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Torque required to power a car

  1. Aug 22, 2009 #1
    How to calculate (I) when a mass is turned by the shaft but it is not the mas of the shaft iself, instead it is a mass attatched to the shaft or mass to be turned by the shaft.

    The confusion I have is that whenever we calculate the Moment Of Inertia for a circular disc of mass M and radius r it is (0.5)(M)(r^2). But this is for a disc of equally distributed mass around the center.
    But consider a disc of mass m and raius r as earlier, which is attatched to a shaft, and we have to turn this shaft to rotate the disc.

    Now, If the disc has the same mass and the same radius but the mass is ditributed diferently. Say, that the disc is slender from the center to a radius r3 and from there there is a thicker protrusion in the disc where greater proportion of the mass is concentrated, ( from r3 to r4) and then again the disc becomes slender upto the final radius r ( as radius and mass of the disc remain the same)

    So how do we now calculate the moment of inertia. Wont it be different in both the cases as the CG changes due to change in radii of mass concentration.
    Last edited by a moderator: Aug 23, 2009
  2. jcsd
  3. Aug 22, 2009 #2


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    Staff: Mentor

    The wiki on moment of inertia talks about how to calculate it: http://en.wikipedia.org/wiki/Moment_of_inertia#Definition

    In short, you calculate it by integrating mass elements with the square of the radius. In practice, there are a number of ways of doing this which depend on the specifics of the problem. For a disc with a hole in it, for example, you can take the moment of inertia of a solid disc of the full radius and subtract the moment of inertia of a solid disc with the radius of the hole. If the disc is non-uniform, you may actually have to construct an equation and solve the integral (or due it numerically with a spreadsheet).

    You can also google for moment of inertia of typical shapes. Here's one hit that may help: http://www.physics.uoguelph.ca/tutorials/torque/Q.torque.inertia.html
    Last edited: Aug 23, 2009
  4. Aug 23, 2009 #3


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    Staff: Mentor

    In an effort to help the OP, I'm going to try something new here. I've "soft" deleted most of the thread and edited the OP to be more concise and readable. "Soft" deleting means the posts are not gone, it's just that only moderators can see them. I'll restore them if they become needed.

    Lets try to address the issues one at a time in a simple and straightforward way. I think I've distilled the OP to the first question(s) that needs to be answered.
  5. Aug 23, 2009 #4


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    Staff: Mentor

    The words "attached" and "mounted", to an engineer means the same as "fixed". In other words, the object is connected in a way that makes it rotate with the shaft, such as being bolted or welded. An example would be a flywheel, large gear or car wheel, all of which would tend to be bolted to a shaft. Is this what you are getting at?

    Finding the moment of inertia of the system simply involves finding the moment of inertia of that object, whatever object it is, and ignoring the shaft.....in fact, it would help us if you would just tell us what the object is that you are looking for a moment of inertia for.
    There's no mystery there: if the object is attached through its center/axis of rotation, then that equation is all you need.
    You haven't given us enough information to calculate the moment of inertia there and in any case, there is no simple equation for that. You'll need to tell us exactly what the shape is (describe it mathematically). Finding the moment of inertia of non-uniform shapes is done via superposition, which is what I described in my other post. You find the moments of inertia of regular objects and then add or subtract them as necessary.

    If you can't express the shape easily, you can always use a numerical method: make an Excel spreadsheet and cut the object by radius from the axis of rotation.
    Nothing you have said so far has suggested a change in CG. What you have said so far implies the objects are radially symmetrical. Are they? If they are, then the CG is at the axis of rotation. But the moment of inertia is different due to the mass distribution.
    Last edited: Aug 23, 2009
  6. Aug 24, 2009 #5
    Let me put it forward in a much clear manner.

    I have got actually three questions all involving the same concept.


    We have to calculate the Torque required to turn a shaft at the end of which is attatched a solid wheel of mass M and radius R, and the mass of the shaft is m and radius r.
    So we are actually supplying torque to the shaft inorder to turn it ( along with the wheel being attatched at the end).
    Hence in this case, it is not merely the shaft alone or the wheel alone which is under discussion. So calculating the MI for the wheel alone ignoring the shaft wont help.
    Thats my question : How would we go about to calculate the MI for the Torque required when the attatched wheel and the shaft both have a mass ( M and m resp) and each with radius R and r resp. The shaft has its own mass m and also has to bear the mass of the wheel M attatched to it, so (0.5)(M)(R^2) alone or (0.5)(m)(r^2) alone wont help.
    So How do we calculate the total value of Torque to turn such an arrangement?

    Number 2 :

    Here, before proceeding let me state the actual question that I had come across. It states " Calculate the Torque to turn shaft of radius r which turns a load of mass M"

    It is not mentioned whether the mass is rectangular or circular or any other random shape. Neither does it state where the mass is placed, it just says that the shaft has to turn the mass.

    This also brought to my mind the case of a vehicle where the driving shaft has to turn the load ie the weight of the car. In this case :

    a) the weight ( and so mass) of the car is distributed all along the vehicle, but to drive it, the driving shaft has to turn this load.
    So in this case the mass is not bolted/welded directly to the shaft . So How do we calculate the Torque to be supplied to the driving shaft. ( dont consider any gearing or anything , just suppose that the engine directly supplies torque/power, just an assumption for simplicity)

    So in both the above cases in question no 2 , what do we do?

    Do we simply assume the masses to be an integral part of the shafts without having a clear picture about the shape and position of these masses w.r.t. the shaft.( as the shape of the masses will influence the MI, THIS IS WHAT IS ACTUALLY TROUBLING ME)


    How else do we calculate the MI and Torque as these masses are not the mass of the shaft and neither are they of a particular shape, but are just carried / turned by the shaft which itself has a mass m and radius r.
  7. Aug 24, 2009 #6
    You're questions are much clearer now.

    For number 1, you just work it out normally. You have to integrate it though if you are trating it as a single unit.

    To work out MOI you are treating a shaft as a 2D circle anyway, so the middle you just add the mass of the shaft and disc together for that dm. After the radius of the shaft you just use the mass of the flywheel per dm.

    I dont know how to make nice sexy integration symbols.

    So I=INT[from centre to shaft radius] r^2 d(m+M) + INT[shaft r to outer r] r^2 dM

    Yellow bit is mass of shaft and mass of disc d(m+M), blue bit is mass of disc only.

    I think in this case you cal also work the MOI of the shaft and mass seperately, then just add them togther.

    Number 2:

    There isnt really enough data without making assumptions to calcualte an answer the question. So i'd simply assume that its a point mass at the end of the shaft. If no data is given on the geometry of the end mass you can just assume that all its mass occurs at radius r.

    Making the calculation simply I=mr^2.

    Car Question:

    To find the torque to drive a car you do not assume it is bolted on to a shaft. The shaft is actually pretty irelevent. This is obviously simplified, becasue real race car dynamics are actaully bloody complicated. But:

    What you have to do is wourk out the force needed to accelerate the car. From F=MA.
    This is the thrust force needed at the contact patch. You can then find the wheel torque needed to generate this thrust by T=Fd.

    This T is the torque needed at the shaft to accelerate the car. Assuming you have 1:1 drive, this the the engine torque needed.

    To keep a car at steady speed, you need to estimate the force on the car due to losses (drag rolling resistance etc). The F at contact patch must then be = to this force. Again using T=Fd you can find the torque necessary.

    Attached Files:

    Last edited: Aug 24, 2009
  8. Aug 25, 2009 #7
    For the car example, the approach of first finidng the Force required to set the car in translational motion seems convincing.

    But then as said, this is the force at path of contact.

    Should'nt it be actually the Force at the CG of the car, rather than the Force at path of contact.

    Secondly, another query, the Torque is calculated a T = F.d , hwere d is the radius of the driving wheel.

    Then does this imply that the Torque required would be directly and solely the function of the wheel radius. So for a certain mass of the car, the torque would vary as the size of the wheel changes.

    Hence what should we deduce, that the power/torque would keep on increasing just because we increase the wheel diameter despite the mas of the car remaining constant. And also that the shaft mass though increasing with the wheel radius is still insignificant as compared to the cars mass.

    So can anybody shed some more light on these two issues namely: the torque being only wheel diameter dependent for a particular mass of the car. and the Force being concentrated at the CG of the car, so torque should be calculated as
    T = F. r , where r = distaance between CG of car and shaft axis.

    Please clarify.

    And also, would there be any other altenative way ( except calculating the force first and then the torque corresponding to it) to find the Torque required to drive the car only given its mass and driving shaft diameter.
  9. Aug 25, 2009 #8
    It doesnt matter where the force acts, it will provide the linear acceleration. You will also get a moment made, but this is why your car pitches backwards under acceleration.

    Correct that D is radius of wheel + tyre. It's also correct that changing the wheel size alters the torque necessary. Larger wheels has the effect of altering the final drive (differential gear ratio).

    Larger tyres go faster for a given RPM, but dont accelerate the car as well.

    Torque at the WHEEL will go up for larger diameters, but the wheel acts as a gear so the ENGINE torque can be different so it doesnt necessarily need more power.

    This is why large wheels accelerat eless well, but have a higher top speed. Small wheels aid acceleration but hurt top speed,

    What?? No... you arent trying to rotate the car about the axis of the shaft.... wehich is what that torque is. You simly want the rotatinos equivilant of the thrust required to proper the car forward.

    Dont get offended at this, but it seems you need to go back and reread engieering mechanics because ther is a findamental misunderstanding of how forces act here. Reread a chapter on free body diagrams and resolving forces.

    Why on earth would shaft diameter have anything to do with the torque needed?

    The shaft doesnt make the torque, it only tansmits it. Therefore so long as it doesnt twist itsself apart under torsion we dont care about the dimensinos of the shafts.

    The reason why the shafts dont matter for linear acceleration of the car is that tey ar eextremely negligable compared to the mass of the car and rotational interia of the wheel.
    So you could have a shaft 1 foot radius (that is hollow) or a 1 inch shaft and it wouldnt amke a bit of difference. It still needs the same torque to actually drive the car.
    Last edited: Aug 25, 2009
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