**Derivation of net torque exerted on a particle under the action of any kind of force**

We shall prove that the equation \(\sum{τ}=Iα\) applies to both a particle moving in a *circular *path and an object of arbitrary shape rotating about a fixed axis. We will start with the simpler case of a particle moving in a circular path about a fixed axis. Let’s say that this particle has mass \(m\) and rotates in a circle of radius \(r\). In general, this particle can experience a net force acting only in the radial direction, only in the tangential direction, or a force which has both radial and tangential components. We can disregard all radial forces acting on the particle because \(\sum{τ_r}=\sum{F_r}rsin(\text{0° or 180°})=0\). Thus, \(\sum{τ_r}\) has no effect on the net torque \(\sum{τ}=\sum{τ_t}+\sum{τ_r}\) and we can rewrite the net torque on a particle as

$$\sum{τ}=\sum{τ_t}.$$

We'll consider circumstances in which a particle can experience a net force with both tangential and radial components; but because the radial components do not contribute to the net torque, we can get away with assuming that only the net force along the tangential direction, \(\sum{F_t}\), is acting on this particle. From Newton’s Second Law we know that \(\sum{F_t}=ma_t\). The net torque exerted on this particle is \(\sum{τ}=\sum{τ_t}=(\sum{F_t})r=(ma_t)r\). Using the relation \(a_t=αr\), one can express this equation as

$$\sum{τ}=(ma_t)r=(mαr)r=mr^2α.$$

Since the rotational inertia is given by \(I=mr^2\), this equation can be rewritten as

$$\sum{τ}=Iα\text{ (for particle under net torque)}.\tag{1}$$

This equation allows us to interpret the dynamics of the rotational motion of a particle of mass \(m\) revolving in a circle of radius \(r\) about an axis through \(O\).

**Derivation of net torque on any rigid object under the action of any net force**

I shall now derive this same equation for an extended, rigid object of arbitrary shape rotating about a fixed axis. This object of mass \(M\) as we know can be modeled as the infinite sum of infinitely many very, very small mass elements—indeed, infinitesimally small. Each mass element \(dm\) will revolve about the axis in a circle with the same angular speed and acceleration. If the angular speed or acceleration of any one of the mass elements differed from that of any of the others, the object would not be absolutely rigid: it would bend and deform a little. As an external net force \(\sum{F}\) is exerted on this object, this net force must distribute itself throughout each of the parts (i.e. mass elements) of the object so that the object remains rigid—if certain mass elements remained unaffected by this net force they would preserve their uniform rotational motion while the other mass elements undergo a change in their rotational motion causing the object to deform or bend.

Since a force acts on each mass elements there will also be a torque acting on each mass element. Thus there is a net torque \(\sum{τ}\) acting on the object as a whole. We can think of this net torque as the combination \(\sum{τ}=\sum{τ_t}+\sum{τ_r}\). Since for each radial torque \(τ_r=F_rrsin(\text{0° or 180°})=0\) acting on each mass element \(dm\), the angle \(ψ\) between the force and the distance \(r\) will be either \(0°\) or \(180°\), each (\τ_r\) acting on each mass element will be zero; thus the sum of radial torques (\sum{τ_r}\) acting on each particle is zero. Therefore, \(\sum{τ}=\sum{τ_t}+0=\sum{τ_t}\). This result means that we only have to consider the tangnetial components, (\sum{τ_t}\), to find an expression for the net torque, (\sum{τ}\). This also means that we only have to consider the tangential components of the forces distributed to each mass element—simply put, the radial components do not contribute to the net torque (\sum{τ}\). For a given mass element \(dm\), we can express the tangential force exerted on this mass element by the equation

$$dF_t=(dm)a_t.$$

Note that given that the mass of the particle is infinitesimally small, the magnitude of the force must also be infinitesimally small because it is the product of a finite variable (namely, the tangential acceleration) and a quantity or number that is infinitesimally small (namely, the mass of the particle). The torque exerted on this mass element about the axis of rotation is

$$dτ=(dm)a_tr,$$

where \(r\) is the lever arm or the perpendicular distance from \(O\) to the line of action of the force \(dF_t\). Since the angular acceleration is common for all mass elements, we can use the relation \(a_t=rα\) to obtain

$$dτ=(dm)(rα)r=αr^2dm.$$

We are motivated to make this substitution by the fact that we wish to obtain the rotational analog of Newton’s Second Law of motion. In order to find the net torque (\sum{τ}\) exerted on the object as a whole, we must take the infinite sum of the tangential torques acting on all the mass elements making up the object to obtain

$$\sum{τ_r}=\int{dτ}=\int{αr^2dm}.$$

Since the angular acceleration is the same for all mass elements (i.e. it does not vary from mass element to mass element) we can pull \(α\) outside the integral to obtain

$$\sum{τ}=α\int{r^2dm}.$$

Since the rotational inertia \(I\) is equal to \(\int{r^2dm}\), we have

$$\sum{τ}=Iα.\tag{2}$$

**What is the qualitative meaning and interpretation of torque?**

Source\(^{[1]}\)

Rotational acceleration is a measure of how much an object *deviates* from the uniform and unchanging rotational motion which, in the absence of torque, would continue to preserve forever. The role played by rotational acceleration in rotational motion is *analogous* to the role played by linear acceleration in linear motion; whereas rotational acceleration is a measure of how much an object deviates from uniform, otherwise unchanging rotational motion, linear acceleration is a measure of how much an object deviates from uniform, otherwise unchanging linear motion. The only difference between these two quantities is that they depend on different things. For an object with uniform linear motion, acceleration depends only on the mass of the object and the force applied to it according to the equation \(\vec{a}=\frac{\vec{F}}{m}\). The more massive an object is the greater is the force required to accelerate that object. The Earth is so massive that the only way to cause a noticeable deviation in its otherwise uniform and unchanging linear motion would require a truly vast force—indeed the Sun is what provides this immense “pull.” A pebble is so unmassive in comparison to the Earth that only a comparatively minute force would be needed in order to change its uniform, otherwise unchanging motion. Newton’s Second Law, \(\vec{F}=m\vec{a}\), very effectively explains the *dynamics* of linear motion. In this equation, \(\vec{a}\) measures how much an object deviates from its uniform, otherwise unchanging motion. The mass of the object, \(m\), measures the inertia of an object (which is to say the amount of “*resistance*” an object has towards changing its uniform motion). The force \(\vec{F}\) plays an opposite role as that of the mass in the equation \(\vec{a}=\frac{\vec{F}}{m}\); the force is a measure of the “tendency” or “inclination” or propensity for an object to change its otherwise uniform motion. You can see this from the equation \(\vec{a}=\frac{\vec{F}}{m}\). The more massive an object is, a greater force is needed to change its uniform motion: hence it is a measure of how much the object “resists” change in its uniform motion.

The equation \(\vec{τ}=I\vec{α}\) explains the *dynamics* of rotational motion entirely analogously to the way \(\vec{F}=m\vec{a}\) explains the *dynamics *of linear motion. Additionally, the roles played by \(\vec{τ}\), \(I\) and \(\vec{α}\) in explaining the dynamics of rotational motion are entirely analogous to the roles played by \(\vec{F}\), \(m\) and \(\vec{a}\) in explaining the dynamics of linear motion, respectively. \(\vec{τ}\) is a measure of the “tendency” or “inclination” of an object to change its otherwise uniform rotational motion; \(I\) is a measure of the amount of inertia or “resistance” an object has towards changing its uniform motion; \(\vec{α}\) is a measure of the actual deviation from the object’s otherwise uniform, rotational motion. As I said before, the only difference between the quantities \(\vec{τ}\), \(I\) and \(\vec{α}\) and the quantities \(\vec{F}\), \(m\) and \(\vec{a}\) is that they depend on different things.

How much a rotating object deviates from its otherwise uniform rotational motion (measured by \(\vec{α}\)) depends not only on how massive the object is and how immense the applied force is (though these things certainly play a role) but, additionally, it also depends on how the mass in the object is distributed and where and in what direction the force is applied on the object. This can be seen in the equations \(I=\sum_{i=1}^{n} m_ir^2_i\) and \(\vec{τ}=\vec{r}×\vec{F}=(Fsinθ)r\vec{k}\) for a system of \(n\) particles. The farther the mass elements (particles) making up the object (or system) are from the axis of rotation, the more inertia \(I\) and resistance the object has towards change in its uniform rotational motion. For example, a hollow ball would have greater rotational inertia than a solid ball (provided they both are of equal mass and volume) because most of the mass elements (or particles) \(m_i\) which make up the hollow ball are farther from the axis of rotation (\(r_i\) is the *perpendicular* distance between a mass element \(m_i\) and the axis of rotation); thus the value of each \(r_i\) is for the most part bigger and therefore \(I\) is bigger.

In the second equation, \(θ\) is the angle between \(\vec{r}\) (the perpendicular distance from the axis of rotation to the point at which the force is being applied) and \(\vec{F}\). The farther away the force is applied (in perpendicular distance \(r\)) from the axis of rotation, the greater is \(\vec{τ}\) and thus the propensity for the body to rotate; also the greater the value of \(Fsinθ=F_{\perp}\) (the *component* of the applied force which is perpendicular to \(r\)), the greater \(\vec{τ}\) will be. Thus for a given and constant force, to maximize \(\vec{τ}\) and the propensity for the body to rotate would require one to maximize \(sinθ\). The maximum value of \(sinθ\) occurs at \(90°\) and \(180°\); this means that the amount of torque applied is maximized when the force is applied at right angles to the lever arm \(\vec{r}\). The closer and closer the applied force \(\vec{F}\) is to being perpendicular with \(\vec{r}\), the greater the value of \(\vec{τ}\) will be as \(sinθ\) gets larger and larger. When the force is applied in a direction parallel to the lever arm, \(sin0=0\). Thus \(τ=(Fsin0)r=0\). Since \(τ=0=Iα\), \(α=0\) and there is no change or deviation in the objects uniform rotational motion. Thus if a force or a net force are applied in a direction parallel to the lever arm \(\vec{r}\) the object suffers no change or deviation in its uniform rotational motion.

A good physical example of these concepts would be to imagine placing the head of a wrench around a screw and suppose that you use your hand to apply a torque to the wrench (not the screw). In this example the axis of rotation will be going through the screw. The farther your hand is from the axis of rotation when applying torque to the wrench and rotating it about the axis, the easier it is to rotate the wrench because as \(r\) increases, \(τ\) increases: this means that the propensity for the body to rotate increases. As your hand applies a force on the wrench over some arclength as the body rotates, you would notice that the force applied by your hand is to being at right angles to the lever arm the more easily the body rotates; whereas if you tried rotating the wrench through some angular displacement with the force applied by your hand very nearly parallel to the lever arm, you would find great difficulty in rotating the wrench.

This article is licensed under a CC BY-NC-SA 4.0 license.

**References**

1. Bozeman Science. "Torque". Online video clip. YouTube. YouTube, 27 November 2015. Web. 22 September 2014.