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What is Torque and what is correct formula?

  1. Sep 19, 2013 #1
    I'm struggling with the concept of torque as I read more I get further confused.

    I read some sources and they say torque and moments are interchangeable, yet other sources make a specific effort to keep the two separate. So question one is , what is the difference between torque and a moment ?

    The formula for torque...so far I have come across τ= r ×F, but also τ = I ×angular acceleration. I'm unsure which I should be using, and if both, which circumstances determine you to use which.

    My overall confusion though is what is torque?

    So moment of inertia is the resistance to the force being applied. So is torque the amount of force I need to apply to rotate an object ? or is it the amount of rotational force a force can put on a body ? what is a Nm (Newton meter) how would one see a Newton meter, I presume its not a rate or ratio.

    thanks for any help.
  2. jcsd
  3. Sep 19, 2013 #2


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    "τ= r ×F"
    This is, correctly applied, the DEFINITION of torque.
    By correctly applying the definition of torque on Newton's second law of motion, F=ma, we gain the RESULT
    "τ=I ×angular acceleration" for the rigid body.

    "ma" in F=ma is often called "the rate of change of (linear) momentum", whereas "I ×angular acceleration" is often called "the rate of change of angular momentum".
  4. Sep 19, 2013 #3

    D H

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    To some, torque and moment (of force) are synonyms. To some engineers, they're not. They distinguish between moment of force and torque. So, yes, there is a potential for confusion. This is physicsforums.com, not engineeringforums.com, so at this site it's best to stick with the nomenclature used in physics.

    By way of analogy, consider a mass attached to a spring. On one hand, Hooke's law describes the force exerted by the spring on the mass is F=-kx, where x is the displacement from the spring's relaxed position. On the other hand, we have F=ma from Newton's laws of motion. Presumably you aren't confused by F=-kx versus F=ma, so why should you be confused by τ=r×F versus τ=Iα?

    In the spring mass system, F=ma is a kinematic description of what transpires. F=-kx is a dynamic description of what's happening. Put the two together and you get a second order differential equation, ##m\frac{d^2x(t)}{dt^2} = -k x(t)##. The same happens with rotational motion. τ=r×F is a dynamic description of torque. τ=Iα is a kinematic description. Put the two together and you get a second order differential equation that describes what's happening rotationally.
    Last edited: Sep 19, 2013
  5. Sep 19, 2013 #4


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    I view the two formulas as cause and effect: If you apply a force [itex]\vec{F}[/itex] at position [itex]\vec{r}[/itex] on a rigid object, it will cause the object to increase its rate of angular velocity [itex]\omega[/itex] according to the formula:

    [itex]\vec{r} \times \vec{F} = I \cdot \dot{\vec{\omega}}[/itex]

    (In general, [itex]I[/itex] is a tensor, so the operator [itex]\cdot[/itex] is a tensor operation.)
  6. Sep 19, 2013 #5
    The first two posts are excellent but may be a bit advanced if your are just starting this
    subject. So study them carefully.

    [Third post made while I was typing this...]

    Some reasonable explanation and details are here:


    You can also check any good introductory physics textbook like Halliday and Resnick.

    that's the idea, but I'd say ...is the ROTATIONAL resistance to the force.....Trying to rotate a body usually takes a different force than trying to move the body linearly.
    The resistance to a linear force is referred to as 'inertia'.

    yes, but at some specified distance 'r' as described in the Wikipedia link.....
    Via Tau [torque] = r x F torque varies both from the force F and the distance of the applied force from the point/axis of rotation.

    refer to the prior equation.
    I assume you know a 'Newton' is a unit of force in the MKS system. A newton meter is such a force applied at a specified distance from the axis of rotation. You can't 'see' such but you may be able to observe the result...rotational motion. Analogously, a Newton [meaning applied linearly] might move an object, say against friction or gravity.
    Last edited: Sep 19, 2013
  7. Sep 19, 2013 #6

    D H

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    Even more generally, the left hand side should be a sum over all external forces, and the right hand side should be the derivative of angular momentum:
    [tex]\sum \vec r \times \vec F = \frac d{dt} (I\cdot \vec \omega)[/tex]
    In the special case that the moment of inertia can be treated as a constant scalar, this becomes
    [tex]\sum \vec r \times \vec F = I\dot{\vec{\omega}}[/tex]
    This is the appropriate equation for introductory level physics.

    The original poster is asking some very introductory level questions, so it's best to ignore the tensorial nature of the moment of inertia. Besides, your equation is not correct if you do look at moment of inertia as a tensor.
  8. Sep 19, 2013 #7

    D H

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    Thread temporarily locked for cleanup.

    Posts which are a bit above the level of a high school / freshman level understanding have been split to a different thread, [thread]711348[/thread].

    Please keep remaining posts on topic and in line with what a young person taking a first year class in physics would understand.

    Thread re-opened.
    Last edited: Sep 19, 2013
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