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Calculate dieter for maximum exit speed

  1. Aug 20, 2009 #1
    During my lab i have to roll a marble down a ramp at different angle. I am required to calculate the angle between the ramp and the table so that the marble will achieve the maximum speed when it slide off the ramp.



    mgh=1/2 mv^2
    V(in x direction) = Vcosx where V is the velocity of the marble and x is the angle of the ramp to the table.




    I equate the 2 equations together and gotten V(in x direction) =(cos x )sqrt 2gh
    where h=LsinX. So i equate (cosX)sqrt(sinX)=y and differentiate once and equate to 0. The answer i got was X=45 which is wrong. However, if i change the equation to V^2=2gL(sinX)(cosX)^2, the angle i got was arcsin (1/sqrt3) which is about 35. Why is this so??
     
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  3. Aug 20, 2009 #2

    kuruman

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    I am not sure I understand the experiment. It seems to me that the steeper you make the angle, the higher the speed will be upon exit. The maximum speed is sqrt(2gh) when the incline is 90o and the ball is in free fall.
     
  4. Aug 20, 2009 #3
    The experiment is releasing a marble from an incline plane and the marble will roll along a straight path after it reaches the bottom of the incline plane. At first i also thought that the steeper the fast the velocity but if the angle is at 90degree then will not be a velocity in the x direction hence the marble will not move once it reaches the bottom of the plane
     
    Last edited: Aug 20, 2009
  5. Aug 20, 2009 #4

    kuruman

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    OK. So you want to maximize the horizontal component of the speed, not the speed itself. The way you phrased the question initially is confusing. Look at it this way, the speed at the end of the incline is going to be

    [tex]v = \sqrt{2gh} =\sqrt{2gLsin\theta} [/tex] where L is the length of the ramp.

    The horizontal component of the velocity is

    [tex]v_{x} = v cos\theta =(\sqrt{2gLsin\theta})cos\theta [/tex]

    So the problem becomes finding the value of θ for which

    [tex]cos\theta \sqrt{sin\theta}[/tex]

    is a maximum. Without the radical it would be 45o. With the radical it's something else. Can you find what?
     
  6. Aug 20, 2009 #5
    Whats radical? I know the angle is arcsin(1/sqrt3) i just don't understand why i cant use
    [tex]cos\theta \sqrt{sin\theta}[/tex] and if i use (sinX)(cosX)^2 i can get the arcsin(1/sqrt3)
     
    Last edited: Aug 20, 2009
  7. Aug 20, 2009 #6

    kuruman

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    Radical means "the square root symbol." As I said before

    [tex]v=(\sqrt{2gL sin\theta})cos\theta[/tex]

    I never said you cannot use this expression, in fact this is the expression you must use. But what is the value for θ for which it has the largest possible value? For θ = 0 it is zero and for θ = 90o it is also zero. So somewhere in between 0 and 90o it must go through a maximum.
     
  8. Aug 21, 2009 #7
    I know what the angle is. I am just asking why is it that when i calculate the maximum value for theta using [tex]v=(\sqrt{ sin\theta})cos\theta[/tex] the answer is 45 but when i square the equation and solve for theta i can get the correct value for theta. Don't mind going through my working in the file attached and tell me which part i did wrongly.
     

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    Last edited: Aug 21, 2009
  9. Aug 21, 2009 #8

    rl.bhat

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    Last edited: Aug 21, 2009
  10. Aug 21, 2009 #9

    andrevdh

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    the marble also have rotational kinetic energy
     
  11. Aug 21, 2009 #10

    kuruman

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    Indeed it does. The kinetic energy at the bottom will be

    [tex]K=\frac{1}{2}*\frac{7}{5}m v^{2}[/tex]

    Clearly, including the 7/5 correction due to rotational energy reduces the linear speed at the bottom but does not change the angle at which the horizontal component of the linear velocity is a maximum.
     
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