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Conservation of energy interpretation of percentage conserved

  1. Feb 9, 2016 #1
    1. The problem statement, all variables and given/known data
    I get that less percentage energy is conserved from potential to kinetic energy by measuring h and v with two balls for the heavier ball. Im trying to sort of why actually is like that!

    Two balls where dropped down from a ramp with different masses and volume. The smaller sphere has a smaller mass, and less volume ofc. Velocity was measured in the end of the ramp!
    3. The attempt at a solution
    It was not rolling 100% perfect so i think that friction will come into account a bit. Since the normal force is bigger for the heavier ball, and friction is reliant on normal force the heavier ball will feel friction easier. The two speheres werent made of the material, maybe the heavier ball was made of a material with bigger friction constant, mu. The heavier ball have bigger area so it would feel air resistance more!

    Questionmarks:
    1. Will rolling friction come into play when a sphere is rolling down a ramp and gaining more kinetic energy?
    2. How can i physically describe that the bigger ball will feel bigger air resistance? (I made an attempt, but cant really explain)
    3. I know that potential energy will transform into to kinetic energy and rotational kinetic energy, however with different h, should less energy be turned into kinetic energy percentagewise, or shouldn't be the same for every h, since everything will just increase, the velocity will increase, so more kinetic energy, it will roll around more times so more rotational kinetic energy, longer distance that friction can make a work, so more friction energy etc.. Should these be propotional to eacother because i weirdly enough got that the trend that the higher h , especially for the heavier ball the more kinetic energy is conserved, higher percentage compared to the starting potential energy. Is it just an accident or is there any way to explain that?
    4. Any general idea why the heavier ball conserve less kinetic energy in percentage!

    Thanks in advance! :)
     
  2. jcsd
  3. Feb 9, 2016 #2

    haruspex

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    You need to provide your own thoughts on answering those questions. Some equations will help.
    I assume the balls always rolled, never slipped, even at the steepest slope.

    Can you derive a theoretical expression for the final velocity based on mass, radius, friction coefficient, and height?
    What about air resistance? How do you think drag should depend on mass, radius, speed?

    Please provide details of how the height of drop and final velocity were measured.
     
  4. Feb 9, 2016 #3
    How they were measured dosen't really matter, it was a sensor and d= diameter and t= measured by sensor, so it wasnt really the 100% ending velocity. Height yeah.. by just measing it! d/t=velocity average

    1. Friction will be taking into account if slippering a bit and not rolling perfectly, then bigger mass => bigger normal force.
    2. I think because it has bigger area, is there any other way to explain it?
    3. I think it was just an accident, because i cannot find any phyiscal reason.
    4. Air resistance doing more work => more energy due to air resistance when bigger area, so less kinetic energy conserved for a ball with bigger area. Dosen't really know what this has to do with mass though, except the friction part, but friction shouldn't be big here i think. So what could mass else do , talking about conserving percentage wise to kinetic energy!

    I wrote a bit more in my first post, but here's the answer to the questions i think!
     
  5. Feb 9, 2016 #4

    haruspex

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    The details of how things are measured do matter. You have a ramp. You place a ball at some point on the ramp and release it. The velocity is measured, where? At some point further down the ramp, but before the ball reaches the level, or after the ball is on the level? If the former, the height drop is measured, I assume, from the point of contact of ball at start to where the velocity was measured. But if the velocity was measured on the level, how was the height measured?
    If slipping? You don't know?
    Greater mass means greater normal force, so greater maximum static friction, but how does that relate to energy loss? What happens to the energy in rolling contact?
    Yes, a larger area means more energy lost to drag, but you want to know about percentage energy loss, yes? The larger ball has greater mass and so more energy is involved. Assuming all balls have the same density, if you double the radius, what happens to the total energy involved, and what happens to the energy lost to drag?
     
  6. Feb 9, 2016 #5
    Thanks for the answers :)
    1. Energy and friction I cant find in my physicsbook. W= F*s so i think the force matter for fricrion. I dont know how to else describe

    2. V was measured at the end of ramp near ground. Height is h over ground- the height ball velocity was measured.

    3. Can area affected by air resistance be calculated ? Its not really the area of a sphere rather the half maybe. And i dont know any formula for air resistance. We learned force air = costant * v^2

    4. Dont really know what u mean by doubeling radius and air resistance energy!
     
  7. Feb 9, 2016 #6

    haruspex

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    Friction is a force between two surfaces parallel to them. In W=Fs, s is the distance the surfaces move relative to each other. In rolling contact, the surfaces do not move relative to each other, so friction does no work.
    What is the effect of friction when a ball rolls down a ramp? To answer that, consider what would happen if there were no friction. What two observations would be different?
    The height needs to be measured as the distance the centre of mass descends from where the ball was released to the point where the speed was measured. You can use the point of contact with the ramp instead of the mass centre if the ball is still on the ramp at both points. From what you write above, it sounds like the height was measured to ground, which is below where the speed was measured,
    Just think about how area and mass depend on radius (for fixed density). If you double the radius, what happens to the cross sectional area, and what happens to the volume and mass?
     
  8. Feb 10, 2016 #7
    Im not english so i cant really defend our measurments for velocity. But they were setup by my physics teacher and im 99 percent sure they were accurate. Notice that i wrote "-" a subtraction sign in my 2nd post which gives a hint that the right h was measured.

    This is the first course i ever read in physics so the friction i can think of is rolling friction but that would prob not come into play. Very intetesting with friction though. But i dont understand what ur seeking !! :/

    Well im not a mathematician and i never heard anything with doubeling radius and something happens with mass. The spheres had different density. So i cant really apply that!..

    One had weight of 60g and the other of 100g. Can i calculate density from that? And how does density come into play them. If i measure the diameter? :)

    And i still dont understand how to explain bigger volume more air resistance. What formula to use for area that is affected by air resistance!!!
     
  9. Feb 10, 2016 #8

    haruspex

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    Air resistance is generally proportional to cross sectional area, i.e. the area of the circle you would see if you cut the ball in half. If the radius of the ball is r, what would that area be? Do you know the radii of the balls?

    Regarding rolling (static) friction, as I wrote, static friction does no work. What it can do is apply both a force and a torque. The force reduces the linear acceleration, while the torque creates rotational acceleration. The net effect is that some of the lost PE goes into generating rotational KE instead of it all going into linear KE.

    You seem to want a qualitative explanation of why the larger ball was slower in your experiment. That is not going to happen. To get an explanation you need to get to grips with some equations (in the first place, to rule out some non-explanations). Do you know any equations related to the matter? I note you deleted that part of the template. It's there for a reason.
     
  10. Feb 10, 2016 #9
    A=4pi r^2

    So the area feeling air resistance would be

    A/2? right?

    So the rotational kinetic energy has to do with torque? How could i exaplin it though, because we havent really started with torque yet, i just know it has something to do with rotational! The only thing i find is this integral! https://gyazo.com/0dfabb462e90e0bef122b5f25ae3b6ff
    But I can't say i understand what it means :/


    What i know i only edit a spelling mistake, This is my first every post on the forum so i dont really know how it works yet! Probably a moderator who deleted something if something was deleted!

    Equations i know is

    Ro = m/V

    Fres=m*a

    PE=mgh
    KE=1/2mv^2

    And what our teacher learned us is

    F_air resistance= constant * v^2 (But i think this constant has something to do with area)



    You can probably do calculations with rotational kinetic energy, but we haven't learned that either :/
     
  11. Feb 10, 2016 #10

    haruspex

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    You wanted to know whether a larger ball would lose a greater fraction of its energy to drag.

    Regarding air resistance, I said cross-sectional area, not surface area, so it's πr2. But for present discussion, all that matters is the r2 part. If you double the radius, how much greater would the area be?

    The energy involved is proportional to the mass, so for constant density that would be proportional to volume, i.e. proportional to r3, if you double the radius, how much greater would the volume be?

    However, you say the two balls were different densities, so we cannot answer the question of whether the larger ball lost a greater fraction of its energy to drag without knowing what the actual radii and masses were.

    Since you have not studied rotational inertia, analysing the fraction of energy that went into that is somewhat difficult. So I will just tell you that for a uniform solid ball 2/7 of the energy will go into rotation, leaving only 5/7 of the lost PE to go into linear KE. But this does not depend on the radius or mass, so both balls should lose the same fraction of energy to rotation, and so end at the same speed.

    So what candidates do we have to explain the larger ball being slower?

    Drag might explain it, but only if the larger ball has a significantly lower density.
    Experimental error might explain it, depending on exactly how things were measured. For example, if the vertical distance from the start point on the ramp to the ground is h, the slope of the ramp is θ, and the radius of the ball is R, then a little geometry shows that the vertical distance travelled by the ball before reaching ground will be h-R(1-cos(θ)). As you can see, a larger R would give less vertical drop.
     
  12. Feb 10, 2016 #11
    Thanks man :)
    "Doubeling area greater would the area be?"

    1. (2r)^2= 4 times greater area! However, i dont know what area of a circle has to do with spehere rolling down. I feel like the air would feel more than just area of a circle!



    2. The ramp was more of an half U so angle dosen't really work! :/

    3.
    I dont understand why energy lost has to do with density!! And why "r^3" , well yeah energy involved is due to mass, but its mass on both sides so it shouldn't matter in velocity and in energy lost...we can just divide both sides with m in theory

    The bigger ball had a bigger density than the smaller ball xD

    4. ACcording to wiki air resistance is F_drag=0,5* density* A*v^2 * Constant, so shouldn't bigger density and bigger area make bigger f-drag!!
    However, maybe the velocity is smaller or is drag a good explanation!?
     
  13. Feb 10, 2016 #12

    haruspex

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    Yes, double the radius would give four times the area and so four times the drag.
    You should have described the ramp shape when you posted the problem. We are not mind readers.

    Twice the radius gives 8 times the volume, right? If the density is the same, that means 8 times the mass and 8 times the gravitational force, and 8 times the normal force, and 8 times the maximum frictional force. On that basis, i.e. ignoring drag, the acceleration should be the same.
    But the drag force is only four times as great, so what can we say about the net force and resulting acceleration?
    All mammals are about the same density. If I fall three metres I'm likely to injure myself, but a mouse would be fine. Why?

    You ask why density matters. Which falls faster, a balloon or a cannonball of the same radius? Why?
     
  14. Feb 10, 2016 #13
    1, Thanks :)
    the thing u last mentioned mouse and elefant is because the f-drag is reliant on density so a cannonball would feel it a lot more?

    Or what is the answer u looking for!!! I think that was a really good similiarty to proof physics though! Im really curious now hehe


    2. My friend also said that about the acceleartion, but why isn't it then..? Because f-drag but how much will that affect f-net?
     
  15. Feb 10, 2016 #14

    haruspex

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    A mouse and elephant would have about the same density, but much different sizes.
    The balloon and the cannonball have different density but the same size.
    Drag does not depend on density, that's the point.

    A body falling througn air is subject to two forces: gravitational force, mg, and drag, kAv2.
    If we assume a fixed density and a ball radius r, m=4πr3ρ/3. A=πr2.
    So what is the net downward force, and what is the resulting acceleration?
     
  16. Feb 11, 2016 #15
    Mass * g?
     
  17. Feb 11, 2016 #16

    haruspex

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    That is the vertical force due to gravity. I asked for the net vertical force, i.e. the sum of them, taking into account that they may be in different directions.
     
  18. Feb 11, 2016 #17
    Sorry but i dont know what force is vertical forware. Only friction is negative and f drag. Postive is acceleration due to g or what else make it accelerate????
     
  19. Feb 11, 2016 #18

    haruspex

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    Yes, and that's what I wrote in post #14: there are two vertical forces on a vertically falling body, mg down and drag (kAv2) opposing it. So what is the net downward force? (Do you understand what is meant by net force?)
     
  20. Feb 11, 2016 #19
    Sigma F = Mg-kAv^2

    So friction dosen't matter?
     
  21. Feb 11, 2016 #20
    The problem is energy has to do with F*s, so i get f-net how does that even help me with energy!
     
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