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Rotational Motion/Projectile motion of a ball on a circular ramp.

  1. Jun 23, 2011 #1
    1. The problem statement, all variables and given/known data

    A small sphere of radius r0 = 1.4 cm rolls without slipping on the track shown in the figure whose radius is R0 = 40.0 cm. The sphere starts rolling at a height R0 above the bottom of the track. Assume that it leaves the track after passing through an angle of 135° as shown.

    Picture: http://imgur.com/7YLlJ

    It asks for speed at the end of the ramp. and what is d.

    2. Relevant equations

    I know all the basic equations. I am just confused how to get the height at the end of the ramp. ie. heres my equation.

    E0=mgh or m(9.8)(.4)

    E1=.5mv2+.5Iw2 +mgh or

    I replace "I" with the moment of inertia for a perfect sphere which is 2/5mr2

    .5mv2+.5(2/5mr2)v2+m(9.8)(h)

    E1 is the energy at the end of the ramp. But i cannot get h which is that height at the end of the ramp.

    I figure the angle of that circle must be 45 degrees. So that arc length is .31m. Since the Radius is 40cm. Arc Length = angle times radius.

    But i cannot get the height. If i get the height i can solve for the speed, then the get d.

    Thanks ahead for any help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 23, 2011 #2

    ehild

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    Try to draw the unknown height into the picture before you say 'I can not get h'. See attachment.

    ehild
     

    Attached Files:

  4. Jun 23, 2011 #3
    how can you say that at the point of the circle will make a full radius above that point?

    Yea you absolutely cant. the triangle would have two lengths of radius 40 cm repeated on the hypotenuse and the sides.
     
  5. Jun 23, 2011 #4

    ehild

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    I certainly have not said that.


    ehild
     
  6. Jun 23, 2011 #5
    then i have no idea what is going on geometrically
     
  7. Jun 23, 2011 #6

    ehild

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    Examine the triangle. What is the length of the hypotenuse?
    What is its angle?

    ehild
     
  8. Jun 23, 2011 #7
    haha i feel like an idiot now. math does that to you.

    but how do i get d. i think my kinematics get me the distance from the point it shot off the ramp. i cant make a triangle underneath to get it.
     
  9. Jun 23, 2011 #8
    But my answers dont seem to be right. here was my math.

    I had h being .1171 m.

    so the velocity turned out to be 1.76 m/s which is wrong.

    I redid the math as I was looking at my work and i dont know what is wrong just wanted to check that with you.


    I get:

    9.8*.4=.5v2+(1.4^2)(.5)(.4)v2+9.8*.1171

    2.77242=.892v2

    v=1.76
     
  10. Jun 24, 2011 #9

    ehild

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    You count the translational KE twice.
    Edit: and included Ro into the rotational energy, why?

    ehild
     
    Last edited: Jun 24, 2011
  11. Jun 24, 2011 #10
    I actually have a question about this, doesn't the centripetal force do work along the arc?
     
  12. Jun 24, 2011 #11

    ehild

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    The centripetal force does not "act" it is just the resultant of other forces. You might have thought of the normal force. But it is normal to the trajectory, so its work is zero.

    ehild
     
  13. Jun 24, 2011 #12
    Is it possible to solve using torque and other newton's laws?

    Feel abd for hijacking (I won't show the steps), but I got roughly 1.98m/s
     
  14. Jun 24, 2011 #13

    ehild

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    You can solve it by other methods than conservation of energy, but it seems the simplest. Your result is about correct.

    ehild
     
  15. Jun 24, 2011 #14
    You count the translational KE twice.
    Edit: and included Ro into the rotational energy, why?

    ehild




    I thought I am supposed to count both the kinetic energy of the ball and the rotational energy, which I thought accounts for the energy total. and what is Ro?
     
  16. Jun 24, 2011 #15

    ehild

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    Yes, but what is the rotational energy? Ro is the radius of the track in the figure, 0.4 m.

    ehild
     
  17. Jun 24, 2011 #16
    I thought the rotational energy is the one that involves the sphere. The 1.4cm
     
  18. Jun 24, 2011 #17

    ehild

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    So 1.4 was cm and the radius of the sphere! Now I understand where is it from, and what is 0.4. But: The rotational energy is not 0.5 I v^2, but 0.5 I ω^2 (ω is the angular velocity).


    ehild
     
  19. Jun 24, 2011 #18
    I use 0.4 m in the beginning of the equation to represent the PE of the initial state before it begins to roll. I know that the rolling energy is .5Iw^2. But i replace I with 2/5r^2 which is the moment of inertia for a sphere. so the .4 here might be the confusion u are talking about.

    hence I get for the rolling energy of the sphere:

    .5*.4*(1.4/100)^2 *v^2

    I also realized ive been using 1.4 METERS as the radius. Derp. but i still do not get the answer. With the 1.4 cm use I get the velocity as 2.35 m/s. Which i know is wrong. Any suggestions Ehild?
     
  20. Jun 24, 2011 #19

    ehild

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    You can not use v instead of w. v is the velocity of the CM, w is the angular speed of rotation. How are they related in case of pure rolling?
    Check the dimension of your formula. Disregarding mass, it has the unit of m^4/s^2 instead of m^2/s^2.

    ehild
     
  21. Jun 24, 2011 #20
    Of course i know that dude.

    but remember Omega = V/R? so the rotational energy is .5IW^2. I was simply canceling out R ahead of time.

    I= .4R^2 and we have (V/R)^2. the R's cancel out So i just get energy.
     
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