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Physics Question - rotational and gravitational energy

Discussion in 'Pre-Medical - MD' started by SailCrazy, Apr 18, 2004.

  1. SailCrazy

    SailCrazy I gotta have more cowbell
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    This is a very general problem, but I'm hoping you can offer some help. I don't need an exceptionally complex answer, just a discussion of the different forces & forms of mechanical energy and what will happen.

    Ok, you have an half-pipe. The left half has friction, the right side does not.

    You release a ball on the left side. It rolls down the pipe without slipping, so that when it reached the bottom it has both translational and rotational Kinetic Energy. It then slides up the right (frictionless) side. It will rise to a height less than the initial height because some of the original U(grav) is still in the form of Rotational kinetic energy. (The ball will still be rotating at point of max height on the right side do to lack of friction.)

    The cylinder then slides back down the half pipe to the bottom. What will happen at the bottom where it reaches the left (friction) side?

    Thanks for your help!
     
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  3. willthatsall

    willthatsall Unretired
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    Just from thinking about it, it seems like when the ball reaches the left side again it will still be spinning in the same direction that it was when it was originally released (no friction on the right so the rotation isn't altered). This means the rotation will cause a force in the opposite direction of the motion when it reaches the left side (friction).
     
  4. IrishOarsman

    IrishOarsman My Rx: Guiness
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    will sounds about right... the rotation of the ball carries some of the g energy but in now the wrong direction. The ball will have picked up more g energy on the way down the frictionless side, but not enough to overcome both the counter-rotation and the constant g force. the ball will go up less high than it did on the frictionless side.

    (edit: fun problem)
     
  5. SailCrazy

    SailCrazy I gotta have more cowbell
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    While the ball is rotating, causing force in the opposite direction, will the contact point between the ball and the half pipe be slipping or not? (on the side with friction)

    I know that when the ball was initially released, the contact point between the ball and the half point was not moving. When the ball has a rotation that is opposite its direction of translational momentum, does that make it slip, or just slow its translational movement?
     
  6. willthatsall

    willthatsall Unretired
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    After I read this question I remembered the concept of rotational kinetic energy from physics class a year ago. However, I am going through Kaplan Comprehensive Review right now and I don't see it as one of the topics. Is it part of the MCAT?

    As far as the slipping on the part of the half-pipe with friction, I am not sure but would it depend on the coefficient of friction between the two surfaces and the weight of the ball (normal force)? More friction = Less slippage?
     
  7. mikeyboy

    mikeyboy Senior Member
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    Just glacing at the problem, I don't see why it would slip on the way up if it didn't slip on the way down. The ball is going at the same speed (translational), and rotating at the same rate on the same surface; just in a different direction. Maybe there is something I'm missing.
     
  8. willthatsall

    willthatsall Unretired
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    Because it is now rotating in the direction opposite the translational motion, unlike when it was on the way down after being released and the translational and rotational motion was in the same direction. It seems like if it doesn't slip, it has to stop moving because it is going to be rotating in a backwards direction.

    Is rotational kinetic energy covered in other books? Kaplan doesn't mention it.
     
  9. bryan45876

    bryan45876 Senior Member
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    Will is correct that the coefficient of friction would come into play when the ball hits the friction side. When this happens the ball is rotating the "wrong" way so it will oppose the motion and dissipate energy. However, the translational motion will overcome this force and the ball will slide up the frictional side of the ramp. As the ball slides up its rotational velocity will begin to decrease and it will momentarily reach zero, then it will then begin to rotate the "correct" way and stop sliding; all this happens while the ball is translating up the ramp.

    If you need any other details pick up a Dynamics book or search the web. I am a junior in Mechanical Engineering at Purdue and we covered this topic last week.
     
  10. willthatsall

    willthatsall Unretired
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    What if the coefficient of friction and normal force were very large, would the ball then completely stop and begin to go back in the other direction because of the rotation?

    Also, can someone give me the equation for rotational kinetic energy? Isn't it E = K + U and K(total) = K(rot) + K (trans)? Are rotational and translational kinetic energy equal, or is the ratio of translational to rotational kinetic energy dependent on the friction or some other factors? Give me some help here.
     
  11. bryan45876

    bryan45876 Senior Member
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    There are 2 different types of motion involved: rotational and translational. In the previous situation we assumed the ball slides. If you were to increase (or decrease) the coefficient of friction or the mass of the ball, it would only affect whether or not the ball slides.

    The ball would never stop or bounce back as a result of the translational motion. What could happen is that the frictional forces overcome the rotational energy and the ball grips the half pipe, doesn?t slide, and begins to rotate in the other direction.

    You could then continue to analyze the problem and determine how high the ball will roll back up. If the ball were to slide all the way to its highest point, that point would be higher than the height the ball would reach if it began to grip the half pipe. This is because the coefficient of static friction is greater than that of kinetic.
     
  12. bryan45876

    bryan45876 Senior Member
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    W = ∆KE + ∆PE
    W = ∆KE(r) + ∆KE(t) + ∆PE
    W = (.5*I*w^2 + .5*m*v^2),f - (.5*I*w^2 + .5*m*v^2),i + mgh,f - mgh,i

    W:Work (also ∆E)
    KE: kinetic energy
    PE:potential energy
    r:rotational
    t:translational
    I:moment of inertia
    w:angular velocity
    m:mass
    v:velocity
    f:final
    i:initial
    g:gravity
    h:height
     
  13. willthatsall

    willthatsall Unretired
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    Thanks, that brought it back... Strange that Kaplan doesn't mention moment of inertia or rotational kinetic energy.
     
  14. bryan45876

    bryan45876 Senior Member
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    In my physics classes I didn't learn rotational kinetic energy and we barely touched on inertia. It wasn't until my Mechanic classes (statics & dynamics) that we covered this.

    I'm curious, what's your major, where do you go to school?
     
  15. willthatsall

    willthatsall Unretired
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    Biological Sciences, relatively small private university in the southeast... We covered this topic and some other rotational kinematics stuff in intro physics (non calc based) without much detail. I know we also talked about moment of inertia in the relationship that it has with how difficult something is to rotate (i.e. when a tightrope walker carries a long stick to increase his moment of inertia and thus make it easier to keep his balance). And the calculation for moment of inertia varies depending on the shape of the object I remember. From looking at Kaplan it appears that none of this is needed for the MCAT though. My physics class was really strange in that we didn't have to memorize equations because they were provided on tests. I think it was taught with the MCAT in mind because most people in there were pre-health majors, but I think it was a little too easy.
     
  16. SailCrazy

    SailCrazy I gotta have more cowbell
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    Can I know whether or not the ball will slip without knowing its mass and the coeficient of friction?
     
  17. liverotcod

    liverotcod Lieutenant Crunch
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    No, you need to know both.

    But let's assume that there is never any slipping on the friction side of the pipe, which is typical for problems of this type (at least in first-year physics). Then it would seem that the ball/pipe interaction at the interface point between friction/frictionless would act as an inelastic collision. If the rotational energy of the ball is greater than its translational at that point (if the ball is light or small relative to the height of the pipe) then it will continue up the friction side at a reduced speed. If the ball is massy or large relative to the pipe height, then it could bounce back up the frictionless side.

    The picture I have in my mind is of an approach shot hitting the green in golf... whether it will roll back depends on its forward velocity.
     
  18. willthatsall

    willthatsall Unretired
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    Good explanation, but isn't that reversed? If the rotational is greater than the translational, shouldn't the ball reverse back towards the frictionless side?
     
  19. liverotcod

    liverotcod Lieutenant Crunch
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    D'oh! You're right, I got it all turned around. That sentence should read "If the rotational energy of the ball is LESS than its translational..." Thanks for the correction.
     
  20. Shrike

    Shrike Lanius examinatianus
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    Oh, crap -- got it backwards. Never mind.
     
  21. Shrike

    Shrike Lanius examinatianus
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    Never mind.
     
  22. liverotcod

    liverotcod Lieutenant Crunch
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    That's pretty harsh for a post that's off the mark. Think about the direction of rotation of the object at the bottom of the pipe, versus its direction of travel.
     
  23. Shrike

    Shrike Lanius examinatianus
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    False.

    The rotational inertia, which is the resistance to spinning, is a function of the geometry of the particle (e.g., solid or hollow, sphere or cylinder) and its mass. It is a linear function of mass. The maximum force acting to spin the particle, the friction force, is a function of the coefficient of static friction and the normal force. The normal force is a function of the angle of the ramp at any given point and a linear function of the mass of the particle. The particle slips if the resistance to rotation exceeds the maximum static friction. There is a mass term on each side of this equation, and the two cancel each other out. Whether the particle spins is a function of the particle's shape shape and coefficient of friction, but not its mass.

    This should look familiar -- mass often cancels out of static friction problems.
     
  24. Shrike

    Shrike Lanius examinatianus
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    Correct -- I screwed that one up. It is a good problem.
     
  25. liverotcod

    liverotcod Lieutenant Crunch
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    I see your point... But what confuses me is the fact that at the point of contact with the friction surface, the object is moving relative to the surface - so it's not static friction that one has to consider. Really the question is if the object's rotational inertia exceeds the kinetic friction at that point. But, if I recall correctly, that too is independent of mass. So maybe I'm making a moot point. It certainly is an interesting problem.
     
  26. bryan45876

    bryan45876 Senior Member
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    The coefficient of static friction is ALWAYS greater than that of kinetic friction for any two materials. That is one of the reasons why you have ABS brakes on your car; the distance that you need to stop will be less if you don?t slide.
     
  27. liverotcod

    liverotcod Lieutenant Crunch
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    Yah, that was kinda my point - that the momentum would have *less* to overcome because it would already be in motion relative to the friction surface, and hence be working against the (smaller) kinetic friction.
     
  28. bryan45876

    bryan45876 Senior Member
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    I'm not exactly sure what you are saying, but when the ball comes back in contact with the frictional side of the ramp it will slide.
     
  29. jedirampage

    jedirampage Senior Member
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    Static friction is what actually causes the ball to rotate. The point of contact between the ball and the surface are stationary relative to each other because in the next instant there will be a new point of the ball in contact with the surface. The same point does not remain in contact, therefore kinetic friction does not apply for the rotational component.
     
  30. liverotcod

    liverotcod Lieutenant Crunch
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    Static friction causes the ball to rotate on its way down the first time. But we were dicussing the transition from the frictionless surface to the frictional surface at the bottom of the pipe, when the object is rotating counter to its direction motion. It may be helpful to read from the beginning of the thread. Then again, we may have beat this horse well past death.
     
  31. SailCrazy

    SailCrazy I gotta have more cowbell
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    I think there are still a few flies buzzing around...

    There is kinetic friction as the ball hits the frictional surface due to the fact that it is rotating the "opposite" direction. That is all we really had to determine for my class.

    I think a more detailed analysis would have to include information on the body - the moment of inertia, in order to determine whether or not the translational energy will "overcome" the rotational energy or vice versa.
     

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