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Can someone help me understand the concept of energy?

  1. Nov 23, 2012 #1
    My teacher assigned us a lab, where we use a resource called "skate park simulator" to come up with a question, and then the resulting formal lab report. It involves a skater on a half-pipe, but I think it could be simplified into the following:

    Basically, an explanation to this question will help me greatly (this is not the lab question)

    Will a ball rolling down a 3m high ramp have the same final kinetic energy as if it were just dropped from a height of 3m? To me, this doesn't make sense because the final velocities will be different in each scenario (E = 1/2mv2). If E = mgh, they should both have the same potential energy, and therefore the same mechanical energy. However, the ball at the bottom of the ramp should have a different velocity than the ball free-falling just before it hits the ground.
     
  2. jcsd
  3. Nov 23, 2012 #2

    lewando

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    If you are making a distinction between a vertical velocity vector and a horizontal velocity vector, then yes the velocity vectors are different, however the v in KE=1/2mv2 is actually the magnitude of the velocity vector (speed).

    If you are making a distinction about the differing speeds of a dropped ball versus the rolling ball, they will be different based on things like air resistance, rolling resistance, and probably more importantly, a form of KE known as rotational KE possessed by the rolling ball.

    If your experiment was done in a vacuum with a frictionless ball and ramp (so the ball does not roll) then the final speeds of the dropped ball and ball sliding down the ramp would be equal.
     
  4. Nov 23, 2012 #3
    "However, the ball at the bottom of the ramp should have a different velocity than the ball free-falling just before it hits the ground. "
    Show your work.
    Give us the expressions you have obtained for the kinetic energy of the ball in both cases.
    Remember that the kinetic energy associated with a spinning ball (such as one that rolls down the half-pipe) is still kinetic energy - rotational rather than translational.
     
  5. Nov 23, 2012 #4
    So let's say we forget about energy for a second. If I had a ramp, elevated at 30o, and a ball at the top of the ramp, the acceleration would be 9.81ms2 30 degrees below horizontal, correct? That is assuming no friction.

    I had a question like this during class once:

    A 50kg child decides to try out a zipline. The zipline is 158m long and the top of the zipline is 19m above the bottom of the zipline. How long does it take the child to go down the zipline?

    opposite side = 19m
    hypotenuse = 158m

    sinθ = 19m/158m
    θ = 6.9°

    Fg = (50kg)(-9.81ms-2)
    Fg = -490.5N
    FNy = 490.5.

    Angle of Normal force is [Up 6.9 Left]
    Therefore, the other component of the normal force (6.9° from the perpendicular)

    sin6.9 = x/490.5N
    x = 59N

    So, a = FNxa
    59N/50kg = a
    acceleration on a 6.9 ° slope of a 50kg child is: 1.2ms-2

    This is what I did, as of now, it makes perfect sense to me, but so does what you described to me.

    My teacher said this was correct, but that doesn't mean much because she has -on multiple occasions- asked me for help (correcting homework, marking tests), so my physics teacher can't help me much more than I can help myself.

    EDIT: The acceleration is not the same, obviously (1.2m/s/s does not equal 9.81m/s/s, but if I was to calculate the final velocity with the given distance, the magnitude of that angular velocity should be the same as if it just free-fell for that same vertical distance. I think I understand now. (I did the calculation, it was close enough to be considered equal).
     
    Last edited: Nov 23, 2012
  6. Nov 23, 2012 #5
    Okay, well if we have a 3m ramp at some arbitrary angle, and a another ball 3 m high, the length at which the ball travels down the ramp is greater than 3 m, and throughout the fall and roll of each of the balls they still travel the same distance vertically (ignore friction, very important). To make it even clearer, work is done by gravity acting on the ball throughout a distance vertically downwards (ignoring friction) if they both travel the exact same distance downwards the work done would be equal (since work is force cross distance since if we apply a force downwards and an object is initially moving sideways it doesn't do any work sideways, only downwards). If this isn't clear enough I can try and chuck in a few more examples.

    It might be quite difficult to think about a ball travelling the same speed at the bottom of the ramp as a ball that was dropped at the same height (initially). Probably due to the sideways motion due to the normal force on the ramp and friction
     
  7. Nov 24, 2012 #6
    Nope I think I get it now. If a 158m long, 19m high ramp is at an angle of 7°, the magnitude of the final velocity vector (just before the ball rolls off the ramp and onto the ground (also point of reference) --assuming no friction) will be the same as the magnitude as the final velocity vector of a ball free falling from a height of 19m just before it hits the ground, correct? That is assuming no air resistance, friction or energy loss of any type.
     
  8. Nov 24, 2012 #7

    lewando

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    If there is no friction on the ramp, then the ball will not roll--it will slide. Otherwise it will roll--then it will have rotational KE along with translational KE, the sum of the two will equal the KE of a dropped ball from the same height.
     
  9. Nov 24, 2012 #8
    Yeah, I'm pretty sure he is talking about the no frictional case without torque...

    Edit: @Fifty - And yes, you are correct
    Just remember that there is horizontal momentum too - PS: there is no energy lost if you're talking about the Earth-ball system without friction,etc,etc
     
  10. Nov 25, 2012 #9
    I understand now. For some reason, our teacher did not teach us about momentum. The only thing I know about momentum is that the unit is a kg-ms-1

    The only topics we have covered are: kinematics, forces, energy/work/power. Based on what is in our text book, it seems like we skipped over a lot of information. I don't remember the words: torque, momentum or average acceleration being used in our class so far. Considering we're almost done with motion (after the work/energy/power unit) I don't think she plans on covering those topics :( Anyways, the point is that I never thought about kinetic energy as the rolling motion of the ball.

    But wait, wouldn't the speed of the outer surface of the ball be the same as the speed of the ball in its entirety?
     
  11. Nov 25, 2012 #10

    lewando

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    Fill in any gaps you think you might have with your own independent study. If your teacher is skipping over parts in your textbook, feel free to read and understand them on your own.

    Yes, for the case of rolling without slipping, the speed of a point on the outer surface of the ball is the same as the overall speed of the ball (the speed of the center point of the ball). Good observation.
     
  12. Nov 25, 2012 #11
    I definitely will read up on topics that I missed, I don't want to be under-prepared for next years physics class, and then be under-prepared for university.

    Thanks for the help, I feel like I understand every concept I've learned so far like the back of my hand.

    If you have a really tough problem in mind that has to do with energy, that would be appreciated, but you have given more than enough help already :).
     
  13. Nov 25, 2012 #12
    Momentum is just a property of a system, it basically arises from inertia. Momentum is just as if you threw a ball in outer space (far away from any other objects) by applying a force to it. So, lets consider the system of the ball and you (ball-you), the original state of the system is you and the ball are still, zero momentum (since momentum = mass * velocity). Once you apply a force to the ball for a small period of time (the length you extend your arm outwards in the direction you're pushing) the ball's momentum changes.

    There's also a thing called conservation of momentum, so going back to the system ball-you, the original state was 0 net momentum (neither you or the ball were moving) so you apply a force to the ball for a certain period of time and the ball goes flying off. But, you also go flying backwards with the exact same momentum as the ball! Therefore, the net momentum of you and the ball is still zero! Therefore we can conclude Newton's third law (there is always an equal and opposite force). So to re-cap momentum is conserved in any system i.e. the momentum is the same no matter what, unless an external force acts on an object within the system. You could imagine a huge lump of dust and rocks in outer space, and if we ignored gravitational effects and external forces, all of the momentum is conserved!

    Just to give you a little bit of an in-sight, even though how basic conservation of momentum is, it is quite interesting! :) I hope you find some beauty and interest in physics, it might seem very dull at school, but you can get quite a rush out of thinking about these sorts of things
     
  14. Nov 25, 2012 #13
    No are you kidding?!?! I LOVE physics. It's my favourite class, even though our teacher isn't the best. I understand the rush you feel.

    I think I understand the concept of momentum. I asked my teacher (after learning that F = ma) why a truck moving a steady speed of 80km/h applies a force to an unfortunate pedestrian. This was before I knew Newton's laws of motion, so I concluded that at the moment of impact, the pedestrian must also be applying a force to the truck, causing it to accelerate.

    Anyways, thanks for the explanation. I'm sure after a bit of thinking I'll understand it better. Now I have to think of a question for the lab that was stated in the Original Post! (It's due in two days!)
     
  15. Nov 25, 2012 #14
    Haha, good luck!
     
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