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Rotational Dynamics Problem - Rod slipping against Block

  1. Dec 13, 2012 #1
    1. The problem statement, all variables and given/known data

    A uniform rod of mass m and length l is pivoted at point O. The rod is initially in vertical position and touching a block of mass M which is at rest on a horizontal surface. The rod is given a slight jerk and it starts rotating about point O. This causes the block to move forward. The rod loses contact with the block at θ = 30°. All surfaces are smooth.

    1. The value of M/m?
    2. The velocity of block when the rod loses contact with the block?
    3. The acceleration of center of mass of rod, when it loses contact with the block?
    4. The hinge reaction at O on the rod when it loses contact with the block?


    Ans 1) 4:3
    2)√(3gl)/4
    3) (3g)/4
    4)(mg/4)j




    2. Relevant equations



    3. The attempt at a solution

    The speed of the tip of the rod in horizontal direction is same as that of the block .

    Let V be the speed of the block when rod loses contact with block.
    So,Horizontal speed of the tip of the rod = V
    Component of speed perpendicular to length of the rod =[itex]V_{\bot}[/itex] = VSin30°=V/2
    Angular speed of the rod at the instant block leaves contact with the rod =[itex]\frac{V_{\bot}}{l}[/itex] = [itex]\frac{V}{2l}[/itex]

    Applying Conservation of Energy ,

    Loss in Potential Energy of Rod = Gain in Rotational Kinetic Energy of Rod + Gain in KE of Block

    [itex]\frac{Mgl}{2}(1-Sin60°) = \frac{1}{2}\frac{ml^2}{3}(ω ^2) +\frac{1}{2}MV^2 [/itex]

    [itex]\frac{Mgl}{2}(1-Sin60°) = \frac{1}{2}\frac{ml^2}{3}\frac{V ^2}{4l^2} +\frac{1}{2}MV^2 [/itex]

    [itex]\frac{Mgl}{4} = \frac{mV^2}{24} +\frac{1}{2}MV^2 [/itex]

    Is my approach correct ? Now ,how am I supposed to proceed further ? one more equation is required to solve the problem ...
     

    Attached Files:

  2. jcsd
  3. Dec 13, 2012 #2

    haruspex

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    No, that's backwards. Which way is the tip of the rod moving?
    Not sin(60), I think. But I see that you actually used the right value below.
    Maybe energy alone won't cut it, and you need to use forces and accelerations. This could be because the energy equation fails to capture the fact that the tip of the rod and the block have moved the same horizontal distance in the same time.
     
  4. Dec 13, 2012 #3
    Thank you very much for the reply...I understand my mistake...

    The component of angular speed of the tip of the rod in horizontal direction is same as that of the block .

    Let V be the speed of the block when rod loses contact with block.
    So,Horizontal component of angular speed of the tip of the rod = ωlcos60°=V

    So,ωl/2=V
    Thus,ω = 2V/l

    Applying Conservation of Energy ,

    Loss in Potential Energy of Rod = Gain in Rotational Kinetic Energy of Rod + Gain in KE of Block

    [itex]\frac{Mgl}{2}(1-Sin60°) = \frac{1}{2}\frac{ml^2}{3}(ω ^2) +\frac{1}{2}MV^2 [/itex]

    [itex]\frac{Mgl}{2}(1-Sin60°) = \frac{1}{2}\frac{ml^2}{3}\frac{4V ^2}{l^2} +\frac{1}{2}MV^2 [/itex]

    [itex]\frac{Mgl}{4} = \frac{2mV^2}{3} +\frac{1}{2}MV^2 [/itex]

    I have thought in terms of forces...but really havent been able to relate things...

    The forces on the rod are

    Hinge force in x direction = Hx
    Hinge force in y direction = Hy
    Force due to gravity = mg
    Normal contact force from the block = N

    The force on the block are

    Normal contact force from the rod = N

    I know how to calculate acceleration and torques but i am unable to relate them so that given condition in the question is satisfied.

    Please can you explain ?
     
  5. Dec 13, 2012 #4

    haruspex

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    Don't worry about the force at the hinge. The critical statement we have to turn into an equation is that the normal force vanishes. In terms of the angular speed and acceleration, what is the acceleration of the block (be very careful with that one), the normal force on the block, and the torque on the rod?
     
  6. Dec 13, 2012 #5
    Net torque on the rod about pivot =[itex]mg\frac{l}{2}cosθ -Nlsinθ[/itex]
    where θ is the angle which rod makes with the horizontal

    The acceleration of the block is a=N/M where N is the normal contact force between block and the rod

    Since ,ωl/2=V

    Therefore , differentiating above we get,
    αl/2=a
     
    Last edited: Dec 13, 2012
  7. Dec 13, 2012 #6

    haruspex

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    Right, so relate that to the angular acceleration of the rod, relate N to the linear acceleration of the block, and relate that acceleration back to the angular acceleration and velocity of the rod.
     
  8. Dec 13, 2012 #7
    Oh...sorry I have edited my earlier post...

    The acceleration of the block is a=N/M where N is the normal contact force between block and the rod

    Since ,ωl/2=V

    Therefore , differentiating above we get,
    αl/2=a
     
  9. Dec 13, 2012 #8

    haruspex

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    No, please work with the forces. Your method above merely expresses the assumption that the two stay in contact. To find when they, in fact, would not we need to look at forces. (You will need the polar coordinate formula for acceleration of the tip.)
     
  10. Dec 14, 2012 #9
    I dont know how to work with polar coordinates...I would be grateful if you could tell me the polar coordinate formula for acceleration of the tip.
     
  11. Dec 14, 2012 #10

    haruspex

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    I expect you do, but maybe not with that terminology. If the rod were rotating at constant angular speed it would nonetheless have an acceleration, right? What's the formula for that and in which direction does it operate? Since the angular speed is not constant it has another linear acceleration. What's the formula and direction of that?
     
  12. Dec 14, 2012 #11

    ehild

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    You know that the coordinates of the tip of rod are y=Lsin(θ) and x=Lcos(θ)
    You can find the components of acceleration in terms of derivatives of θ. The tip of the rod is at the same x coordinate as the left side of the block, so the second derivative of x is the acceleration of the block. What force acts on the block? What is the force when the rod loses contact with the block?

    ehild
     
    Last edited: Dec 14, 2012
  13. Dec 14, 2012 #12
    ehild....Thank you

    [itex]\frac {d^2x} {dt^2} = -Lcosθ[/itex]

    What does -ve sign signify??

    Force on the block = N
    Thus, N=-MLcosθ . I am not sure about the -ve sign.

    There is no force horizontally on the block when it loses contact with the rod i.e N=0
     
  14. Dec 14, 2012 #13

    ehild

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    Try again.

    x=Lcos(θ), but θdepends on t.

    dx/dt = -Lsin(θ)(dθ/dt)

    d2x/dt2=...?


    ehild
     
  15. Dec 14, 2012 #14
    Do we have to take θ = ω0t + (1/2)αt2 where ω0 =0 ?
     
  16. Dec 14, 2012 #15

    ehild

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    NO, why do you think that the angular acceleration is constant? θ is a function of t, ω is its first derivative and α is the second derivative with respect to t.

    ehild
     
  17. Dec 14, 2012 #16
    x=Lcos(θ)

    dx/dt = -ωLsin(θ)

    d2x/dt2= -L(ωcosθ + αsinθ)
     
  18. Dec 14, 2012 #17

    ehild

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    You know product rule and chain rule of differentiation? x is function of θ but θ is function of t. The velocity is function of θ and ω, both depend on t.
    ωsin(θ) is a product of two functions.
    Apart from Physics, how do you calculate the first and second derivative of a function f(g(t))?

    ehild.
     
  19. Dec 14, 2012 #18
    x=Lcos(θ)

    dx/dt = -Lsin(θ)(dθ/dt)

    d2x/dt2= -L [cosθ(dθ/dt) + sinθ(d2θ/dt) ]
     
  20. Dec 14, 2012 #19

    ehild

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    Not quite... d2x/dt2= -L [cosθ(dθ/dt)2 + sinθ(d2θ/dt2) ]= -L(cosθ ω2+sinθ α).

    What force acts on the block?

    What is the acceleration of the block when it loses contact with the rod?

    How are ω and α related at that moment?

    ehild
     
  21. Dec 14, 2012 #20
    The first term in the bracket on the right side should be cosθ(dθ/dt) .Here we keep dθ/dt constant and differentiate sinθ .
     
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