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Velocity of end points of the rod

  1. Dec 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Rigid uniform Rod 'AB' of mass 'M' and Length 'L' is pulled slightly ( Gently ) at the bottom at time t=0 when it just reaches the horizontal ground, at that time find the Velocities of End Points of the rod.

    All the surfaces are friction less.
    7AC9Klw.jpg


    2. Relevant equations


    3. The attempt at a solution

    I first tried to find the value of ##\theta## at which the rod left contact with the vertical wall.
    mpyfVwc.png
    From the figure

    ##{ V }_{ BC }=\frac { l\omega }{ 2 }##

    ##{ V }_{ BC }=\frac { l\omega }{ 2 } sin\theta \hat { i } +\frac { l\omega }{ 2 } cos\theta \hat { j } ##

    and \({ V }_{ AC }=-\frac { l\omega }{ 2 } sin\theta \hat { i } -\frac { l\omega }{ 2 } cos\theta \hat { j } \)

    As ##{ V }_{ BC }={ V }_{ B }-{ V }_{ C }## and ##{ V }_{ AC }={ V }_{ A }-{ V }_{ C }##

    or ##{ V }_{ C }={ V }_{ B }-{ V }_{ BC }##

    So ##{ V }_{ C }\hat { j } =-\frac { l\omega }{ 2 } cos\theta \hat { j } ## (considering velocity in vertical direction only. We are doing this because ##{ V }_{ B }\hat { j } =0##)

    Similarly we can get

    ##{ V }_{ C }\hat { i } =\frac { l\omega }{ 2 } sin\theta \hat { i } ##............(4)

    So ##{ V }_{ C }=\frac { l\omega }{ 2 } sin\theta \hat { i } -\frac { l\omega }{ 2 } cos\theta \hat { j } ##

    Also from conservation of energy

    ##mg\frac { l }{ 2 } (1-sin\theta )=\frac { 1 }{ 2 } \left( m{ v }^{ 2 }+{ I }_{ C }{ \omega }^{ 2 } \right) ##......................(1)

    Also ##{ V }_{ C }={ \left( \frac { l\omega }{ 2 } \right) }^{ 2 }## (we can get this from eq(4))

    Putting this in eq-1 we get

    ##{ \omega }^{ 2 }=\frac { 3g }{ l } (1-sin\theta )##............(2)

    When the rod looses contact with the vertical rod the Normal force acting on Rod becomes 0. So no horizontal force act on the rod after the moment the rod left contact with vertical wall. So acceleration of the CoM of the rod in horizontal direction is 0.

    So ##{ \frac { d }{ dt } (V }_{ C }\hat { i } )=\frac { d }{ dt } \left( \frac { l\omega }{ 2 } sin\theta \hat { i } \right) =0##

    So ##\frac { d }{ d\theta } \left( \sqrt { \frac { 3g }{ l } (1-sin\theta ) } sin\theta \right) \frac { d\theta }{ dt } =0\\ ##

    or ##\frac { d }{ d\theta } \left( \sqrt { (1-sin\theta ) } sin\theta \right) =0##

    or ##cos\theta \sqrt { (1-sin\theta ) } =\frac { sin\theta cos\theta }{ 2\sqrt { (1-sin\theta ) } } ##

    So ##sin\theta =\frac { 2 }{ 3 } ##

    So at ## \theta=arcsin(2/3)## ##\omega=\sqrt{\frac{g}{L}}##

    So ##{ V }_{ C }=\frac{\sqrt{gl}}{3} { i } -\frac{\sqrt{5gl}}{6} { j } ##

    After loosing contact with the vertical wall no horizontal force acts on the CoM of the rod so the vertical component of the rod will not change.

    So ##V_{A}=\frac{\sqrt{gl}}{3}##. Where ##V_{A}## is the velocity of the lower end of the rod when the rod becomes horizontal.

    Let the rod make an angle ##\phi## withe the horizontal after loosing contact.
    feEWeZl.png
    So ##-\frac { Nlcos\theta }{ 2 } =\tau ##

    or ##\frac { Nlcos\phi }{ 2 } =-I\frac { d\omega }{ d\phi } \omega ##

    or ##\frac { Nlsin\phi }{ 2 } =-I\frac { { \omega }^{ 2 } }{ 2 } +C##

    At ##sin\theta=2/3 {\omega}^{2}=\frac{g}{l}##

    So ##C=\frac { l(8N+mg) }{ 24 } ##

    So ##\frac { Nlsin\phi }{ 2 } =-I\frac { { \omega }^{ 2 } }{ 2 } +\frac { l(8N+mg) }{ 24 }##

    When the rod becomes horizontal ##\theta=0##
    So ##N=\frac { m(l{ \omega }^{ 2 }-g) }{ 8 } ##

    Now the acceleration of the CoM of the rod(##a##) is ##\frac{mg-N}{m}##

    vertical velocity of CoM when rod left contact with the wall is ##\frac{\sqrt{5gl}}{2}##

    distance covered by the CoM =##\frac{l}{3}##


    So vertical velocity of the CoM of the rod when it becomes horizontal(##v##) is

    ## v^{2}=\frac { 5gl }{ 36 } +\frac { 2l(mg-N) }{ 3m } ##


    Also when the rod becomes horizontal ##\frac{l\omega}{2}=v##. As the vertical velocity of point A is 0.

    So on putting values of ##N## and ##\omega##

    I got ##4v^{2}=\frac{13gl}{6}##

    Also the ##V_{B}=2v##

    So ##V_{B}=\sqrt{\frac{13gl}{6}}##.

    Is my approach correct? Are my answers correct?:confused:
    Please help.:)
     
    Last edited: Dec 26, 2014
  2. jcsd
  3. Dec 26, 2014 #2

    haruspex

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    I guess you mean
    At [itex]\sin \phi = \frac 23 [/itex], [itex]\omega^2=\frac gl [/itex]
    But that N is the normal force at the moment it leaves contact with the wall, no?
     
  4. Dec 26, 2014 #3
    No. ##\theta## is the angle(between rod and horizontal) at which the rod leaves contact with the vertical wall. Where as ##\phi## is a general angle(between rod and horizontal) after the rod leaves contact with the vertical wall.

    So [itex]\sin \theta = \frac 23 [/itex], [itex]\omega^2=\frac gl [/itex]

    The rod leaves contact with the vertical wall but it remains in contact with the horizontal wall. In the FBD in #post1 I have mentioned it.
     
  5. Dec 26, 2014 #4

    haruspex

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    Sure, but you wrote "At sinθ =2/3" in the context of a general equation involving ϕ. So I think what you intend is along the lines of "At ϕ = ϕ0 = θ". There are a couple of other places where it seems to me you wrote θ where you meant ϕ. E.g.
    Yes, I understand N is the normal force from the ground here. You use N to substitute for a constant C in a general equation for the period after losing contact with the wall, and the basis for the substitution is the instant of losing contact with the wall. For that to be valid, the N substituted must be its value at that instant. Maybe I'm misreading your algebra, but you appear to treat this N later as the normal force at any later time.
    It might help clarify things if you make more use of suffixes, like ϕ0 and N0 for the values at the instant of losing wall contact. Indeed, with that you could have used theta throughout, but with different suffixes/embellishments for the critical values.

    I do have another concern I forgot to mention before. The question says "when it just reaches the horizontal ground". You calculated on the basis of its becoming horizontal, but are you sure it will be on the ground at that time? Maybe it becomes airborne at some point? If so, the question is not well posed.
     
  6. Dec 27, 2014 #5
    Oh! I missed that. I assumed ##N## to be constant but it is a function of ##\phi##. How should I proceed?
    I tried to find out the ##\theta## at which the lower end of the rod leaves contact with the ground.

    The moment at which the rod leaves contact with ground the acceleration of the CoM of the rod will be ##g##.

    So ##\frac { d }{ dt } \left( \frac { l\omega cos\theta }{ 2 } \right) =g##

    or ##\theta =arcsin\left( \frac { \sqrt { 129 } +3 }{ 12 } \right) ##
    But the value of number inside the brackets is coming out to be greater the 1. So, can I assume that the rod will not loose contact with the horizontal ground?

    How should I proceed to solve the question?
     
  7. Dec 27, 2014 #6

    haruspex

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    Once it leaves the wall, horizontal speed of the mass centre is fixed. So put that to one side and treat it as the CoM descending vertically at speed ##v = v(t) = (l/2)\omega \cos \phi##.
    What energy equation do you now have?
    I don't know how you got that last equation, but I confirm that it will not become airborne.
     
  8. Dec 27, 2014 #7

    The initial velocity of the rod is ##\frac { l\omega cos{ (\phi }_{ 0 }) }{ 2 } =\frac { \sqrt { 5gl } }{ 6 } ##
    where ##\phi_{0}## is the angle b/w rod and horizontal at the moment the rod looses contact with the vertical wall.

    Initial potential of the rod is ##\frac{lsin\phi_{0}}{2}=\frac{l}{3}##

    Let ##v'## be the velocity of the CoM of the rod when it becomes horizontal.


    So by the conservation of mechanical energy---

    ##\frac { m }{ 2 } { \left( \frac { \sqrt { 5gl } }{ 6 } \right) }^{ 2 }+\frac { mgl }{ 3 } =\frac { m{ v' }^{ 2 } }{ 2 } ##

    ##v' =\sqrt { \frac { 29gl }{ 36 } } ##

    So we have found the velocity of CoM of the rod when it becomes horizontal. When the rod becomes horizontal then the vertical velocity of point A(the point which was in contact with the ground) will be zero.

    So ##\frac{l\omega}{2}=v'##. So can we find the velocity of CoM and the angular velocity of the rod about the CoM in this way?
    Then velocity of the points A and B can be find by adding the velocity of the CoM and velocity about the CoM. right?

     
  9. Dec 27, 2014 #8
    The 1st equation in #post5 can be written as --

    ##\frac { d }{ d\theta } \left( \frac { l{ \omega }^{ 2 }cos\theta }{ 2 } \right) =g##

    Substituting the value of ##\omega## I got

    or ##\frac { 3 }{ 2 } \frac { d }{ d\theta } (1−sinθ)cos\theta =1##

    or ## 2{ sin }^{ 2 }\theta -sin\theta -1=\frac { 2 }{ 3 } ##

    on solving this I got ##\theta =arcsin\left( \frac { 3\pm \sqrt { 129 } }{ 12 } \right) ##

    Thanks for helping haruspex.:)
     
  10. Dec 27, 2014 #9

    ehild

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    It is correct so far.

    You meant that the horizontal component of the velocity of the CoM will not change. But VA will. XA=XC+L/2 cosθ, and θ changes with time. A is accelerating.
    If you solve rotational motion of a rigid body, do it with respect to a fixed axis or with respect to the CoM. And apply conversation of energy. The frame of reference fixed to A is not inertial.
     
  11. Dec 27, 2014 #10

    haruspex

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    You've left out rotational energy (both sides).
     
  12. Dec 28, 2014 #11
    Sorry.:s

    So
    ##\frac { { I }_{ c }{ \omega }_{ 0 }^{ 2 } }{ 2 } +\frac { m }{ 2 } { \left( \frac { \sqrt { 5gl } }{ 6 } \right) }^{ 2 }+\frac { mgl }{ 3 } =\frac { m{ v' }^{ 2 } }{ 2 } +\frac { { I }_{ c }{ \omega }_{ f }^{ 2 } }{ 2 } ##
    where ##\omega_{ 0 }## is the angular velocity of the rod when it looses contact with the vertical wall and ##\omega_{ f }## is the angular velocity of the rod when it become horizontal.

    or ##\frac { m{ l }^{ 2 } }{ 12 } \times \frac { { \omega }_{ 0 }^{ 2 } }{ 2 } +\frac { m }{ 2 } { \left( \frac { \sqrt { 5gl } }{ 6 } \right) }^{ 2 }+\frac { mgl }{ 3 } =\frac { m{ v' }^{ 2 } }{ 2 } +\frac { m{ l }^{ 2 } }{ 12 } \times \frac { { \omega }_{ f }^{ 2 } }{ 2 } ##

    or ##\frac { 32gl }{ 36 } ={ v' }^{ 2 }+\frac { { l }^{ 2 }{ \omega }_{ f }^{ 2 } }{ 12 } ##.

    When the rod becomes horizontal then the vertical component of velocity of point A(the point which was in contact with the ground) will be zero.

    So is ##\frac { l{ \omega }_{ f }^{ 2 } }{ 2 } ={ v' }##?
     
  13. Dec 28, 2014 #12
    Thanks for replying ehild!:)
    Oh! yes I meant horizontal component.:s

    Please clear my confusion.
    rod5.png
    Here let red arrow represent vertical and horizontal component of the velocity of the CoM and green arrow represent the angular velocity of the point A about CoM. Total velocity of point A is vector sum of these three vectors.
    When the rod becomes horizontal point A will not have any vertical velocity. right?
    So vertical component of velocity of CoM is equal to angular velocity of point A about CoM. right?
    So being in opposite direction they cancel out. So remaining vector is the horizontal velocity of CoM. So isn't velocity of pointA equal to the component of the horizontal velocity of the CoM?:confused:
     
  14. Dec 28, 2014 #13

    ehild

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    The horizontal component of the velocity of the CoM is equal to VA if the rod is horizontal.
     
  15. Dec 28, 2014 #14
    So, is my expression of ##V_{A}## in post 1 correct?

    What are your comments on my post #11. Should I find ##v'## and ##\omega_{f}## from the two equations?

    then velocity of point B (when the rod becomes horizontal) is ##\frac { l{ \omega }_{ f } }{ 2 } +{ V }_{ CoM }##
    where Vcom is the total velocity of the CoM of the rod when it becomes horizontal.
     
  16. Dec 28, 2014 #15

    ehild

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    What is v'? is not it the vertical velocity of the CoM?
     
  17. Dec 28, 2014 #16

    haruspex

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    Almost... I think you left in something by mistake.
     
  18. Dec 28, 2014 #17
    Yes it is a vertical velocity of CoM of the rod when it becomes horizontal.
     
  19. Dec 28, 2014 #18
    Did I commit mistake in the conservation of energy equation?
     
  20. Dec 28, 2014 #19

    ehild

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    How are ω and v' related? And what is v' finally?
     
    Last edited: Dec 28, 2014
  21. Dec 28, 2014 #20
    You meant ωf and v'. right?:oops:
    0.5lωf=v'.
    I have mentioned that at the end of #post 11 but by mistake I typed ω2. I think haruspex was pointing that in post#16.
     
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