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Several questions regarding Torque.

  1. Aug 18, 2015 #1
    Question 1:
    What exactly is torque? I know this seems like a very vague question the answer of which can be found easily online, but I don't think any website online defines torque properly. The definition of force is a proper definition because it allows us to make predictions, if we know F and M we can find the acceleration. As for torque, all the definitions given are explaining the value of the torque, T = r cross F.
    I thought maybe torque was the rotational equivalent of linear Force, in the sense that it is directly proportional to angular acceleration, this turned out to be wrong. If the perpendicular force and the mass are constant, Torque is directly proportional to the r, while the rotational acceleration is inversely proportional to r. So obviously T is not directly proportional to rotational acceleration. What is it then? why is it that things with a higher torque rotate faster? or do they?
    I understand that torque determines whether or not something rotates, but how does torque govern the rotation?

    Question 2:
    Are there any intuitive explanations as to why a smaller force is needed to produce the same torque further away from the axis of rotation? I know the vector equations and the proofs, yet they all seem to come back to the magical equation of T = r x F. My question is, why is this equation true? why is it easier to push a door at the far end? after all, all the particles of the door move the same distance when you open it irregardless of where the point of action of the force is.
     
  2. jcsd
  3. Aug 18, 2015 #2
    For a constant rotational inertia it is

    $$τ=Iα$$

    Which is analogous to Newton's second law for rotational motion

    Would you consider the principle of the lever to be an intuitive explanation?
     
  4. Aug 18, 2015 #3
    If you break down I into mr^2, it seems like the further away the force is applied form the fulcrum, the slower the rotation if the force is constant. Now this is not true because the torque should be greater. Unless, r in the I = mr^2 is not the r the distance vector but only the absolute length of the lever.
    The principle of the lever goes back to the vector equation, but why is the vector equation true?
     
  5. Aug 18, 2015 #4
    Even still, because t = mr^2 * angular acceleration. t = F * r, so F= mr * angular acceleration.
    so increasing the length of the lever decreases the angular acceleration. It this true or am I wrong somewhere?
     
  6. Aug 18, 2015 #5
    Consider a rigid rod pivoted about its center. The moment of inertia is ##1/12 ML^2## where L is the length of the rod. Now let a force perpendicular to the rod at a distance R act on the rod
    $$RF=1/12ML^2α$$
    Because the geometry is fixed if you move the point of application of the force this will result in a greater torque and hence a greater angular acceleration.

    The r in the moment of inertia formula is the distance of each point mass from the axis of rotation.

    You may be interested in this thread:
    https://www.physicsforums.com/threads/can-you-explain-the-lever-using-newtons-laws.823974/
     
  7. Aug 18, 2015 #6

    SteamKing

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    Your analysis is incorrect. Torque in rotational motion is analogous to force in rectilinear motion. If you compare the SUVAT equations for the latter type of motion to their rotational equivalents, you'll see that the structure of the two sets of equations is identical.

    You're also getting sloppy with terminology. Angular acceleration is not the same as angular velocity, just like linear acceleration is different from linear velocity.

    It's a definition of torque in vector form. There is an equivalent formula for the magnitude of torque, which is M = F⋅d, where F is the applied force and d is the perpendicular distance between the line of action of the force and the axis of rotation.

    For a given force, F, the magnitude of the torque M increases as the moment arm d gets longer.

    Do they? Most doors are hung from hinges, so that the motion of the door is constrained. The motion of that part of the door further away from the axis of the hinges is quite a bit greater than the motion of that part of the door next to the hinges.
     
  8. Aug 18, 2015 #7
    How is my analysis incorrect?
    T = r x F.

    T =
    I.beta
    I = m r^2
    T = m r^2.beta
    r * F = m r^2 . beta
    F = m r . beta
    beta = F / m r
    you can see here that the angular acceleration is inverely proportional to the length of the lever. So if I increase the length of the lever, or increase the distance from the point of application of torque to the axis of rotation, the angular acceleration is decreased. On the other, torque is obviously increased as T = r x F.
     
  9. Aug 18, 2015 #8

    SteamKing

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    As was mentioned in an earlier post, r in the formula for mass moment of inertia, I, is not the length of the "lever"; it is the distance of a differential portion of the mass of the object from the axis of rotation.

    The moment of inertia I is different for bodies with different shapes and distributions of mass:

    https://en.wikipedia.org/wiki/List_of_moments_of_inertia

    Deriving the mass moments of inertia for various bodies is an exercise in integral calculus. In general, about a given axis,

    I = ∫∫∫ r2 dm, where r is the cartesian distance of the element of mass dm from a particular axis of rotation. The distance r may not be related at all to the torque applied to the body. The integration takes place over the mass of the body for which the moment of inertia is calculated.

    The same body can have different moments of inertia depending about which axis it is rotated. For example, take a long, slender bar. If you rotate this bar about an axis which runs lengthwise, it is very easy to rotate, i.e. it doesn't take much torque and you can do it with your fingers. On the other hand, if you grasp one end and want to swing the rod like a bat or a club, you can't do it with your fingers very easily, if at all. You must grasp the end of the rod in both hands and swing it with your body.
     
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