# Why is the concept of torque necessary?

In my physics textbook (“Physics for Scientists and Engineers” by Serway and Jewett) it says that “the cause of changes in the rotational motion of an object about some axis is measured by a quantity called torque.”

Where $$\tau = Fr \sin\phi$$.

When $$\phi$$ equals $$90^\circ$$,

$$\tau = F r = (m r \alpha) r = m r^2 α$$

Why not use $$F = m r \alpha$$ to explain the change in an object’s rotational motion instead of torque?

• Because applying the same force at different radii has decidedly different outcomes. Have you ever used a cheater bar on a stuck nut/bolt? Nov 24 at 19:00
• @JonCuster, I've used a cheater before. And note - for lug nuts, if a cheater doesn't work, check for left-handed threads (rare, but they DO exist). Nov 24 at 19:23
• @DavidWhite They exist but only for specific purposes, I believe...
– Gert
Nov 24 at 21:02
• @Gert, yes, and in the distant past, I owned a car with left handed lug threads on the left side of the car. No doubt this was to prevent loose lug nuts from "rolling themselves" all the way off the bolt. Nov 24 at 21:06
• @JonCuster Wouldn't that be explained by F = m r α? Assuming constant mass and force, the angular acceleration varies as the radius varies. Nov 24 at 22:26

The concept of torque is not necessary as such.

But it is useful.

When things are useful, someone might eventually start to use them. Then, sadly if you will, everyone else will have to understand them as well in order to follow what is being done.

In the case of torque, it is useful because it is a "rotational version" of force. You can't simply tell me that a force causes a rotation - I need more information than that to understand the resulting rotation since different forces can cause the same rotation and vice versa the same force can cause different rotations depending on where (distance from the rotational centre) and how (angle of the force) the force is applied. But if you told me the torque, then I understand exactly how the resulting rotation turns out. Only one torque can cause a specific rotation (a specific rotational, angular, acceleration). Torque is a measure that takes into account both force magnitude, force angle and distance so that all the confusingly varying parameters are accounted for.

So, unfortunately if you will, torque as a concept and property is here to stay. And quite useful to get the hang of if you ask me.

• Why the sentiment that it's unfortunate?
– AnoE
Nov 25 at 8:00
• @AnoE So it speaks to the (assumed) frustration of the OP of having to learn yet another physical property... Nov 25 at 8:14
• resulting rotation turns out Pun fully intended, I presume? Nov 26 at 13:17
• Sometimes the frustration of the learning curve is an extremely worthwhile investment of mental effort. 2 days ago
• If learning makes you $\textit{super}$ frustrated then you're probably doing it wrong. yesterday

In principle you could always use force and not use torque. But, the use of torque greatly simplifies the analysis of the motion of a system of particles.

For example, for rotation of a rigid body about a fixed axis the rotational motion is simply described as $$\tau = I \alpha$$ where $$\tau$$ is the net torque, $$I$$ is the moment of inertia, and $$\alpha$$ is the angular acceleration.

The complicated general rotation of a rigid body is best described using torque; such as using the Euler equations. For example, see the text Classical Mechanics by Goldstein for examples of the general rotation of a rigid body.

• Thank you for your post. τ =I α, so F r = m r^2 α. What difference does the r factor make? If you have any idea as to what fundamental concepts I am missing, please let me know. Nov 24 at 22:28
• @BrunoNowak - You know how in many (translation-related) problems you don't have to worry about the exact point where the force is applied, but can instead treat the whole object as pointlike? It's kind of a "summary description" of how the whole thing moves. Well, this is something like that, but for rotation (you don't have to worry about how the constituent parts at different radii behave, it's all "summed up" in $\tau$ & $I$, in a way). Nov 25 at 1:27
• For a point mass the force analysis suffices; I = m r^2 so τ =I α is the same as F r = m r^2 α. For a rigid body it is more complicated: I is over the entire body and incorporates the r of each differential mass element in the body. See any basic physics test such as one by Halliday and Resnick. You need the concept of torque for evaluating the motion of a system of particles; just try to evaluate a sphere rolling and slipping down an incline and you will see. Nov 25 at 1:37
• I think the rigid body bit is really crucial to torque. When you want to replace all of the complex molecular interactions on an object with a simple rigid body, you need to account for all of the ways a rigid body can move, including rotation Nov 25 at 18:43

One big reason for using torque is that it creates a simple way of understanding when an object is in equilibrium.

When the sum of the FORCES on an object is zero, then we know that the object does not have translational acceleration.

Similarly, when the sum of the TORQUES on an object is zero, then we know that the object has no rotational acceleration.

Example: Imagine that you are trying to push a door closed, but your friend is trying to push the same door open. You are pushing normal to the door ($$\phi = 90^\circ$$), but they are pushing at an angle ($$\phi = 70^\circ$$). You are both applying this force at radius $$r$$ from the hinge, but your friend is applying their force at $$2r$$. What has to be true for the door to be stationary? The easiest way to express this is that the torques should add to zero. $$\tau_{you} + \tau_{friend} = 0\\ F_{you} \times r \times \sin(90)-F_{friend} \times 2r \times \sin(70)=0$$ Say you want to figure out how much force your friend has to apply to keep the door stationary if you apply 100 N. Just plug in $$F_{you} = 100$$N and solve for $$F_{friend}$$. That's a lot easier than trying to draw out a free body diagram!

If I understand correctly, the crux of your question is:

Why not use $$F = mr\alpha$$ to explain the change in an object’s rotational motion instead of torque?

## Reason #1: A more general case of concentrated mass

There are a few reasons, but the first good reason is that the $$r$$ in $$\tau = Fr$$ and the $$r$$ in $$mr^2\alpha$$ are not always the same $$r$$! For example, consider this pendulum, shown in the figure below, whose mass is mostly concentrated at the point $$r_1$$ from the pin (the point about which the pendulum rotates). A real-world example would be a carbon fiber pendulum rod with a lead ball embedded in the middle. Lead has a significantly higher density than carbon fiber, so we may sometimes be able to discount the inertia of the rod and only consider the inertia of the lead ball. The force could be applied at a different point on the pendulum ($$r_2$$ in the drawing). In the above case, the moment of the force (or torque, if you prefer) is $$F r_2 \sin(\theta)$$. The rotational inertia is $$mr_1^2\alpha$$. So we cannot simplify the expression to $$F\sin(\theta) = mr\alpha$$. Instead we have $$Fr_2\sin(\theta) = mr_1^2\alpha$$ This expression is often written in terms of torque $$\tau = Fr_2\sin(\theta)$$ and the mass moment of inertia $$I = mr_1^2$$: $$\tau = I\alpha$$

## Reason #2: Distributed mass

Unlike the pendulum example above, most real bodies have distributed mass across their entire volume. In that case, their mass moment of inertia must be calculated with volumetric integrals. Without getting into the math-y details, we can split a body into many tiny pieces, each with their own small mass $$dm$$ and is located some distance $$r$$ from the center of rotation. That value of $$r$$ will be different for each piece! We need to account for the moment of inertia of every piece: $$r^2dm$$. Performing the summation of the moment of inertia of all those tiny pieces yields the mass moment of inertia $$I$$ of the object about the center of rotation. If in our pendulum example above the entire thin rod was made of lead instead of having a concentrated mass at the center, the mass moment of inertia would be $$I = \frac{1}{3}mL^2$$, where $$L=r_2$$ is the length of the rod.

If we insert that expression into the equation $$\tau = I\alpha$$ we get $$\tau = \frac{1}{3}mr_2^2\alpha$$ or $$Fr_2\sin(\theta) = \frac{1}{3}mr_2^2\alpha$$ In this specific case where the force is applied at $$r_2$$ (the end of the pendulum), we could simplify to $$F\sin(\theta) = \frac{1}{3}mr_2\alpha$$ but that does not capture the general idea of torques and moments of inertia. It only describes the dynamics of this particular system at this particular instant.

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Why not use F = m r α to explain the change in an object’s rotational motion instead of torque?

There are three major conserved mechanical quantities: energy, momentum, and angular momentum. Each of them are separate quantities and they are separately conserved. Because they are conserved, we are interested in the rate of transfer of each of them.

Force is the rate of change of momentum.

Power is the rate of change of energy.

Torque is the rate of change of angular momentum.

These are distinct quantities. If you wish to calculate the change in angular momentum then you must use torque, neither force nor power will provide that information. It is also important to know that it is possible to have a torque without a net force, a force without a power, and so forth.

Because from $$F = m r \alpha$$, it can be seen that for knowing what magnitude of force was applied to a rotating object with given angular acceleration - you have to know distance $$\vec r$$ vector, i.e. position where this force was applied. And in the second (linear) Newton law, force does not depend on distance. So it must be that angular acceleration has nothing to do with force alone.

The second point is that, as you noted yourself, when multiplying both sides of equation by $$r$$, we get:

$$rF = m r^2 \alpha = \tau$$

But ... $$mr^2 = I$$ is nothing more but a moment of inertia of rotating point about a given axis. So you assume that all given angular rotating systems will be point-like, which is not always the case. In general, a system can have an arbitrary moment of inertia. You can check some at common list of moments of inertia. So in the end, your equation becomes:

$$\tau = I \alpha$$

The so-called second Newton law for angular rotation systems.

• I think I agree with Agnius, but I'm sure that Post lacks too much in English 2 days ago
• We mostly speak in Physics language here, which doesn't care about vocal or writing skills, nor does nature cares about it. 2 days ago

Consider a stiff but massless bar with a frictionless axle at one end. A mass (m) is mounted on the bar at a radius (r) from the axle and a force (F) pushes on the bar at a radius (R). The work done by the tangential component of the force pushing the bar through an arc length (S) will be transferred by the bar, accelerating the mass through an arc length (s): $$F_t$$S = mas or $$F_t$$Rθ = m(rα)(rθ) where θ is the angle of rotation of the bar. dividing both sides by θ gives τ = (m$$r^2$$)α. This logic can be applied to any number of forces and masses. Note that the radius to the force can be different from the radius to the mass. The sum of torques tries to cause an angular acceleration, and the sum of terms (m$$r^2$$) (the rotational inertia) opposes that acceleration. This approch gives you the best definition of a torque: A torque is the work per unit angle of rotation (as in Joules per radian) that can be done by a force which is acting in a manner which would tend to cause a rotation.

The concept of torque is not merely "necessary". It simply is. Torque is axiomatic, in the same way as "left" and "up" or "beneath" and "more than."

Argue all night about the specific words used in any given language and still, how could "necessary" come into this, any more than it does for other basic concepts such as breadth or height, length or weight or volume?