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Homework Help: Generating Heat

  1. Feb 29, 2008 #1
    This isn't one of those textbook problems, but it's still homework, so I thought it'd be okay to post this here.

    1. The problem statement, all variables and given/known data
    I have a wheel with brake pads that generate heat through friction. Everything except the velocities are constant, and I'm trying to figure out equations to test if my idea is a viable concept.

    How I worked it is I took the difference in KE of the wheel with and without the brake pads active and used that as my heat transfered from the brakes. Assuming that is right, as is all my work, I came up with the final equation and my statements to use the equation. I need to know if these are right. I tried to make it as readable and easy to follow...

    http://img149.imageshack.us/img149/975/frictionbz2.jpg [Broken]

    2. Relevant equations

    3. The attempt at a solution
    If this is right, or even if it isn't, is there an easier way to test how practical my concept is? Like taking the average force of a person and compare it to the force needed to generate the heat? If thats possible, how would I go about doing so?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Feb 29, 2008 #2
    I think q=mcdt isnt the right way to go about it. Lets say the total loss in kinetic energy is transferred as heat to the brakes and the wheel (which is what you've assumed).

    Btw, the kinetic energy of the brake pad is zero, only the wheel has kinetic energy, and in this case, as the kinetic energy is rotational as well as translational, the net KE will be [tex]\frac{1}{2}I_{cm}w^2+\frac{1}{2}mv_{cm}^2[/tex].

    If, according to our assumption, the total loss in kinetic energy is equal to the heat transferred, then we must calculate the heat transfer based on heat current and the thermal resistances of the brake pad and wheel.

    Another assumption you're making is that there are no losses due to friction taking place anywhere in the system. If frictional losses need to be taken into account, then you will have to calculate the loss in translational and rotational velocities because of them, or exclude the work done by friction on the system.

    How complex do you want to make this problem?
    Last edited: Feb 29, 2008
  4. Feb 29, 2008 #3
    It doesn't have to be very complex. I am ignoring frictional losses except for the brakes, and any other inefficiencies.

    Do I need to change my equation and use [tex]\frac{1}{2}I_{cm}w^2+\frac{1}{2}mv_{cm}^2[/tex]? If so, How do I get the rotational inertial and the angular speed? Assuming it's an ideal environment, can I ignore the rotational KE?
  5. Feb 29, 2008 #4


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    Is the wheel actually rolling on a surface or is it held fixed? In other words is the center of the wheel moving or not? If it is moving you need both terms an dthere is a simple relation between omega and v. If the center is held fixed you just need the kinetic energy of rotation.
  6. Feb 29, 2008 #5
    The center of the wheel would be moving.
  7. Feb 29, 2008 #6


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    Ok. Then yes, you must use kinetic energy of translation 1/2 mv^2 plus kinetic energy of rotation. The moment of inertia depends on whether it is a full wheel or a hollow wheel (with most of the mass a the rim) I assume you know the equations for the moment of inertia. As for omega, it will simply be [itex] \omega = v/R [/itex] with R the radius of the wheel.
  8. Mar 1, 2008 #7
    For most wheels (unless the wheel is a disc), the moment of inertia will be [tex]mr^2[/tex].
  9. Mar 1, 2008 #8
    Heat current is given by the expression,


    where [tex]\frac{dQ}{dt}[/tex] is the rate of flow of heat,
    K is the thermal conductivity,
    A is the cross sectional area, and
    [tex]\frac{dT}{dx}[/tex] is the temperature gradient.

    Heat resistance is given by [tex]R=\frac{l}{KA}[/tex]

    The above expression is basically Ohm's Law for Heat Current.

    In this case, we can apply Kirchoff's Current Law (algebraic sum of all currents at a junction must be zero), and therefore, the current flowing through the brake pad should be equal and opposite to the current flowing through the wheel. First though, are you familiar with any of this?
  10. Mar 1, 2008 #9
    Unfortunately, I can only follow about half of that. My problem isn't exactly restricted to these equations. I just need to figure out if I can generate enough heat by pushing a wheel. If all this is too complicated to figure out, would it be easier to solve if I used a generator as opposed to friction?
  11. Mar 1, 2008 #10
    For what do you want to generate heat exactly? And how much heat do you need to generate? What do you mean by a generator? This is a very interesting problem, and one part of it is giving me a little problem. If someone who's fluent in this field could look into it, it would really help.

    I would like to get to the bottom of this, going with the assumptions we have already made. The one part that is giving me a bit of trouble is the flow of heat in the wheel.
  12. Mar 2, 2008 #11
    Well, by generator I meant something along the lines of those hand crank radios/flashlights, but on a little bit of a bigger scale. Based off another post, I know boiling water isn't very probable, but I'm just exploring other possibilities for hand generated energy. Originally I was considering friction/heat, but electrical energy might be more efficient, I'm not sure. I guess at this point I'm just trying to find an equation that can tell me how much energy can be generated if I plug in certain numbers (radius, mass, etc).

    My knowledge in physics pretty much covers high school classes (AP B/C) and a few refreshment courses in college but that was a few years ago and I'm a little rusty.
  13. Mar 2, 2008 #12
    You could use joules heating effect. Get a heating coil. The heat generated is given by [tex]h=i^2rt[/tex], i is current, r is resistance, and t is time.
  14. Mar 2, 2008 #13


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    What is your goal? What are you trying to do? you do not provide a good enough description of your problem to know whether your equations have any meaning. Please tell us what you are doing.. in detail.
  15. Mar 3, 2008 #14
    I think the problem is that we have to find the amount of heat generated during braking.
  16. Mar 4, 2008 #15
    My problem isn't very specific (I don't have any numbers for my constants, etc) and I said pretty much everything I do know. I think it's specific enough to know what I'm looking for though. I have a pretty good idea from everybody's replies, and enough work to present that even if I can't find it at the end, I should be okay. Worst case scenario, I can just use electric generator and compare it to the average watts people are putting out on pedal generators.

    Its always great to get more insight, but at this point, it's fine either way. Thanks a lot for the help, to those who replied!
  17. Mar 4, 2008 #16
    One thing you can do is this:


    dq/dt= du/dt+ dw/dt

    as du=0, du/dt=0 (internal energy is constant)

    w=U+T (potential + kinetic),

    dw/dt=dU/dt + dT/dt, again dU=0


    [tex]T=\frac{1}{2}mv^2 +\frac{1}{2}I_{cm}w^2[/tex]

    Differentiating the above expression with respect to time,

    [tex]dT/dt= 2mva[/tex] as v=rw, [tex]a=r\alpha[/tex] and [tex]I_{cm}=mr^2[/tex]

    From above,


    This is the net rate of flow of heat generated. From post #8, this rate of flow of heat is equal to the heat flowing in the brake pad and the wheel.

    If the thermal conductivity of the brake pad is k1, its initial temperature is the ambient temperature, [tex]\theta _0[/tex], and its cross sectional area is A1 and l1 is its width, then the heat current flowing through the brake pad is:

    [tex]\frac{dq_1}{dt}=k_1A_1(\frac{\theta (t)-\theta _0}{l_1})[/tex] ................(1)

    Similarly, if the thermal conductivity of the wheel is k2, area of cross section is A2, initial temperature is equal to the ambient temperature, and its width is l2, then the current flowing through the wheel's cross section at a particular instant of time may be given by,

    [tex]\frac{dq_2}{dt}=k_2A_2(\frac{\theta (t)-\theta _0}{l_2}) +\phi[/tex]...........(2)

    The factor [tex]\phi[/tex] is what I cant figure out. If the wheel moves at a velocity v, at at a particular time t, the area under contact is A, then at an instant t+dt, a small area dA will have left the contact region and another area dA will have entered it. If we do assume that there are no losses due to radiation or convection of heat, then that particular area dA that just left will be at a higher temperature when it comes back after one revolution of the wheel. The new area dA will be at a temperature lower than the area A-dA (through which heat current still flows). How do I figure out the rate of heat flow in this small area dA? Please help me here, or just tell me how to analyze the situation. How do I write out this second differential equation?

    The sum of equations (1) and (2) will be equal to the heat flow found by differentiating the kinetic energy expression, 2mva, where v may be written as [tex]v=v_0+at[/tex], getting an expression dependent only on acceleration.

    But how do I figure out [tex]\phi[/tex] ?
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