To begin our analysis of rotational mechanics one should first remark that the basic equations of translational kinematics (forming the “starting point” for a description of translational mechanics) cannot be used to describe the motion of a rotating body. This is because in our analysis of translational kinematics we assumed that all of the mass elements composing the object travel, more or less, the same distances in a given amount of time with the same velocities and accelerations. The very first idealizations that we made in our analysis of translational kinematics was that the object remains absolutely rigid (which is to say that it does not deform) and that all parts (mass elements) of the object undergo equal displacements in a given amount of time with the same velocities and accelerations; this is what allowed us to treat the whole object as a single particle. But with a rotating object, this is not the case. As a body rotates, different mass elements travel different linear distances in a given amount of time with different linear velocities and accelerations. For example if a solid ball were spinning on the ground, in a given amount of time (say one second), the mass elements farther away from the axis of rotation will travel a greater linear distance than mass elements closer to the axis of rotation and thus the farther away mass elements will also be moving with greater speeds.

In rotational kinematics we begin at a similar starting point. We begin by first only analyzing the motion of rotating objects that are absolutely rigid. We then recognize that if we treat the whole object as a collection of particles, even though the three basic linear quantities of translation kinematics (linear displacement, linear velocity, and linear acceleration) are different for different mass elements, there are three angular/rotational quantities which remain the same for all mass elements—namely, angular displacement, angular velocity, and angular acceleration.

Suppose that we are watching an object rotate about a fixed axis (we will investigate the more complicated case where the object “wobbles” later on). Suppose that \(O\) indicates the center of the object and that the axis of rotation (which we can think of as the z-axis) passes perpendicularly through the object and \(O\). Let \(P\) be an arbitrary point in the object (where some particle is located) that moves as the object rotates. Also let the x-axis be a fixed reference line in space. As the object rotates, the line segment \(\overline{OP}\) will describe an angle \(θ\) which will be taken to be measured *counter-clockwise* from the reference line. This is to say that the angle \(θ\) will be positive if the particle rotates counter-clockwise from the reference line and negative if it rotates clockwise from the reference line. Note that the particle at \(P\) and indeed all other particles making up the object will travel in circles as the object rotates. It is therefore convenient to indicate the position of a given particle in polar coordinates as \((r,θ)\). As the particle at \(P\) traverses the arc length \(s\) (describing the angle \(θ\)) in a given period of time \(Δt\), the arc length and the angle are related by the equation

$$s=rθ.\tag{1}$$

As the particle at \(P\) moves from location \(A\) to location \(B\) it sweeps out the angle \(Δθ=θ_f-θ_i\). This quantity \(Δθ\) is called the *angular displacement* of the rigid object and is defined as

$$Δθ\equiv{θ_f-θ_i}.\tag{2}$$

Since all other particles making up the object undergo the same angular displacement this quantity is indicative of the angular displacement of the object as a whole. This is analogous to when a rigid object undergoes linear displacement; since every part of the object undergoes the same linear displacement this quantity can be used to describe the linear displacement of the object as a whole. The rate at which this angular displacement occurs can vary. If this rigid body spins rapidly, this angular displacement will occur in a very short time interval. If this body rotates very slowly, then this angular displacement will occur over a longer time interval. To quantify the average rate-of-change of the angular displacement with respect to time, we define the quantity call average angular speed:

$$ω_{avg}\equiv{\frac{θ_f-θ_i}{t_f-t_i}}=\frac{Δθ}{Δt}.\tag{3}$$

The instantaneous angular speed is defined as the limit of this quantity as \(Δt\) approaches zero:

$$ω\equiv{\lim_{Δt\to0} \frac{Δθ}{Δt}}=\frac{dθ}{dt}.\tag{4}$$

The instantaneous angular velocity is positive when the object is spinning counter-clockwise because \(Δθ\) is increasing (since it is measured counter-clockwise from the reference line) and it is negative when the object is spinning clockwise because \(Δθ\) is decreasing.

If a particle moves along not just a circle but, more generally, any curve (with a radius of curvature, \(r\))from any point \(P\) to \(Q\), it will traverse an arclength \(Δs\) and sweep out an angular displacement, \(Δθ\), which satisfies equation (1). The ratio \(\frac{Δs}{Δt}\) gives the average tangential speed, \(v_t\), of the particle as it moves from \(P\) to \(Q\). Let's divide both sides of Equation 1 by \(Δt\) to obtain

$$v_{avg-t}=\frac{Δs}{Δt}=r\frac{Δθ}{Δt}=rω_{avg}.\tag{5}$$

By taking the limit on both sides as \(Δt→0\), we get

$$v_t=\frac{ds}{dt}=r\frac{dθ}{dt}=rω.\tag{6}$$

If the instantaneous angular speed is changing, then the rigid body is undergoing angular acceleration. We define the average angular acceleration as

$$α_{avg}\equiv{\frac{ω_f-ω_i}{t_f-t_i}}=\frac{Δω}{Δt}\tag{7}$$

and the instantaneous angular acceleration as

$$α\equiv{\lim_{Δt\to0} \frac{Δω}{Δt}}=\frac{dω}{dt}.\tag{8}$$

The angular acceleration \(α\) is positive when a rigid body rotating counter-clockwise is speeding up (\(ω_f>ω_i\) so \(ω_f-ω_i>0\) and \(α>0\)) and when a rigid body rotating clockwise is slowing down (\(ω_f>ω_i\) since it is less negative and thus \(ω_f-ω_i>0\) and \(α>0\)).

Since the rotation axis is fixed, there are only two possible directions associated with the vectors quantities \(\vec{ω}\) and \(\vec{α}\). In the case of fixed axis rotation we do not need to use vector notation to express the directionality of these two quantities; the directionality can simply be expressed by a plus or minus sign. We will, however, briefly cover the directionality of these vectors using vector notation since it will be very useful later on. We can specify the direction of these quantities using the right-hand rule. If the object is rotating counter-clockwise, the direction of \(\vec{ω}\) is out of the page; if the object is rotating clockwise, the direction is into the page. If the object is rotating counter-clockwise then \(\vec{ω}\) is positive; if it is rotating clockwise then \(\vec{ω}\) is negative. From the definition \(\vec{α}\equiv{\frac{d\vec{ω}}{dt}}\), if the body is speeding up when \(\vec{α}\) acts in the same direction and slowing down when it acts in the opposite direction.

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