apr 30, 2010 #2

    rl.bhat

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    To calculate T, we have to consider torque acting on the cylinder due to the tension.
    Torque τ = r*T = Iα = 1/2*Mr^2*α....(1)
    The tangential acceleration a = r*α. so α = a/r.
    Substitute this value in eq(1). find T.
    From that find a.
     
  4. Apr 30, 2010 #3
    Hey im not sure on some of the variables you used what is big T?.. is little T torque
     
  5. Apr 30, 2010 #4
    ok so i think i know what everything is, small T is torque(we use tau) , big T is Tension.. so i tried to solve for a but using Ia/r=1/2mr^2a/r but the a's just cancel out. i cant use the T*r because i dont know the tension... Please help.
     
  6. Apr 30, 2010 #5

    rl.bhat

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    r*T = Iα = 1/2*Mr^2*α....(1)
    Angular acceleration α = a/r
    So r*T = Iα = 1/2*Mr^2*a/r....(2)
    T = 1/2*M*a.
    Substitute the value of T in
    mg-T=ma
    So a = g/(1+M/2m)
    Now proceed.
     
  7. Apr 30, 2010 #6
    Hey thankyou that was very helpful, did you get a =7m/s^2 sound right to me - less than free fall... just to let you know the reason i was having so much trouble from your 1st post was that i had alpha=r/a not a/r
     
    Last edited: Apr 30, 2010
  8. May 1, 2010 #7
    Hey, after i find the acceleration (7) i just put that straight back in the mg-T=ma to get T don't i. i got 42.15N can you please check that and let me know... thanks
     
  9. May 1, 2010 #8

    rl.bhat

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    Your calculation of a and T are correct.
     
  10. May 1, 2010 #9
    Thankyou for d) the force the axle exerts on the cylinder would be the tension + the force due to gravity wouldn't it. which i got 159.87N for.
     
  11. May 1, 2010 #10

    rl.bhat

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    No.
    The the force the axle exerts on the cylinder is equal to the tangential force on the rim of the cylinder and is equal to m*a = 15*7 = 105 N
     
  12. May 1, 2010 #11
    Hi, just a quick question. why do they give the radius of the cylinder if it is not needed, is there a way to solve these problems that does require the radius?
    I have another very similar problem to this that i didnt use the radius for either.
     
  13. May 1, 2010 #12

    ehild

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    Sorry,rl.bhat, the tangential force at the rim of the cylinder is not equal to the tension?

    And m*a = 7*15 is not the net force on the bucket, which is mg-Tension?

    ehild
     
  14. May 1, 2010 #13
    Hey can you tell me how to do it if thats not right please, and about the radius? the thing that he said was F=ma was the force the axle exerts of the cylinder. Is the rest of my answers correct? you said that they looked good in my other post, sorry about the amount of posts but i really need the correct answers.
     
  15. May 1, 2010 #14
    I take it form the times when yous reply that none of you are in Australia?
     
  16. May 1, 2010 #15

    ehild

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    My opinion was that your answer was correct. rl.bhat said it was wrong. Both of us can be mistaken. You have to decide.

    ehild
     
  17. May 1, 2010 #16
    You meant my answer form my other post for d). You are both in agreement about the rest aernt yous??
     
  18. May 1, 2010 #17
    HEy just interested to know, i tried solving for the final velocity with the energy method you showed me(ehild) and it came out at about 9m/s but with kinematic equations using the found acceleration of 7m/s^2 it comes out to 11.8m/s. why would this be?
     
  19. May 1, 2010 #18

    ehild

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    I haven't show you any method to get the final velocity. Maybe, it was somebody else in an other thread you started. Show your full work, how do you get the final velocity if a body moves 10 m with acceleration a=7 m/s^2 ?


    ehild
     
  20. May 2, 2010 #19
    i dont have any trouble with normal kinematics, v=sqrt(2*10*7) since u is 0. that comes out to 11.9. The other way i was trying was total energy initial-E=mgh+1/2mv^2+1/2Iw^2 and then letting w=v^2/r^2 and solving for v.... i think that i know why its not working, the bucket would still have some unknown mgh when it hits the water not 0?????
    this is the one that i was talking about: https://www.physicsforums.com/showthread.php?t=399445
     
  21. May 2, 2010 #20
    If you set the surface of the water to h=0 then there will not be any remaining potential energy when the bucket hits the water.
     
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