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Angular acceleration, velocity, momentum of a door?

  1. Oct 29, 2015 #1
    Skjermbilde 2015-10-29 kl. 21.45.27.png

    1. The problem statement, all variables and given/known data

    A door of width l = 1.00 m and mass M = 15.0 kg is attached to a door

    frame by two hinges. For this problem, you may ignore gravity, as we are

    interested in rotational motion around the vertical axis. There is no friction

    of any kind.

    An angry mother-in-law slams the door shut, by pushing at the middle of

    the door (l=2 from the hinges) with a force of F = 100N, lasting a time

    t = 0:.00 s. The door is initially not rotating. The door can be taken to be

    a uniform rod, for the purpose of this exercise.

    a) What is the angular acceleration of the door while it is being pushed?

    b) What is the resulting angular velocity, angular momentum and rotational

    kinetic energy from this push?
    c) Assuming that she lets go of the door (leaving it to slam shut) at the

    moment in the motion when the door is perpendicular to the wall, how long

    does it take for the door to close?

    d) What would be the result of a), b), c) and d) if she had pushed not in the

    middle of the door but at the edge (l from the hinges)?

    In addition to the force of the mother-in-law, the hinges also provide force,

    both radial (centripetal) and tangential force while the mother-in-law pushes

    (they also compensate for gravity to keep the door upright, but ignore that

    for now).

    e) Combining angular acceleration and linear acceleration considerations, find

    the tangential force Fh supplied by the hinges as a function of d, the distance

    between the hinges and the point of application of the mother-in-law force

    (d was l=2 in part a) and b), and l in part c); now use a general d). Don’t

    put in explicit numbers, just find the equation.

    f) For which d do the hinges not need to provide any tangential force?

    3. The attempt at a solution
    Can someone check on my solutions, All answers appreciated!!

    a)
    τ= l/2*F=50
    τ=M*(l/2)^2*α
    α=τ/M*(l/2)^2= 13.33 rad/s^2

    b)
    ω=ωi+αt=2.7 rad/s
    I=(1/12)*Ml^2=1.25 kg*m^2
    L=I*ω=3.375 kg*m^2/s
    K=(1/2)*I*ω^2=4.56 J

    c)
    θ=θ+ωt+(1/2)αt^2
    π/2=2.7t+(1/2)13.33t^2
    used the abc rule
    t=0.323 s
    d)
    τ= l*F=100
    τ=M*l^2*α
    α=τ/M*l^2= 6.7 rad/s^2

    ω=ωi+αt= 1.34 rad/s
    I=(1/3)*Ml^2=5 kg*m^2
    L=I*ω=6.7 kg*m^2/s
    K=(1/2)*I*ω^2=4.489 J

    θ=θ+ωt+(1/2)αt^2
    π/2=1.34t+(1/2)6.7t^2
    used the abc rule
    t=0.512 s


    e)
    This solution i am sure is wrong
    Fh(d)= τ(d)/I + F(d)/M

    d)
    d=l
     
  2. jcsd
  3. Oct 29, 2015 #2

    SammyS

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    It appears that you have some typos in the statement of the problem

    I assume she pushes the door at a point located L/2 from the axis of rotation. (Lower case L should be banned as a variable.)

    The time interval over which she pushes is not given. It looks like you used t = 0.20 seconds.

    What did you use for the moment of inertia of the door ?
     
  4. Oct 29, 2015 #3

    haruspex

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    ... and about what axis?
    Haveagoodday, you seem to have used ML2/4 in a) and ML2/12 in b).
     
  5. Oct 31, 2015 #4
    I have same answers in a and b, but you have a typo in b. In c you have to calculate again, your answer is wrong.
     
  6. Oct 31, 2015 #5

    haruspex

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    None of the answers given to a) and b) in the OP are correct.
     
  7. Oct 31, 2015 #6
    And why is that so? My only guess is that I should take Fsin0 instead of pure F... Then it will change all the answers
     
  8. Oct 31, 2015 #7
    Wait, it will not change a thing.
     
  9. Oct 31, 2015 #8
    A) ΣFt=mat->at= ΣFt/m=6,7
    α=at/r=6,7/0,5=13,4
    Where did I go wrong?
     
  10. Oct 31, 2015 #9
    at is tangential acceleration
     
  11. Oct 31, 2015 #10

    SammyS

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    My guess is that haruspex did not guess.
     
  12. Oct 31, 2015 #11

    haruspex

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    This is rotation. Use torque and moment of inertia.
     
  13. Nov 1, 2015 #12
    Now I got 40 in a.
     
  14. Nov 1, 2015 #13

    haruspex

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    Better, but as Sammy pointed out in post #2 the L=2 must be a typo, probably for L/2. Does that change your answer?
     
  15. Nov 1, 2015 #14
    L/2 is correct. And that is what I used.

    All answers are new and angry mother-in-law is slaming the door faster, when she holds at the edge. Seems reasonable.
    But I'm so stuck at e) now...
     
  16. Nov 1, 2015 #15

    haruspex

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    Then you are probably using the wrong moment of inertia formula. What did you use?
    Would you like them checked?
    How far do you get? What equations do you have?
     
  17. Nov 2, 2015 #16
     
    Last edited: Nov 2, 2015
  18. Nov 2, 2015 #17
    i now got 40 in a
    and 8 rad/s, 10 kgm^2/s, 40J in b
    are these answers correct?
    And what equations can i use in c?
     
  19. Nov 2, 2015 #18

    haruspex

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    As I implied in post #15, I get a different answer. Please post your working.
     
  20. Nov 2, 2015 #19
    how did you calculate c?
     
  21. Nov 2, 2015 #20
    in a:
    I=Ml^2/12=1.25
    α=T/I=40
    in b:
    w=w+αt=8
    L=Iw=10
    K=Iw^2/2=40
     
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