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Rod slipping against wedge

  1. Aug 9, 2014 #1
    1. The problem statement, all variables and given/known data

    A thin rod of mass m = 1 kg and length l=1m is placed along the hypotenuse side of a wedge lying on a frictionless floor having mass M =5kg and angle 60° . At t= 0 the system is released from rest. Both the objects are free to move .Find the speed of the wedge when the rod makes an angle 30° with the floor.

    2. Relevant equations



    3. The attempt at a solution

    Let V be the speed of the wedge and v be the speed of the CM of the rod .

    Applying conservation of momentum MV = mvx

    Applying energy conservation (mgl/2)(sin60° – sin30°) = (1/2)MV2+ (1/2)mv2+ (1/2)Iω2

    Not sure how to proceed. I guess we need to find some constraint relation .

    I would be grateful if somebody could help me with the problem.
     

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    Last edited: Aug 9, 2014
  2. jcsd
  3. Aug 9, 2014 #2

    TSny

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    Right.

    Hint: The upper end of the rod remains in contact with the surface of the wedge. That should give you a relation between a certain component of the velocity of the wedge and the same component of the velocity of the upper end of the rod.
     
  4. Aug 9, 2014 #3
    Thanks for the response ,TSny .

    Component of velocity of the upper end of the rod perpendicular to the hypotenuse face = Vcos30° .

    But how do I find the component parallel to the face ? How does that help us in determining the velocity of the CM of the rod ? I guess I need to find the x-component of velocity of CM of the rod .
     
    Last edited: Aug 9, 2014
  5. Aug 10, 2014 #4

    TSny

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    Good.

    I'm not sure you need to find the component parallel to the face. But you do have a second constraint at the lower end of the rod that you can use.
     
  6. Aug 10, 2014 #5
    Sorry for interrupting in between.
    I understood the first equation.I have also understood LHS of eq(2) but I am not able to understand why you have used expression [itex]\frac{Iω}{2}[/itex] here.
    The rod is not rotating here I think.:confused:
     
  7. Aug 10, 2014 #6

    TSny

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    The angle that the rod makes to the horizontal continually decreases until the rod becomes horizontal when it hits the floor. So, at the instant the rod makes a 30o angle, the rod is rotating.
     
  8. Aug 10, 2014 #7
    I have thought about it but not sure how we would use the constraint at the lower end . Does this help us in determining the velocity of the CM or the angular velocity of the rod ?
     
    Last edited: Aug 10, 2014
  9. Aug 10, 2014 #8
    The velocities of wedge and rod perpendicular to the to the contact surface should be same.
    So Vcos60=vcos30
    Is it right?
     
  10. Aug 10, 2014 #9

    TSny

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    Can you use the constraint at the lower end of the rod to help you express the y-component of the CM in terms of ω? It is helpful to remember the relative velocity formula ##\vec{V}_A = \vec{V}_C + \vec{V}_{A/C}## for any two points A and C of the rod. Think about the points you might choose for A and C.
     
  11. Aug 10, 2014 #10
    Yes.

    If I consider the upper end of the rod and the Center then the only useful information we have is the component of velocity of upper end perpendicular to the face of the wedge .

    Also , not sure how the component of velocity of the upper end we have calculated is useful for us ?
     
  12. Aug 10, 2014 #11

    TSny

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    Combine these. That is, if you let A be the upper end of the rod and C be the CM, what do you get if you combine ##\vec{V}_A = \vec{V}_C + \vec{V}_{A/C}## and the constraint equation for A?
     
  13. Aug 10, 2014 #12
    ## \vec{V}_A = (\frac{M}{m}V+\frac{l}{2}ωsinθ)\hat{i} ##

    Is it correct ?
     
    Last edited: Aug 10, 2014
  14. Aug 10, 2014 #13

    TSny

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    ## \vec{V}_A ## should have both x and y components. Are you taking ##\hat{i}## to point to the right?

    We've got a lot of variables. For now, just express the Cartesian components of ## \vec{V}_C## as ##v_x## and ##v_y##. Express the Cartesian components of ##\vec{V}_{A/C}## in terms of ##\omega## and ##\theta##. Then you can express ## \vec{V}_A ## in terms of ##v_x## , ##v_y## , ##\omega## and ##\theta##.
     
  15. Aug 10, 2014 #14
    ##\hat{i}## to the right and ##\hat{j}## upwards .

    ## \vec{V}_C = \frac{M}{m}V\hat{i} + \frac{l}{2}ωcosθ\hat{j}##

    ## \vec{V}_{A/C} = \frac{l}{2}ωsinθ\hat{i} + \frac{l}{2}ωcosθ\hat{j}##
     
  16. Aug 10, 2014 #15

    TSny

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    Close. But I think there are several sign errors. Does the CM have an upward y-component of velocity? Reconsider the direction of ##\vec{V}_{A/C}##.
     
  17. Aug 10, 2014 #16
    Doesn't the negative value of ω make the y-component of velocity negative ?
     
  18. Aug 10, 2014 #17

    TSny

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    OK. I didn't notice that you were taking clockwise as positive rotation.

    [But it seems less confusing to me to take CCW as positive rotation for this problem.]
     
  19. Aug 10, 2014 #18
    So, post#14 okay ?

    But the angle which the rod makes with the horizontal , automatically makes CW positive ?
     
    Last edited: Aug 10, 2014
  20. Aug 10, 2014 #19

    TSny

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    Yes, post 14 is OK with ω a negative number.

    I was considering ω as just the angular velocity of the rod in the CCW direction, rather than defining it as ω = dθ/dt. But, your way is good.
     
  21. Aug 10, 2014 #20
    So , ## \vec{V}_A = (\frac{M}{m}V+\frac{l}{2}ωsinθ)\hat{i} + lωcosθ\hat{j} ##

    Here y-component is negative .

    But how do I determine whether the x-component is positive or negative , because that will determine the angle which the x-component makes with the direction perpendicular to hypotenuse face ?

    Edit : I am trying to find the component of these components in the direction perpendicular to the face .
     
    Last edited: Aug 10, 2014
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