I'm studying rigid body dynamics lately. I came across the definition of torque, and though I've found a lot of explanations as to why there is an r there (the moment), all of them are mathematical (equating work and so on). None of them explained physically and I still couldn't figure out why the distance from axis of rotation increases the net effect, or torque.

So I thought about this and came to this line of thought -

Rotation can be thought of as a rigid body have all the infinitesimal masses performing circular motion about a fixed axis. There is pure rotation, hence angular velocity is constant. Thus velocity which is omega times r increase with the distance from the rotation axis. So if force is applied at more distance, this implies more velocity of the point of the application, and since the body is rigid, all the other mass connected to the point of application goes along through inter-atomic interactions and hence more rotational effect. Is this line of thought correct?

So what happens inside a body when it rotates? Do the rest of the atoms go along due to electromagnetic attraction and if so, can someone explain exactly what happens inside the body when it rotates and where does that r come from from an inter-atomic point of view?

  • $\begingroup$ Torque is just a way to express force at a distance. A purely rotating rigid body needs to torque so I am failing to see the connection between the definition of torque and the material acceleration of the atoms in the body. $\endgroup$ – ja72 Sep 16 '14 at 20:09
  • $\begingroup$ It has a connection. Farther is the point of application of force from axis of rotation, greater is the angular acceleration in a pure rotation. Since we are assuming the body is rigid, thus relative velocity between particles is zero, and the rest of the body moves along. The greater angular acceleration you give, the whole body will trace out a greater arc in space with the same amount of force, hence more rotational effect or torque. $\endgroup$ – geekbuddy Sep 16 '14 at 20:16
  • $\begingroup$ Torque is related to spatial acceleration and not material acceleration. Spatial acceleration is zero for a constantly rotating body. $\endgroup$ – ja72 Sep 16 '14 at 21:23
  • $\begingroup$ I'm sorry, I meant tangential velocity not the angular acceleration. The tangential velocity for circular motion of point of application increases with distance from rotation axis (v=omega*r), which implies greater angular displacement by the point of application, hence greater rotational effect of the rigid body. $\endgroup$ – geekbuddy Sep 17 '14 at 9:07

Why would there be no radius in torque? A real torque is a real force that acts on a real rotating body (rigid or not) at a real radius. If you look at e.g. rotating machine parts, they all have a finite diameter. That diameter is of enormous importance for the design of a part, because together with the material constants it determines just how much torque can be translated trough that part. A thin axle will break much easier than a thick one, for instance. A thin axle will deflect far more than a thick one for the same amount of torque. The radius matters, big time!

Now, if a force moves an object a certain distance, what do you get? Work. That definition holds, whether that movement is linear, or not. So if a force moves an object around a perimeter (equal to $2\pi r$), then you still get a work $W=2\pi r F$ performed, right? Well, in practice we then proceed to hide that $2\pi$ in the definition of the angle and the angular velocity.

Rigidity of bodies and point masses/mass distributions are just approximations to make life easier. None of that really exists. Don't waste too much time on getting used to them, because these are useless concepts outside of Newtonian mechanics and almost all of physics lives outside of Newtonian mechanics.


The physical explanation for why torque increases with r is that the longer the lever arm is the greater angular acceleration you can cause for a given force F. If a screw is stuck because it was screwed in too hard (ie with too much torque), you need to get a longer wrench. With the longer wrench (ie, larger r_w) you can generate greater force at the edge of the screw r_s by the ratio r_w/r_s and thus start to accelerate the screw out of the hole. This is also the concept of the fulcrum or a seasaw, (ie, if you sit farther back on a seasaw than the other person you can overcome their weight even if they are heavier than you.)

"Assuming the lever does not dissipate or store energy, the power into the lever must equal the power out of the lever. As the lever rotates around the fulcrum, points farther from this pivot move faster than points closer to the pivot. Therefore a force applied to a point farther from the pivot must be less than the force located at a point closer in, because power is the product of force and velocity" [Uicker, John; Pennock, Gordon; Shigley, Joseph (2010). Theory of Machines and Mechanisms, 4th ed.. Oxford University Press, USA].


It's part and parcel of the definitions of angular momentum and torque.

Consider a system of particles, not necessarily a rigid body. Without loss of generality, we can use an inertial frame that is instantaneously co-located and co-moving with the system center of mass. The angular momentum of the system with respect to the origin of this frame is defined as $$ \mathbf L = \sum_\alpha \mathbf r_\alpha \times m_\alpha \dot {\mathbf r}_\alpha $$

Differentiating with respect to time yields $$\frac{d\mathbf L}{dt} = \sum_\alpha \dot {\mathbf r}_\alpha \times m_\alpha \dot {\mathbf r}_\alpha + \mathbf r_\alpha \times m_\alpha \ddot {\mathbf r}_\alpha $$ The first term inside the sum, $\dot {\mathbf r}_\alpha \times m_\alpha \dot {\mathbf r}_\alpha$, is identically zero, leaving $$\frac{d\mathbf L}{dt} = \sum_\alpha \mathbf r_\alpha \times m_\alpha \ddot {\mathbf r}_\alpha = \sum_\alpha \mathbf r_\alpha \times {\mathbf F}_{net,\alpha} $$ where $ m_\alpha \ddot {\mathbf r}_\alpha \equiv {\mathbf F}_{net,\alpha}$ by Newton's second law. This net force acting on the $\alpha^{th}$ includes internal forces from other particles in the system and external forces that come from outside the system. Splitting this net force into internal and external forces, we have

$$\frac{d\mathbf L}{dt} = \sum_\alpha \mathbf r_\alpha \times \bigl({\mathbf F}_{ext,\alpha} + \sum_{\beta\ne\alpha} \mathbf F_{\alpha,\beta}\bigr) $$ where $\mathbf F_{\alpha,\beta}$ denotes the force exerted by the $\beta^{th}$ particle and the $\alpha^{th}$ particle. If all internal forces follow the strong form of Newton's third law (forces between pairs of particles are equal but opposite, and act along the line connecting the particles) then $\sum_\alpha \sum_{\beta\ne\alpha} r_\alpha \times \mathbf F_{\alpha,\beta}$ vanishes. In other words, the strong form of Newton's third law says internal torques do not contribute to change in angular momentum. The end result is $$\frac{d\mathbf L}{dt} = \sum_\alpha \mathbf r_\alpha \times \mathbf F_{ext,\alpha} $$

This is true for any system of particles that obeys Newton's laws of motion, not just a rigid body.

So why are angular momentum and torque defined this way? The answer is simple: These concepts keep appearing over and over again. Physicists tend to give those key concepts that keep appearing over and over again names and standard definitions.

  • $\begingroup$ Okay. I get this. Can you explain how the initial angular momentum equation appeared from scratch? $\endgroup$ – geekbuddy Sep 17 '14 at 9:11
  • $\begingroup$ That was an effort than spanned centuries. There's a hint of it in Kepler's equations; Kepler's second law is a statement of conservation of angular momentum. Newton did some work on the concept. It took physicists and mathematicians a couple centuries to refine those initial concepts. Treating quantities such as displacement, velocity, acceleration, force, etc. as vectors was a very late addition, dating to the late $19^{th}$ / early $20^{th}$ century. Packaging these concepts (and more) in a manner understandable by beginning students was no mean feat, either. $\endgroup$ – David Hammen Sep 17 '14 at 12:13
  • $\begingroup$ Physics educators are largely of the opinion that presenting the torturous path that eventually resulted in how introductory physics is presented today would hinder rather than aid in learning. (You can always take history of science classes if that stuff interests you.) You need to just take some things such as why momentum and angular momentum are defined the way they are for granted. $\endgroup$ – David Hammen Sep 17 '14 at 12:22
  • $\begingroup$ A couple of other examples: You've been taught that $F=ma$ is Newton's second law. Newton never wrote that, or anything like that, in his Principia. Maxwell's equations as written by Maxwell are 21 scalar equations. You will not be taught those. You'll instead be taught four vector equations that Maxwell did not write; vectors post-date Maxwell by a quarter of a century. Later, after you learn more mathematics, you'll be able to reduce those four equations to just one. $\endgroup$ – David Hammen Sep 17 '14 at 12:24
  • $\begingroup$ Okay David, I got you. Perhaps some things are obtained purely from experiments and can't be thought of theoretically maybe. I got this intuition which explains torque and works for me, let me know if I am right? The tangential velocity for circular motion of point of application increases with distance from rotation axis (v=omega*r), which implies greater angular displacement by the point of application, and since we take a rigid body everything else goes along at that speed, hence greater rotational effect of the rigid body. $\endgroup$ – geekbuddy Sep 17 '14 at 13:29

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