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Finding Kinetic Energy on a Vertical Circle with Friction

  1. Feb 11, 2010 #1

    jhae2.718

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    1. The problem statement, all variables and given/known data
    An object of mass m=200g slides down a frictionless ramp from a height H= 60 cm. Near the bottom of the ramp, the path changes into an arc of a circle with radius r= 20 cm. The arc has coefficient of kinetic friction μk=0.4. The angle θ=120° and points P1 and P2 which are the ends of the arc, are at the same height.

    Questions:
    *What is the speed of the object at point P1?
    *What is the kinetic energy of the object at point P2?

    diagram.PNG

    2. Relevant equations
    [itex]W_{ext}={\Delta}U+{\Delta}K+{\Delta}E_{therm}
    [/itex]
    [itex]
    {\Delta}E_{therm}=F_ks_{rel}
    [/itex]
    [itex]
    K=\frac{mv^2}{2}
    [/itex]
    [itex]
    U=mgh
    [/itex]
    [itex]
    F=ma
    [/itex]

    3. The attempt at a solution
    Finding the speed at point P1 is done through a simple application of the Work-Energy Theorem (letting U0 be at the bottom of the arc):
    [itex]
    {\Delta}H=H-r+r\sin{60^{\circ}}
    [/itex]
    [itex]
    mg{\Delta}H=\frac{mv^2}{2}
    [/itex]
    [itex]
    v=\sqrt{2g{\Delta}H}
    [/itex]
    [itex]
    v=\sqrt{2\left(9.80665 \,\mathrm{m/s^2} \right)\left(0.4+0.2\sin{60^{\circ}} \,\mathrm{m} \right)}
    [/itex]
    [itex]
    v \approx 3.35298\,\mathrm{m/s}
    [/itex]

    Finding the kinetic energy at P2 is much more difficult. Here, the surface transforms from a frictionless one to one with coefficient of kinetic friction 0.4. Since the height of P1 is the same as at P2, ΔU=0. Furthermore,
    [itex]
    {\Delta}K=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}
    [/itex]
    and
    [itex]
    {\Delta}E_{therm}=F_ks_{rel}=\mu_kF_ns_{rel}
    [/itex]
    The problem is that Fn is dependent on the angle at which the force of gravity acts relative to the surface of the curve. Additionally, the net force on the object is equal to the centripetal force on the object, [itex]\frac{mv^2}{r}[/itex]. Thus, to calculate the force of friction it becomes necessary to account for a variable force which is dependent on position and velocity. Velocity is also dependent on the force of kinetic friction.

    The normal force is
    [itex]
    F_n=mg\cos{\phi}+\frac{mv^2}{r}
    [/itex], where phi is the angle between the force of gravity and its component parallel to the normal force. Thus, the thermal energy is equal to:
    [itex]
    {\Delta}E_{therm}=F_ks_{rel}=\mu_k \left( mg\cos{\phi}+\frac{mv^2}{r} \right) s_{rel}
    [/itex]
    The total distance srel is the arc length, [itex]\frac{2\pi r}{3}[/itex].

    Thus, the problem becomes deriving an expression for Fk in terms of position or angle which can be integrated to find the work done by the force.

    I've tried writing a calculator program in TI-Basic to find the force at different positions, but it hasn't yielded the answer.
    U is μk, M is the mass of the block, G is acceleration due to gravity, X is the angle, and R is the radius of the curve. Z represents Fn at the point, and T is just used to start the list.
    Code (Text):

    For (X,-2pi/3,2pi/3,.01)
    U(MGcos(X)+MV^2/R)(XR)->{Z}
    sqrt(V^2-Z)->V
    If T=0
    Then
    {Z}->L1
    1->T
    Else
    augment({Z},L1)->L1
    End
    End
     
    This generates a list of various frictional forces, which I summed to get total work done, and then used to solve for Kf.

    The book says the answer is 0.71 J.
     
    Last edited: Feb 12, 2010
  2. jcsd
  3. Feb 11, 2010 #2

    turin

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    You never mentioned what the problem is to solve. You just gave some information.
     
  4. Feb 11, 2010 #3

    jhae2.718

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    Trying to find the KE at P2. I updated the original post.

    Seems I forgot to actually write the problem statement as I was fixing some LaTeX mistakes...
     
  5. Feb 11, 2010 #4
    I could be mistaken but I think your delta H for the velocity at P_1 is wrong?
     
  6. Feb 11, 2010 #5
    I'm mistaken!
     
  7. Feb 11, 2010 #6

    jhae2.718

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    The answer shown for the first question is correct; if I didn't make that clear I apologize.

    The first part is pretty straightforward; the latter is where it gets complicated. No one in the class is quite sure where to start, and there's no explanation offered as far as I know.

    Also, for reference, I've defined U0 as at the bottom of the arc.

    Thanks for looking at the problem.
     
    Last edited: Feb 11, 2010
  8. Feb 12, 2010 #7

    jhae2.718

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    For calculating the total energy dissipated as friction, is the following a correct method:

    srel remains constant as the arc length of the change in angle.
    Calculate Fk for the initial velocity. Use this and srel to find the next velocity as the position changes. Use this as the next velocity and repeat.

    Then, I should be able to use K=0.5m*v^2 to find the final kinetic energy, correct?
     
  9. Feb 12, 2010 #8

    turin

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    I would say, "yes." Are you supposed to use a computer program (or calculator program) to solve this, or you just want to do it that way? What you are describing is a simple numerical solution to a integro-differential equation.

    Sure, where I'm sure that you understand v to be the final resulting velocity. However, if you know differential equations, you can set up an inhomogeneous first order ordinary differential equation for K directly, and then solve it exactly. One of the tricks is to recognize that the centripetal force term can be expressed in terms of kinetic energy. Another trick is to choose a good independent variable (I used the angle φ that you are using in your expression for the normal force reaction to the weight).

    I'm sorry that I can't give you a simpler approach. Let me know if you are neither expected to determine a numerical solution nor use differential equations. Then, I will call upon the gods.
     
  10. Feb 12, 2010 #9

    jhae2.718

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    The question is for an AP Physics C class, and, as far as I am aware, neither a numerical nor diff eq is intended. Only basic diff eq's are covered, but we haven't done that yet, which is why I was attempting a numerical solution.

    I'd love to us a differential equation to solve it, as I have been trying to teach myself differential equations on my own time.

    So, would [itex]F_c=\frac{2K}{R}[/itex]?

    Thank you for your response.
     
    Last edited: Feb 12, 2010
  11. Feb 12, 2010 #10
    Isn't [tex] \Delta H = H - (r-rcos\theta)[/tex]?
    Not sure how do figure out the second part, I'd like to know though =\
     
  12. Feb 12, 2010 #11

    turin

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    That's what I get.

    Excellent. Let's do it! I will describe my approach:

    1) The first step that I suggest is to get an equation for K as a function of φ from energy principles. There are two things that will change the kinetic energy (as you have already recognized): a change in potential energy, which I'll call U, and the work due to friction, which I'll call Q. This will give you an integral equation for K, because K shows up in the integrand for Q. (You have basically already done this, you just need to interpret it with an independent variable φ and dependent variable K.)

    2) Next, differentiate the integral equation w.r.t. φ. The integral will depend on φ through the upper limit, so the derivative of the integral w.r.t. φ is just the integrand evaluated at φ. (You can actually prove that to yourself, or you can just accept it.) This will give you the 1st order ODE. Such a diff. eq. has an exact solution, meaning that you can express K as an integral of a known function (with some integration constant).

    3) In order to solve the diff. eq., I used an integrating factor. I think that this is a standard procedure. You can look up integrating factor on wikipedia to learn more about it. It is basically using the idea of the product rule for differentiation, except that the result is known and one of the factors is known and you want to find the other factor that makes it true.

    4) Once you get the integrating factor, you basically integrate a known function to find the solution.

    Maybe you can post your results for each of these steps so that we can see where you are getting stuck.
     
  13. Feb 12, 2010 #12

    jhae2.718

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    This is all fairly new to me, so I'm probably wrong, but this is what I've got:
    [itex]
    K_f=Q-K_i
    [/itex]
    [itex]
    K=\mu_ks_{rel} \left( mg\cos\phi + \frac{2K}{r} \right) - 1.1242\,\mathrm{J}
    [/itex]
    [itex]
    K=\mu_ks_{rel}mg\cos\phi + \mu_ks_{rel}\frac{2K}{r} - 1.1242\,\mathrm{J}
    [/itex]
    Letting [itex]s_{rel}=r\phi[/itex]...
    [itex]
    \frac{\mathrm{d}K}{\mathrm{d}\phi}=\mu_krmg\sin\phi + 2\mu_k\frac{\mathrm{d}K}{\mathrm{d}\phi}
    [/itex]
    [itex]
    \frac{\mathrm{d}K}{\mathrm{d}\phi} \left( 1-2\mu_k \right) = \left( \mu_krmg \right) \sin \phi
    [/itex]
    [itex]
    \frac{\mathrm{d}K}{\mathrm{d}\phi} = \frac{\left(\mu_krmg\right) \sin \phi}{1-2\mu_k}
    [/itex]
    [itex]
    \mathrm{d}K = \frac{\left(\mu_krmg\right) \sin \phi}{1-2\mu_k} \mathrm{d}\phi
    [/itex]
    [itex]
    K = \frac{\mu_krmg}{1-2\mu_k} \int_0^{\frac{2\pi}{3}}{\sin \phi\,\mathrm{d}\phi}
    [/itex]
    [itex]
    K = \frac{\mu_krmg}{1-2\mu_k}\left. {\cos \phi } \right|_0^{\frac{{2\pi }}{3}}+C
    [/itex]
    [itex]
    K = 0.7845\,\mathrm{J} \left. {\cos \phi } \right|_0^{\frac{{2\pi }}{3}}+C
    [/itex]
    I'm guessing that C=1.124 J...

    Somewhere I went wrong, as I end up with -0.05275 J..
     
  14. Feb 12, 2010 #13

    jhae2.718

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    Actually, I think you're right...

    That would make [tex] K_i = 0.980665 \,\mathrm{m/s} [/tex].
     
  15. Feb 15, 2010 #14

    turin

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    I figure that the initial decent for the first part of the problem is

    Δy = - H + r( 1 - cos(θ/2) )

    I guess y'all are calling the change in height ΔH rather than Δy. Anyway, other than that sign reversal, I figure that the angle should be half of the angle between P1 and P2.
     
  16. Feb 15, 2010 #15

    turin

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    This is backwards.

    This is where the calculus should come in. K is reduced by adding up all of the little energy losses from one little increment to the next along the path through the arc. These energy losses are due to friction (Q). You have written it as one big whammy (i.e. not an integral). This is incorrect because the friction force changes along the path.

    You must write Q as a "sum" (integral) of little contributions along the path. That means replacing s with ds, and then integrating. Do you know what an integral is?

    Again, this would be ds = r dφ. These are increments in the path, not the entire path.

    By this point, it is all messed up. I am willing to continue to help you with the calculus, but perhaps a numerical approach would better suit you. A numerical approach will require the same basic concepts as the analytical approach, but without the extra "formality" and notation. For example, instead of ds and dφ, you would have Δs and Δφ, and instead of an integral (which is denoted by the stretched "s" symbol, ∫), you would just have an ordinary sum (of a lot of terms). I think that both approaches will provide good simple exercises for starting a physics career. For your level of math, the calculus approach may be detrimentally distracting from the meaning of the physical process, and you may want to wait until you've gone through a college semester worth of calculus.

    BTW, this seems rather involved for an AP problem (that's advanced highschool, right?) Seeing your difficulty with the calculus, I am starting to think that there must be an easier approach that I am not seeing. I will call upon the gods to help you.
     
  17. Feb 15, 2010 #16

    jhae2.718

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    Here, I must apologize for sloppy notation. I was using theta to represent both theta and theta/2.

    I see where I made a mistake. I'm not very familiar with setting up differential equations...
    I am familiar with both differential and integral calculus. But, to answer the question, an integral is a Riemann sum with the limit as Δx approaches zero.

    I'll admit I should have known this.

    The basic calculus should not be a problem; really, my difficulty is from having never had to set up a problem like this before. I've already taken AP Calculus BC, but the only differential equations we've had to solve were for logistic growth, if I remember correctly. That said, I plan on retaking calculus in college, as I don't want to go into Calc II having not regularly used calculus for over a year, especially studying engineering.

    AP is high school. Yes, the problem does seem rather complicated, which is why we (including the teacher) thought that there had to be a simpler way to result in a solution. That said, I can't really think of a way that would result in using either a calculus or numerical approach.

    I really appreciate your help, and I'm sorry if I'm not catching on fast enough. I'll try to rework the problem and post back later.
     
  18. Feb 15, 2010 #17

    jhae2.718

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    Okay, so I think I've gotten the correct expression for my diff eq:
    [itex]K_f=K_i-Q[/itex] //Corrected original algebraic error...
    [itex]
    K_f=K_i-\int_0^{\frac{.4\pi}{3}}{ \left( \mu_kmgcos\phi+\frac{2K}{r} \right)\mathrm{d}s
    [/itex]
    [itex]
    K_f=K_i-\int_0^{\frac{2\pi}{3}}{ \left( \mu_kmgcos\phi+\frac{2K}{r} \right) r\mathrm{d}\phi
    [/itex]
    [itex]
    \frac{\mathrm{d}K}{\mathrm{d}\phi}=-\left( \frac{-\mu_kmg}{2}+\frac{2K}{r} \right) r
    [/itex]
    [itex]
    \frac{\mathrm{d}K}{\mathrm{d}\phi}=\frac{\mu_kmgr}{2}-2K
    [/itex]
    [itex]
    \frac{\mathrm{d}K}{\mathrm{d}\phi}+2K=\frac{\mu_kmgr}{2}
    [/itex]
    This is the 1st order ODE I end up with.
     
  19. Feb 15, 2010 #18

    turin

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    Firstly, I want to say that you seem to be very bright, and I appologize if I sounded condescending. I don't want to sound cheesy; I mean that sincerely.

    For example, you realized that you needed to include the centripetal force as part of the normal force. I would have carelessly neglected that part, and therefore solved the problem incorrectly (and I'm not the only one)! In fact, I suspect that the author of the problem may have even made this mistake, because it is precisely this ingredient of the problem that makes it so involved (i.e. requiring a diff. eq. instead of just an integration).

    Also, congratualations for learning some latex. I believe that will serve you well.

    I just hope that you don't get too frustrated by my feable attempt to help, and I appologize if I sound repetitive. I just want to make sure to explain exactly where you are making mistakes. Anyway, on to the problem:

    Correct. (However, see the following.)

    This is almost correct. First of all, just a little nitpicky detail: the φ and K in the integrand should be funtcions of s. However, regarding the K, this little detail becomes quite important. Since the rate of change in K (w.r.t. s or φ) depends on K, you must account for every change in K along the path. This means that you also must include the small changes in gravitational potential energy (U) that contribute to a change in K along the path, because this will in turn change the centripetal force. So, you really need to generalize your previous equation a bit by using a general K instead of Kf on the l.h.s.. For instance, make K (and U and Q) functions of s (or φ):

    K(s)=Ki-Ui+U(s)-Q(s)

    or

    K(φ)=Ki-Ui+U(φ)-Q(φ)

    or just remember that they are implicitly functions of some variable along the path, and then write in the appropriate expressions in terms of that variable. This means that you need to replace the upper integration limit with the path variable (e.g. s or φ), and change the integration variable to a "dummy" variable (e.g. s' or φ').

    Ignoring the problem mentioned in the previous step, this is the correct idea. It is just a simple change of variables (something that you will learn more about in calculus 2 and 3 where it gets a lot more complicated).

    At this point you cannot do a straightforward integration, because the K in the integrand depends on the (dummy) path variable, and you don't yet know how. So, this step is where you should get the diff. eq.. The upper integration limit would be a variable (e.g. s or φ) instead of the constant 2π/3, and don't forget to use the "dummy" integration variable (e.g. s' or φ').

    In order to get the diff. eq., you will need this "rule" (which is actually not too difficult to prove using the definition of a derivative in terms of a limit, and is closely related to the fundamental theorem of calculus):

    [tex]\frac{d}{dx}\left(\int_{x_0}^{x}dx'f(x')\right)=f(x)[/tex]

    It's probably important to point out that this "rule" requires f to be a continuous function, at least in a neighborhood of the point x.

    (Actually, after writing this definition out for you, I realize that the integral equation can be skipped. But, it's a bit more abstract, and I don't want to confuse you any more than I already have.)
     
  20. Feb 15, 2010 #19

    jhae2.718

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    Okay. Try #3...
    [itex]
    K\left(\phi\right)=K_i-U_i+U\left(\phi\right)-Q\left(\phi\right)
    [/itex]
    Define U(φ) and Q(φ) as:
    [itex]
    U\left(\phi\right)=mgr\left( 1-\cos\phi \right) \, d\phi
    [/itex]
    [itex]
    Q\left( \phi \right) = \left(\mu_kmg\cos\phi+\frac{2K\left( \phi \right)}{r} \right) rd\phi
    [/itex]
    Substituting in and using the dummy variable φ':
    [itex]
    K\left(\phi'\right)=K_i-U_i+mgr\int_0^\phi{\left( 1-\cos\phi' \right) \, d\phi'}-r\int_0^\phi{\left(\mu_kmg\cos\phi'+\frac{2K\left( \phi' \right)}{r} \right) d\phi'}
    [/itex]
    Taking the derivative dK/dφ:
    [itex]
    \frac{dK\left(\phi\right)}{d\phi}=mgr\left( 1-\cos\phi \right)-r\left(\mu_kmg\cos\phi+\frac{2K\left( \phi\right)}{r} \right)
    [/itex]
    [itex]
    \frac{dK\left(\phi\right)}{d\phi}=mgr-mgr\cos\phi-\mu_kmgr\cos\phi-2K\left( \phi \right)
    [/itex]
    [itex]
    \frac{dK\left(\phi\right)}{d\phi}+2K\left( \phi \right)=mgr-mgr\cos\phi-\mu_kmgr\cos\phi
    [/itex]
    [itex]
    \frac{dK\left(\phi\right)}{d\phi}+2K\left( \phi \right)=U \left( \phi \right)-\mu_kmgr\cos\phi
    [/itex]

    Thanks again for all your help. I really appreciate it...
     
    Last edited: Feb 15, 2010
  21. Feb 16, 2010 #20

    turin

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    Sorry, there was a typo in my previous post, similar to your typo regarding K=Q-Ki. The signs on the potential energies are backwards. This should be

    [tex]K\left(\phi\right)=K_i+U_i-U\left(\phi\right)-Q\left(\phi\right)[/tex]

    That typo should have stood out as a red flag. We are talking about two states, the initial state indicated by the subscript "i" and the (generalized) final state indicated by the dependence on φ. (The final state for our purpose is the state of the object when it reaches the angle φ, which we allow to vary.) The initial state energy is

    Ei=Ki+Ui

    and the final state energy is

    E(φ)=K(φ)+U(φ)

    The energy in any given state is always the sum of the kinetic and potential energies. In a lossless (i.e. frictionless) system, you would have

    E(φ)=Ei

    However, in the case of our lossy system, the difference between the initial energy and final energy is the accumulated energy loss due to friction, Q(φ). So

    E(φ)=Ei-Q(φ)

    I hope that all such typos will be obvious now.

    These energies are not differential energies, but cummulative energies. In particular, you do not need an integral to express U(φ); it is just the value of the gravitational potential energy when the object is at φ. Q(φ) needs to be expressed as an integral, because it is nonconservative (not a state property).

    BTW, I just noticed another detail. The second term in your integrand for Q is just the normal force, not the friction force, so you need to fix that (a simple fix).

    I'm assuming that you're defining φ as the angle from the downward vertical. If so, then the lower limit on the integral should not be φ'=0 unless you somehow know the kinetic energy at that point. Rather, you know the initial kinetic energy at a different value of φ, and that should be your lower limit.

    Actually, this doesn't matter from a practical standpoint, since the derivative of the integral does not depend on the constant lower limit, but you should try to understand the expressions that you are writing. Think of the integral as a mathematical description of an accumulated effect from some starting point to some ending point. In our case, the starting point is when the friction begins to act, and the ending point is the variable position, φ.

    OK, you correctly used the "rule" for differentiating the integral, but the you need to go back and fix your implementations of U and Q (and the minus signs - sorry again for that typo). You are getting closer.
     
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