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Energy, Force, and Inertia Problem. Ball down ramp to loop to loop

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data
    A solid sphere of mass m and radius r rolls without slipping along the track shown below. It starts from rest with the lowest point of the sphere at height h above the bottom of the loop of radius R, much larger than r. (Consider up and to the right to be the positive directions for y and x respectively)


    What are the force components on the sphere at the point P if h = 3R? (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)


    2. Relevant equations
    Isphere=(2/5)mr2
    K=.5mv2
    K=.5Iω2
    PE=mgh
    ac=v2/R

    3. The attempt at a solution
    Fx
    First, solving for v2
    mg(3R)=.5mv2+(.5)(2/5)(mr2)(v2/r2)
    mg2R=.7mv2+mgR
    v2=20gR/7

    Centripetal acceleration is the acceleration in the x direction
    a=v2/R=20gR/7R
    Fx=ma=-20mg/7 <=== Correct!

    Fy
    Force in y direction is only the force from gravity
    F=ma
    Fy=-mg <=== Wrong!

    please help! What am I missing?
     

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    Last edited: Nov 10, 2012
  2. jcsd
  3. Nov 10, 2012 #2

    TSny

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    Hello, and welcome!

    Note that you are measuring heights from the "floor". Does point P have a height?

    EDIT:
    OK, I saw that you left out the potential energy at P in the first equation, but it looks like you took care of it in the second equation.
     
  4. Nov 10, 2012 #3

    TSny

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    I don't see anything wrong with your answer for Fy. Wait! Is the force of gravity the only force with a y component acting on the ball at P?
     
    Last edited: Nov 11, 2012
  5. Nov 10, 2012 #4
    Well, i guess there is friction to? But there wasn't any coefficients given for kinetic friction
     
  6. Nov 10, 2012 #5

    TSny

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    Right. Is the friction kinetic or static?
     
  7. Nov 10, 2012 #6
    Actually, static correct? Because the object is rolling.
     
  8. Nov 10, 2012 #7

    TSny

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    That's right- rolling without slipping. So, even if they did give you a coefficient of static friction it wouldn't be of much help. The static coefficient only let's you find the maximum force of static friction, and there is no reason why the static friction would need to be at its maximum value at P.

    So, you'll need to find the friction force some other way. Will the ball have any angular acceleration at P?
     
  9. Nov 10, 2012 #8
    The ball itself? Well angular acceleration will be the derivative of ω with respect to s? EDIT: s as in seconds, from g
     
  10. Nov 10, 2012 #9

    TSny

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    That's a correct statement if s denotes time. [Edit: Ah, I see that you are using s for time!] But it's not of much help.

    Do you think the ball will be speeding up or slowing down at P?
     
  11. Nov 10, 2012 #10
    Definitely slowing down, because v will be decreasing
     
  12. Nov 10, 2012 #11

    TSny

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    Will ω also be decreasing?
     
  13. Nov 10, 2012 #12
    Yes, because v=ωr
     
  14. Nov 10, 2012 #13

    TSny

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    Good. So, looks like you have an unknown y-component of acceleration, an unknown angular acceleration, and an unknown friction force. You'll need three independent equations that relate these unknowns. How many equations can you think of?
     
  15. Nov 11, 2012 #14
    1) [itex]\sumτ[/itex]=I(alpha)
    2)v=ωr
    3)fs=μFn
    4) alpha=aT/r
    5)aT=(-mg-fs)m
    Fn= -20mg/7
    I=2/5mv2/r2

    I can't seem to relate it using these, i'm probably missing one of the equations.
     
    Last edited: Nov 11, 2012
  16. Nov 11, 2012 #15

    TSny

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    Oooo. Shotgun approach. Need to narrow the field. Eq. #1 looks like a keeper. #4 is also good except the subscipt T is suspicious. What does aT stand for?

    Finally, you'll need another equation that relates at least two of the three unknowns: ay, [itex]\alpha[/itex], and the friction force. (Hint: We don't want to upset Newton - which is easy to do.)
    [EDIT: your #5 equation looks like it might be the third equation. Not sure the signs are correct though and there appears to be another mistake in it also.]
     
    Last edited: Nov 11, 2012
  17. Nov 11, 2012 #16
    at=ay

    Okay, thanks! I'll try to work them out.
     
  18. Nov 11, 2012 #17

    TSny

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    Ok, good. (I was going to have to quit for now anyway - it's late and I need my beauty rest.)
     
  19. Nov 11, 2012 #18
    My attempt:

    Equation 1 expands to:
    mg+fs=(2/5)mr2(ay/r)

    plugging the right side of this equation into equation 3:

    ay=(2/5)mray/m

    and this is where i get stuck, because the acceleration terms cancel out.
     
  20. Nov 11, 2012 #19

    TSny

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    What you have written on left side doesn't represent the type of quantity you wrote earlier on the left side of equation 1. (The dimensions of what you have on the left don't match the dimensions you have on the right.)

    Also, it might help to think about the direction of the friction force. If the angular speed of the ball is slowing down at P, what must be the direction of the net torque on the ball (clockwise or counterclockwise)? What direction does the friction force need to act in order to produce that direction of torque about the center of the ball?
     
  21. Nov 11, 2012 #20
    EDIT: From previous, i've changed torque to only static friction in the positive direction, as mg acts through the axis. Also, the friction force needs to counteract the clockwise movement, therefore needing to be positive (counterclockwise).

    So i'm pretty sure i've worked this out correctly. I just want to double check, as this is my last attempt for points. Thanks!

    Equations:
    1) [itex]\sum[/itex]τ=Iα
    fsr=Iα
    α=fsr/I

    2)α=ay/r
    fsr/I=ay/r
    ay=fsr2/I

    3)may=(-mg+fs)
    mfsr2/((2/5)mr2)=-mg+fs
    5/2fs-fs=-mg
    3/2fs=-mg
    fs=-2/3mg

    Plug back into 1)
    α=(-2/3)mgr/I
    α=(-10g/6r)

    Plug into 2)
    (-10g/6r)=ayr
    ay=(-10g/6)

    Fy=may
    Fy=(-10mg/6)
     
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