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Pulley problem, acceleration, angular velocity, inertia

  1. Jul 11, 2015 #1
    1. The problem statement, all variables and given/known data
    In the figure, block 1 has mass m1 = 450 g, block 2 has mass m2 = 530 g, and the pulley is on a frictionless horizontal axle and has radius R = 5.3 cm. When released from rest, block 2 falls 71 cm in 5.0 s without the cord slipping on the pulley. (a) What is the magnitude of the acceleration of the blocks? What are (b) tension T2 (the tension force on the block 2) and (c) tension T1 (the tension force on the block 1)? (d) What is the magnitude of the pulley’s angular acceleration? (e) What is its rotational inertia? Caution: Try to avoid rounding off answers along the way to the solution. Use g = 9.81 m/s2.
    20150711_150601_zpstxicvulv.jpg


    2. Relevant equations
    F=ma

    3. The attempt at a solution
    20150711_145435_zpsvsyxmx0t.jpg

    im on part a, and i dont know why i keep getting a=-9.81m/s/s, it should be less than that.
     
  2. jcsd
  3. Jul 11, 2015 #2
    I will give a hint, what do you think of the acceleration of the two boxes, can you compare them ? can the acceleration be g ? Apply newton's second law for both masses, T1 - m1g = m1a, T2 - m2g = m2a, Oh now I've noticed that the two acceleration have different (opposite) ditection !,
     
  4. Jul 11, 2015 #3
    a=(m1g-m2g)/(m1+m2) =-.80082m/s/s where am i going wrong?
    I see that a is in opposite directions but that is the variable im solving for. g is still 9.81 for both.

    The tensions cancel each other out right?

    edit; i have so far; m1a=T-m1g and m2(-a)=T-m2g I must not be doing the algebra correctly when solving for a here.
     
    Last edited: Jul 11, 2015
  5. Jul 11, 2015 #4
    The T's have the same magnitude, if the cancel them selves out, the objects would be free falling right?, the acceleration you've got seems correct to me !
     
  6. Jul 11, 2015 #5
    If the tensions canceled eachother out then the gravitational forces of each mass would be working against eachother.
    w1-w2=-0.7848N
     
  7. Jul 11, 2015 #6
    a = (m-M)g/(M+m) is right, I've edit my last post, they don't cancel each other out !,
     
  8. Jul 11, 2015 #7
    0.80082 is incorrect also.
     
  9. Jul 11, 2015 #8
    Did you try both signs ?, try avoid rounding up as the exercice proposes,
     
  10. Jul 11, 2015 #9
    hehe i tried .800816327 should be positive since its asking for magnitude, but i tried negative also with no prevail.
     
  11. Jul 11, 2015 #10

    Nathanael

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    That should be all you need to find the acceleration. (And the reasonable assumption that acceleration is constant.)

    You can't find the acceleration with newton's laws because you don't know the mass of the pulley.
     
  12. Jul 11, 2015 #11
    I know that its velcoity is 0.142m/s. How do you get acceleration from that? do i take the derivative of it?
     
  13. Jul 11, 2015 #12

    Nathanael

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    That's the velocity the whole time? Or the initial or final velocity? Or what?

    Derivatives of a number are meaningless. Remember the definition of derivatives, it is a limiting process which involves the behavior around a point.
    (If you only have a single point then the derivative is meaningless.)
     
  14. Jul 11, 2015 #13
    .142 is the velocity at 5s. the velocity changes at a constant rate starting from 0m/s. if acceleration is the change in velocity over time, a=dv/dt. if i took a derivative of .142 its = 0 because its a constant. I need to express velocity as a function of time and then can take a derivative of that. its change in position is -0.71/5s
    how do you find a function for velocity? only way i can think of is to graph the position vs time and get a position function then take two derivatives of that. Is that the quickest way?
     
  15. Jul 11, 2015 #14

    Nathanael

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    If you know the final velocity you can simply say that Vf=at (because as I said, you can assume acceleration is constant).

    But 0.142 is not the final velocity. You took the total distance divided by the total time, why would that be the final velocity? What does the total distance divided by the total time represent?
     
  16. Jul 11, 2015 #15
    Aww, I've assumed that the pully is massless, If that is not provided, It would be better to follow Nathaneal's method !!
     
  17. Jul 11, 2015 #16
    its average velocity.
     
  18. Jul 11, 2015 #17

    Nathanael

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    Right. Keep going :wink: (How is average velocity related to final velocity?)
     
  19. Jul 11, 2015 #18
    final velcoity =average velocity multiplied by the time at final position so .142tf
    assuming speed is linear
     
  20. Jul 11, 2015 #19

    Nathanael

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    No I think you're mixing something up. (The dimensions are wrong.) The final distance is the average velocity times the change in time.
    (If you used symbols instead of numbers (!!) then you would see at once it is not right: (Δx/Δt)Δt ≠ Vf)

    (For constant acceleration,) the average velocity only depends on the initial and final velocities.

    (Solving problems with symbols has another advantage, rounding errors are not exaggerated. You really should get in the habit of using symbols until finished.)
     
  21. Jul 11, 2015 #20
    The initial velocity is 0, right ? So Vavg = Vf/2
     
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