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Work-energy problem (units!)

  1. Apr 25, 2013 #1
    1. The problem statement, all variables and given/known data
    The 8 lb uniform rod AB shown in the figure rotates in the vertical plane about a pin at C. The spring attached to the rod at D has a spring constant of 1.25 lb/in., and its unstretched length is 24in. If the rod is released from rest in the position shown, determine its angular velocity when it is in the vertical position.

    phpBgvMlL.png

    2. Relevant equations
    U = mgh
    U = .5*k(x^2)
    U = .5*Iω^2
    I = I_cm + mr^2


    3. The attempt at a solution
    Initial length = sqrt(36^2+20^2) = 41.183 in
    Initial stretch = 41.183 - 24 = 17.183 in

    Since the movement will tend towards lower energy (i.e. the spring will compress), the spring will become less stretched as the rod goes towards its vertical position (as opposed to stretching further into the opposite vertical position).

    Final length = sqrt(26^2+10^2) = 27.857 in
    Final stretch = 27.857 - 24 = 3.857 in

    As the spring becomes less stretched, elastic potential energy is converted into other forms of energy (rotational or gravitational potential).

    Initial elastic potential energy = (.5)(k)(x)^2
    = .5(1.25 lb/in)(17.183^2)
    = 184.53 lb*in

    Final elastic potential energy
    = .5(1.25 lb/in)(3.857^2)
    = 9.298 lb*in

    Gravitational potential energy is gained as the rod goes into its vertical position.
    Final U = mgh
    = (8 lb)(386.4 in/s^2)(10 in)
    = 30912 lb*in

    Rotational energy is gained = .5*Iω^2
    Final I = ml^2/12 + mr^2
    = (8 lb)(40 in)^2/12 + (8 lb)(10 in)^2
    = 13600 lb*in^2

    Initial energy = final energy
    184.53 lb*in = 9.298 lb*in + 30912 lb*in + .5(13600)ω^2

    Solve for ω
    ω = imaginary...?

    So, obviously, I'm screwing up somewhere. Probably with the units, but I don't know which is wrong. Is there anything I'm not accounting for? What value should I be using for gravity?

    Thank you so much.
     
  2. jcsd
  3. Apr 25, 2013 #2

    CWatters

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    Your sums look correct but does the rod swing up or down?
     
  4. Apr 25, 2013 #3

    SteamKing

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    For the bar AB, is 8 lb the weight of the bar or its mass?
     
  5. Apr 25, 2013 #4

    PhanthomJay

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    Just a note on USA units: The weight W of the rod is already given as 8 lbs., a force unit. Do not multiply it by 'g'. For example, for h = 10 inches, gravitational potential energy is mgh = Wh = (8)(10) = 80 in-lb.

    Similarly, when using mass units, you must divide the weight by g to get the mass. W = mg, or m = W/g. When using inches for the length unit, then in this problem the mass of the rod is 8/386.4.
     
  6. Apr 25, 2013 #5
    I think it swings upward as per my explanation in my original post; since the spring is already stretched, it'll want to compress, and thus, it'll swing upward.

    Not sure - the problem just states that it's an 8lb. uniform bar... I dislike imperial units :(
    If I do my calculations assuming 8 lbs is the weight, I get something that makes sense, if you'll look below. Still not sure if it's correct though.

    Thank you for this!


    Here are my revised calculations

    Initial elastic PE = .5*k*x^2 = .5*1.25*17.173^2 = 184.53 lb*in
    Final elastic PE = .5*k*x^2 = .5*1.25*3.857^2 = 9.298 lb*in
    Final grav. PE = m*g*h = w*h = 8*10 = 80 lb*in
    Final rotate PE = .5*I*ω^2
    = .5*((1/12)*m*l^2 + mr^2)*ω^2
    = .5*[(1/12)*(8/(12*32.2))*40^2 + (8/(12*32.2))*10^2]*ω^2
    = .5*4.831*ω^2

    Initial = Final
    184.53 = 9.298 + 80 + .5*4.831*ω^2
    ω = 6.279 rad/s

    Someone else mentioned that I should be careful with the gravitational potential energy because not all of the rod moves upward. Technically, that is true; 10 inches worth moves down and only 30 inches moves up. Should I consider that, or should I only consider the mass center, which moves up 10 inches, like I did in my revised calculations above?


    I found a similar problem with a confirmed answer.
    HVLRaXW.png
    The answer is ω=19.04 rads/s.
    Whenever I work this one out though, I can't get that.

    Initial length = sqrt(.9^2+.5^2) = 1.029563 m
    Initial stretch = 1.029563 - .6 = .42956 m

    Final length = sqrt((.9-.25)^2+.25^2) = 0.6964194 m
    Final stretch = 0.6964194 - .6 = 0.0964194 m

    Initial elastic PE = .5*k*x^2 = .5*220*.42956^2 = 20.2974 N*m
    Final elastic PE = .5*k*x^2 = .5*220*0.0964194^2 = 1.022637 N*m
    Final grav. PE = m*g*h = 4*9.81*.25 = 9.81 N*m
    Final rotate PE = .5*I*ω^2
    = .5*((1/12)*m*l^2 + mr^2)*ω^2
    = .5*[(1/12)*4*1^2 + 4*.25^2]*ω^2
    = .5*.58333*ω^2

    Initial = Final
    20.2974 = 1.022637 + 9.81 + .5*.58333*ω^2
    ω = 5.69656 rad/s

    This is not equal to the book-provided answer of 19.04 rads/s. I don't know where I'm messing up.
     
    Last edited: Apr 25, 2013
  7. Apr 26, 2013 #6

    CWatters

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    My mistake. I'm used to working in SI units and assumed 8lbs was the mass rather than the weight so I incorrectly multiplied by g. That would mean the force excerted by the rod would be greater than the spring and it would swing downwards stretching the spring even more.
     
  8. Apr 26, 2013 #7
    So, to confirm, the rod does indeed swing upwards, right?
    Would you mind checking over my calculations for the alternate problem I posted? I can't seem to get the supposed correct answer (according to the book). The problem's in SI units :)
     
  9. Apr 26, 2013 #8

    CWatters

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    Have you got the moment of inertia right? I worked it out as follows..

    The moment of inertia about the end of a rod is

    (1/3)ml2

    http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html

    You have two rods of mass m/4 and 3m/4 and lenghts L/4 and 3L/4 so..

    I = Ileft + Iright
    I = (1/3)(m/4)(L/4)2 + (1/3)(3m/4)(3L/4)2
    = (m/12) L2/16 + (3m/12) 9L2/16
    = (m/192) L2 + (27m/192) L2
    = (28m/192) L2
     
  10. Apr 26, 2013 #9

    CWatters

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    Yes I believe so.
     
  11. Apr 26, 2013 #10
    (28m/192) L2 resolves out to .58333 when m=4kg and L=1m are inputted. .58333 is the same as my calculations for I.

    Doesn't look like moment of inertia is the problem. =/

    I suppose the book-provided solution could be incorrect...
     
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