Topic 2B-01: Torque
In the problems encountered so far, objects are frequently drawn as an extended object, like a block. However, we have treated such objects as point-like particles. In this section we are going to consider extended objects specifically. We assume that the extended objects are not deformable; those non-deformable objects are called the "rigid bodies". Two point-like particles connected by a mass less but non-deformable bar is the simplest kind of rigid body. The number of such particles and mass less bars can be increased to infinity and then into continuum, and thus a picture of a rigid body emerges. To consider the balance of rigid bodies that is the title of this section, two concepts, torque and the center of mass, are essential. In this topic we will discuss torque and defer the center of mass until next topic.
A reference point O and a position vector r that originates from O and
ends at point P are drawn in Fig. 1. A force F originates from point P.
Torque τ is defined as the vector product between r and
τ = r x F . (1)
τ, mathematically defined in Eq. (1), is a vector that originates from point O. It is perpendicular to the plane that contains both r and F. The direction, or the orientation of τ is defined by the right-hand rule. That is to consider vectors r and F are painted on the head of a right-hand screw as if both origninate from a same point. As the head of the screw is rotated from r to F, the direction of the advancement of the screw is the direction of τ. The angle spanned by r and F is denoted as θ as shown in Fig. 1. The magnitude of τ is
|τ| = |r|·|F|·sinθ. (2)
The next question is what this torque means in physics. In Fig. 1 suppose a solid rod is installed between point O and point P. Let us lightly nail down the rod at point O into the plane containing F and r so that it takes some efforts to rotate the rod around the nailed point O. If the magnitudes of F and r are fixed but the angle θ is allowed to be changed, we will find that the power to roatate the bar is proportional to sinθ. When sinθ = 0, that is, when F and r line up, then the bar cannot be rotated at all, just as our intuition tells us. If the angle θ is fixed, then the power to ratoate the bar is proportional to the product between the magnitude of F and the magnitude of r. In other words the magnitude of the torque τ as given in Eq. (2) indicates the rotational power created aroung point O if points O and P are connected by a solid rod, and the dirction of the torque vector ndicates the axis of this roation created by the force applied at point P.
Equation (1) is a mathematical statement; the vectors involved can originate from any point in a 3-dimensional space. We have imposed stricter conditions that F must originate from point P, and both r and τ must originate from point O as shown in Fig. 1.
There are two addition rules for a torque vector. Assume
F = F1 + F2.
as in Fig.2. Eq. (1) becomes
τ = r x F1 + r x F2.
τ1 = r x F1,
τ2 = r x F2,
τ = τ1 + τ2. (4)
There is no problem to adopt this addition rule in physics, since F1 and F2 both originate from point P, and torque vectors τ1 and τ2 both originates from point O.
The second rule of addition requires more careful analysis. Let us assume
r = ra + rb.
According to the pure mathematical approach as outlined in Topics 2-04, the above equation can be substituted into Eq. (1) and get
τ = ra x F + rb x F. (5)
It is tempting to call ra x F as τa and rb x F as τb. However, there are some difficulties. The situation is depicted in Fig. 3. Force vector F originates from point P and the position vector ra ends at point R. The cross product ra x F is not defined as a torque. We need to reinterpret Eq. (5) as that there are force vectors Fa and Fb originating from points R and Q respectively. Furthermore |F| = |Fa| = |Fb| , and F // Fa // Fb. It is ra x Fa and rb x Fb that can be defined as torque vectors τa and τb respectively. Eq. (5) should be interpreted as saying that
τ = τa + τb . (6)
It should be warned that most physics textbooks and physics teachers do not make the distinction between the mathematical vectors and vectors in physics the originating points of that must be kept distinct. They use Eq. (5) in physics without any explanation like given here, causing confusions in the minds of logically thorough students. However, it is often tedious to write out all physics considerations like distinguishing between F and Fa. We may sometimes misuse the details of physics and indulge into the same ignorance like other textbooks and teachers to save the time of writing, and use Eq. (5) without any further warning. Readers, when encountering such occasions, should always recall the arguments here and supplement different force vectors originating from different points but with the same magnitude and the direction as the original force vector in their own mind just like we did in Fig. 3.