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KE, v squared and a constant of proportionality

  1. Feb 20, 2005 #1

    DB

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    I was reading about Kinetic Energy and why it is 1/2 mv^2. I couldn't really understand what it was saying, it was mentioning a constant of proportionality being:
    [tex]\frac{kg * m^2/s^2}{m^2/s^2}[/tex] and then of course you're left with mass. It was very confusing, so basically what I'm asking is for a basic mathematical explanation of why Ke=1/2mv^2

    Thanks in advance.
     
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  3. Feb 20, 2005 #2

    arildno

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    [tex]m\vec{a}\cdot\vec{v}=\frac{d}{dt}(\frac{1}{2}m\vec{v}^{2})[/tex]
    where the mass has been assumed constant.
     
  4. Feb 20, 2005 #3

    dextercioby

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    Assume for simplicity mass to be constant...Then the mathematical form of the second principle becomes:

    [tex] m d\vec{v}=\vec{F} dt [/tex]

    Multiply with the vector [itex]\vec{v} [/itex] and take into account that

    [tex] \vec{v} dt=d\vec{r} [/tex]

    ,u get:

    [tex] d(\frac{1}{2}m\vec{v}^{2})=\vec{F}\cdot d\vec{r} [/tex]

    Define:

    [tex] T=:\frac{1}{2}m\vec{v}^{2} [/tex]

    and differential work:

    [tex] \delta W=:\vec{F}\cdot d\vec{r} [/tex]

    And u obtain the law of variation of KE (due to G.W.Leibniz) in differential form:

    [tex]dT=\delta W [/tex]...

    This is basic stuff...

    Daniel.
     
  5. Feb 20, 2005 #4

    Andrew Mason

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    I gather that you are interested in knowing WHY kinetic energy is important as well as why it is 1/2mv^2.

    In case dexter has totally confused you, what he is saying is that kinetic energy is the measure of the amount of Work required to give an object its motion (or an object's ability to do Work). Work, which is force applied over a distance, is considered a useful quantity. Can you think of why that is? Hint: the units of work or energy are Joules, named after James Joule and he is famous for working with heat. He discovered a relationship between 'work/energy' or force x distance and the amount of heat produced.

    If you apply a constant force F to an object of mass m over a distance d, and the only resistance to that force is the inertia of the object, then the object will experience a constant acceleration a = F/m so at time t, v = at and d = vt/2.

    So [itex]Fd = mad = \frac{mvd}{t} = \frac{mv(vt)}{2t} = \frac{1}{2}mv^2[/itex]

    AM
     
  6. Feb 20, 2005 #5

    DB

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    why is it that distance=vt/2, why divided by 2?
     
  7. Feb 20, 2005 #6

    dextercioby

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    He would have written [itex] d=\frac{at^{2}}{2} [/itex],but he found more useful to use velocity.

    Daniel.
     
  8. Feb 20, 2005 #7

    cepheid

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    Remember that the object is moving along with constant acceleration. So its change in velocity is the rate of change of velocity with time (i.e. the acceleration) multiplied by the total time interval:

    [tex] \Delta v = at [/tex]

    We're assuming the object started from rest, and then achieved a final velocity 'v' after accelerating for t seconds:

    [tex] \Delta v = v - 0 = v = at [/tex]

    So since v is the final velocity achieved, and since the acceleration is constant, the vt graph is a straight line (with slope a) starting from v = 0 and going to v = v. The total distance travelled during this time is the area under that graph, which is the area of the triangular region beneath that vt graph. It has base t and height v, so it has area vt/2 (= d).

    Note that if the object had been travelling at velocity v all along (draw a horizontal line v = v from t = 0 to t = t), then you have the total distance travelled is vt, and the area underneath that flat line is a rectangle with area vt. But since you started from zero and only ended up at v at the end, your distance travelled is half that (the area of the triangle is half the area of the rectangle).

    Without all of this reference to area, what this is saying is that if you accelerate smoothly from v = 0 to v = v, you will travel the same distance as you would have if you had been travelling at velocity v/2 for the entire time. The distance covered is the average velocity (v/2) times the time interval. This is a consequence of the mean value theorem in calculus.

    Phew! I hope that answers it.
     
  9. Feb 20, 2005 #8

    Andrew Mason

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    If it starts at 0, accelerates (v = at), and if it ends up with speed v, what is the distance covered? If it was travelling at speed v all the time, the distance would be d=vt. But it starts at 0. The distance is the average speed multiplied by time:

    [tex]d = \frac{v_f + v_i}{2}t = vt/2[/tex]

    Edit: I see that cephid has answered this quite well.
    Edit: Corrected - sign in numerator

    AM
     
    Last edited: Feb 21, 2005
  10. Feb 21, 2005 #9

    dextercioby

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    It matters that the accleration is constant (namely that the graph of the velocity is a straight line).Else,an integration would have been required to compute the average velocity.

    Daniel.
     
  11. Feb 21, 2005 #10

    Tom Mattson

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    Ya gots ta add the velocities to get the average.
     
  12. Feb 21, 2005 #11

    dextercioby

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    And that follows from the definition:
    [tex] v(t)=at [/tex] (1)
    [tex] v(0)=0 [/tex] (2)
    [tex] v(T)=aT [/tex] (3)
    [tex] \langle v\rangle_{[0,T]}=:\frac{1}{T}\int_{0}^{T} v(t) dt=...=\frac{aT}{2} [/tex] (4)

    Therefore:[tex] S_{[0,T]}=\langle v\rangle_{[0,T]}}T=\frac{aT^{2}}{2} [/tex]

    Daniel.

    P.S.To prove the "sum/2" part,substitute the limits of integration:

    [tex]0\rightarrow \tau [/tex] (5)
    [tex]T\rightarrow \tau+T [/tex] (6)
     
  13. Feb 21, 2005 #12

    DB

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    Many thanks for all your help guys, I clearly see why KE = 1/2mv^2. Just one more question. When I'm looking at a linear graph of constant acceleration, and I want to solve for the distance d, am I looking a a triangle (as cepheid said) with a maximum point of when v=v, and then that side connecting too the time v reached v? Thanks
    PS I'm seeing t as the x axis, v as the y.
     
  14. Feb 21, 2005 #13

    dextercioby

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    Then,the graph should be a parabola...Because linear (in time) acceleration,requires,by integration,a parabolic dependence of time for the instantaneous velocity...

    Daniel.
     
  15. Feb 21, 2005 #14

    Andrew Mason

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    Yes. d is the area under the graph. v is the rate of change of the area (d) with time. Since that area is a triangle (if it starts at v,t = 0,0), its area is 1/2 base x height: [itex]A = d = 1/2 vt = 1/2(at)t = \frac{1}{2}at^2[/tex]. Or you could just integrate: [itex]\int vdt = a \int tdt = \frac{1}{2}at^2[/itex].

    AM

    PS. I am not sure what dexter's last post is about...
     
  16. Feb 21, 2005 #15

    dextercioby

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    I missed the word "constant" (on the acceleration) and read just "linear (in time) acceleration"...That's how i justify my post.

    Daniel.
     
  17. Feb 22, 2005 #16

    cepheid

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    Yes. One vertex of the triangle is the origin (lower left end of vt graph). Now go to the maximum v...this point is the second vertex (upper right end of vt graph). Now project a line vertically down from that point until it intersects the x-axis (at time 't', which is the time at which you reached 'v', just as you said). That point on the x-axis is the third vertex. I guess a diagram would have been helpful here.
     
  18. Feb 22, 2005 #17

    DB

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    Thanks a bunch
     
  19. Feb 22, 2005 #18
    DB I read all the answers here, and nowhere did I see what I would have put. I would have answered you by explaining to you an experiment that Galileo did several hundred years, and that anyone can do, and then tell you the answer to your question lies somewhere in this experiment.

    Of course all the approaches above are nice too.

    But, if I had to develop a mind based on all the knowledge that can be gleaned from just one experiment, this would be it:

    Physics Classroom

    Galileo and Inertia

    And this site is my favorite:

    Newtons First Law

    This experiment gives you motivation to define total system energy E as the sum of the kinetic energy of the ball, with the potential energy of the ball, and figure out that if there was no 'friction' that the total system energy would be time invariant, and hence conserved, and that you would have built a perpetual motion machine.

    This would mean that something is invariant. But what?

    The answer is the system's total energy E.

    Define: E=T+U

    dE/dt=0 (total system energy constant in time; invariant)

    T=kinetic energy
    U = potential energy = mgh

    Define the speed of the center of inertia of the ball in the rest frame of the experimental device v. The following quantity is conserved:

    mgh + (1/2mv^2)

    where the inertial mass m of the falling ball is also treated as invariant. In other words dm/dt=0

    Your goal is to formulate true statements that are frame independent.

    One of the many things that Galileo inferred from this experiment, is that if he reduced what he called 'friction' (I don't know the italian word he used), the ball almost rose to the same height it was released from, all else constant. He knew that air friction slowed it down, but he also noticed that
    if he smoothed the surface down (imagine the surface is made out of ice instead), it went higher. The key thing to note, is that if what are called frictional forces are removed from the set up, so that they don't act on the ball after it is relased, the only external force that acts upon the ball is the force called gravity. So that brings us to gravity.

    We cannot remove gravity now can we.
    Gravity is not a frictional force now is it?

    The earth is under the experiment, and in fact is what gets the ball to fall in the very first place.

    The weight of the ball is given by:

    [tex] W= Mg [/tex]

    Where g is the local acceleration due to gravity (which can be measured very easily), and M is the 'mass' of the ball.

    If we mix up Galileo's thoughts with Newton's we get the following:

    [tex] F = \frac{GMm}{r^2} = mg [/tex]

    Where M is the mass of the earth, m is the mass of the ball which we watch roll back and forth, g is acceleration due to gravity, r is the distance between the center of inertia of the earth, and the center of inertia of the ball, and G is the Newtonian Gravitational constant.

    F is the 'force' of gravity.

    But, if Newton's law is true, the g isn't actually a constant now is it. It's value really would depend on just what h is. The higher you dropped the ball, the lower would be its rate of free fall, so that if you released it high enough, it wouldn't fall at all, it would float.

    So really, that brings us to the fact that the force of gravity above is a conservative force, meaning that it can be written as the gradient of a scalar function U, and mathematically, the curl of the gradient of U must be zero for a conservative force.

    Let F denote the force of gravity.

    We have:

    [tex] \vec F = -\nabla U [/tex]

    But where on earth did this come from?

    It is a purely mathematical fact that:

    Theorem:

    [tex] \frac{\hat R}{R^2} = \nabla(\frac{-1}{R}) [/tex]

    You prove this.

    In order to formulate gravitational field theory, you start off defining gravity as a central force.

    This means that if two bodies interact gravitationally, that the direction of the force of object one on object two is along the straight line path from the center of inertia of object one to the center of inertia of object two.

    That is the symbol [tex] \hat R [/tex] that you see above. That is a unit vector which points from the CM of body one, to the CM of body two.

    Define the gravitational field of an object of mass M as follows:

    [tex] \vec \Gamma = GM\frac{\hat R}{R^2} = GM \nabla (\frac{-1}{R} )[/tex]

    Then an object of mass m will experience a gravitational force given by:

    [tex] \vec F = m \Gamma [/tex]

    So that the gravitational force o an object with mass M, on an object with mass m is given by:


    [tex] \vec F = m \vec \Gamma = GMm\frac{\hat R}{R^2} = GMm \nabla (\frac{-1}{R}) [/tex]

    The only problem now is that the inertial mass of the earth is a strictly positive quantity, and the inertial mass of the falling body is a strictly positive quantity, and the way we have the force formula above set up, the force between the two objects is repulsive instead of attractive, which is wrong since gravity is assumed to always pull things together not push them apart.

    So all we have to do now, is just throw in a minus sign right?

    Instead define the gravitational field as follows:

    [tex] \vec \Gamma = -GM\frac{\hat R}{R^2} = -GM \nabla (\frac{-1}{R} )[/tex]

    Now we have the gravitational force on an object with a mass of m, given by

    [tex] \vec F = m \vec \Gamma [/tex]

    And the force is attractive.

    Let's step back a moment, to the gravitational field.

    [tex] \vec \Gamma = -GM\frac{\hat R}{R^2} = -GM \nabla (\frac{-1}{R} )[/tex]

    Suppose that the density of an object is represented by [tex] \rho [/tex]

    Density is mass divided by volume, so we can express the field using the integral calculus as follows:

    [tex] \vec \Gamma = -G \int \rho d\tau \frac{\hat R}{R^2} [/tex]

    Where d tau is a differential volume element, so that this is a volume integral.

    Now use the mathematical theorem above, to write the gravitational field created by a body with inertial mass M as follows:

    [tex] \vec \Gamma = -G \int \rho d\tau \nabla (\frac{-1}{R}) [/tex]

    For mathematical reasons alone, you can pull the gradient symbol all the way out front and write this as:

    [tex] \vec \Gamma = \nabla -G \int \rho d\tau (\frac{-1}{R}) [/tex]

    And now lets pull out that stray minus sign which we inserted ad hoc, to make the force of gravity purely attractive, and write:

    [tex] \vec \Gamma = - \nabla G \int \rho d\tau (\frac{-1}{R}) [/tex]

    Thus, we have:

    [tex] - \vec \Gamma = \nabla G \int \rho d\tau (\frac{-1}{R}) [/tex]

    So...

    [tex] \vec F = m\Gamma = - m (-\vec \Gamma ) = - (-m\vec \Gamma ) [/tex]

    And -m times Gamma is:

    [tex] - m\vec \Gamma = m\nabla G \int \rho d\tau (\frac{-1}{R}) [/tex]

    Now, if we assume that m isn't affected by the gradient operator, we can allow the gradient operator to act upon it, so that we have this:

    [tex] - m\vec \Gamma = \nabla G m\int \rho d\tau (\frac{-1}{R}) [/tex]

    Now we can finally see what U is.

    [tex] U = G m\int \rho d\tau (\frac{-1}{R}) [/tex]

    Notice that U is a scalar function, not a vector function.
     
    Last edited: Feb 22, 2005
  20. Feb 22, 2005 #19

    Tom Mattson

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    Physicsguru, what on Earth are you going on about? That doesn't answer his question in the slightest.

    There is no experiment that tells answers his question, because his question is about the theory. That's what it means when someone asks for a mathematical explanation. Having said that, I too have not seen the answer that I would have given. He is in essence asking for a proof of the work-energy theorem.

    It starts from the definition of work. For simplicity, I'll stick to a particle moving in 1D.

    [tex]
    W=\int_{x_1}^{x_2}Fdx
    [/tex]
    [tex]
    W=m\int_{x_1}^{x_2}\frac{dv}{dt}dx
    [/tex]
    [tex]
    W=m\int_{x_1}^{x_2}\frac{dv}{dx}\frac{dx}{dt}dx
    [/tex]
    [tex]
    W=m\int_{x_1}^{x_2}\frac{dv}{dx}vdx
    [/tex]
    [tex]
    W=m\int_{v_1}^{v_2}vdv
    [/tex]


    Note that in the last step I changed the limits of integration, which is because I changed the variable of integration.

    Finally:

    [tex]
    W=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2
    [/tex]

    If you start from rest then v1=0 and you can simply set v2 equal to v, since there will no longer be any need to distinguish the two. Then you have the formula for KE.
     
  21. Feb 22, 2005 #20

    DB

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    Thank you both Physiscsguru and Tom, I apreciate your answers and I found that once I read Tom's response it helped me understand more clearly Physiscsguru's, though I must say my calculus is very very rusty, but I'm doing my best, and will be looking back to your posts throughout my education.
     
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