Chapter 4. Rotation and Conservation of Angular Momentum Notes: • Most of the material in this chapter is taken from Young and Freedman, Chaps. 9 and 10. 4.1 Angular Velocity and Acceleration We have already briefly discussed rotational motion in Chapter 1 when we sought to derive an expression for the centripetal acceleration in cases involving circular motion (see Section 1.4 and equation (1.46)). We will revisit these notions here but with a somewhat broader scope. We reintroduce some basic relations between an angle of rotation θ about some fixed axis, the radius, and the arc traced by the radius over the angle θ . Figure 1 shows these relationships. First, the natural angular unit is the radian, not the degree as one might have expected. The definition of the radian is such that it is the angle for which the radius r and the arc s have the same length (see Figure 1a). The circumference of a circle equals 2π times the radius; it therefore follows that 360 1 rad = = 57.3. (4.1) 2π Second, as we previously saw in Chap. 1, the angle is expressed with s θ = (4.2) r and can be seen from Figure 1b. We can define an average angular velocity as the ratio of an angular change Δθ over a Figure 1 – The relations between an angle of rotation about some fixed axis, the radius , and the arc traced by the radius over the angle . - 62- time interval Δt Δθ ω = z . (4.3) ave,z Δt For example, an object that accomplishes one complete rotation in one second has an average angular velocity (also sometimes called average angular frequency) of 2π rad/s. If we make these intervals infinitesimal, then we can define the instantaneous angular velocity (or frequency) with Δθz ω z = lim Δt→0 Δt (4.4) dθ = z . dt That is, the instantaneous angular velocity is the time derivative of the angular displacement. The reason for the presence of the subscript “ z ” in equations (4.3) and (4.4) will soon be made clearer. It should be noted that an angular displacement Δθ can either be positive or negative; it is a matter of convention how the sign is defined. We will define an angular displacement as positive when it is effected in a counter-clockwise direction, as seen from an observer, when the fixed about which the rotation is done is pointing in the direction of the observer. This is perhaps more easily visualized with Figure 2. 4.1.1 Vector Notation Since a rotation is defined in relation to some fixed axis, it should perhaps not be too surprising that we can use a vector notation for angular displacements. That is, just as we can define a vector Δr composed of linear displacements along the three independent Figure 2 – Convention for the sign of an angle. - 63- axes in Cartesian coordinates with Δr = Δx ex + Δyey + Δzez , (4.5) we can do the same for an angular displacement vector Δθ with Δθ = Δθ xex + Δθ yey + Δθzez . (4.6) It is understood that in equation (4.6) Δθ x is an angular displacement about the fixed x-axis , etc. The notation used in equations (4.3) and (4.4) is now understood as meaning that the angular displacement and velocity are about the fixed z-axis . An example is shown in Figure 3, along with the so-called right-hand rule, for the angular velocity vector dθ ω = . (4.7) dt The introduction of a vector notation has many benefits and simplifies the form of several relations that we will encounter. A first example is that of the infinitesimal arc vector dr that results for an infinitesimal rotation vector dθ of a rigid body (please note that we have intentionally replaced s for the finite arc in equation (4.2) with dr and not ds ). Let us consider the special case shown in Figure 4 where an infinitesimal rotation dθ = dθzez about the z-axis is effected on a vector r = r ex aligned along the x-axis . As can be seen from the figure, the resulting infinitesimal arc dr = dr ey will be oriented along the y-axis . We know from equation (4.2) that dr = rdθ, (4.8) Figure 3 – Shown is the vector representation for an angular velocity about the , along with the so- called right-hand rule. - 64- y dθz dr x (+z) r Figure 4 - Infinitesimal rotation of a rigid body about the . but how can we mathematically determine the orientation of the infinitesimal arc from that of the rotation and radius? To do so, we must introduce the cross product between two vectors. Let a and b two vectors such that a = axex + ayey + azez (4.9) b = bxex + byey + bzez . Then we define the cross product a × b = (aybz − azby )ex + (azbx − axbz )ey + (axby − aybx )ez . (4.10) It is important to note that a × b = −b × a. (4.11) It is then straightforward to establish the following ex × ey = ez ey × ez = ex (4.12) ez × ex = ey , and ei × ei = 0, (4.13) where i = x, y, or z . Coming back to our simple example of Figure 4, and considering equations (4.8) and (4.12) we find that dr ey = dθzez × r ex . (4.14) Although equation (4.14) results from a special case where the orientation of the different vectors was specified a priori, this relation can be generalized with - 65- dr = dθ × r, (4.15) as could readily verified by changing the orientation of r and dθ in Figure 4. We therefore realize that the infinitesimal arc dr represents the change in the radius vector r under a rotation dθ ; hence the chosen notation. Moreover, we can find a vector generalization of equation (1.44) in Chapter 1 that established the relationship between the linear and angular velocities by dividing by an infinitesimal time interval dt on both sides of equation (4.15). We then find v = ω × r, (4.16) where v = dr dt and ω = dθ dt . Just as we did for the angular velocity in equations (4.3) and (4.4) (but using a vector notation), we can define an average angular acceleration over a time interval Δt with Δω α = (4.17) ave Δt and an instantaneous angular acceleration with Δω α = lim Δt→0 Δt (4.18) dω = . dt Combining equations (4.7) and (4.18), we can also express the instantaneous angular acceleration as the second order time derivative of the angular displacement dω α = dt d ⎛ dθ ⎞ = (4.19) dt ⎝⎜ dt ⎠⎟ d 2θ = . dt 2 4.1.2 Constant Angular Acceleration We have so far observed a perfect correspondence between the angular displacement dθ , velocity ω , and acceleration α with their linear counterparts dr , v , and a . We also studied in Section 2.1 of Chapter 2 the case of a constant linear acceleration and found that - 66- a(t) = a (constant) v(t) = v0 + at (4.20) 1 2 r(t) = r0 + v0t + at , 2 with r0 and v0 the initial position and velocity, respectively. We further combined these equations to derive the following relation 1 2 2 ⎡v (t) − v0 ⎤ = a ⋅ ⎡r(t) − r0 ⎤. (4.21) 2 ⎣ ⎦ ⎣ ⎦ Because the relationship between θ, ω, and α is the same as that between r , v , and a , we can write down similar equations for the case of constant angular acceleration α (t) = α (constant) ω (t) = ω 0 + αt (4.22) 1 2 θ (t) = θ 0 + ω 0t + αt 2 and 1 2 2 ⎡ω (t) −ω 0 ⎤ = α ⋅ ⎡θ (t) −θ 0 ⎤ (4.23) 2 ⎣ ⎦ ⎣ ⎦ without deriving them, since the process would be identical to the one we went through for the constant linear acceleration case. 4.1.3 Linear Acceleration of a Rotating Rigid Body We previously derived equation (4.16) for the linear velocity of a rotating rigid body. We could think, for example, of a solid, rotating disk and focus on the trajectory of a point on its surface. Since this point, which at a given instant has the velocity v , does not move linearly but rotates, there must be a force “pulling” it toward the centre point of the disk. This leads us to consider, one more time, the concept of the centripetal acceleration discussed in Chapter 1 (see Section 1.4). It is, however, possible to use equation (4.16) to combine two types of accelerated motions. To do so, we take its time derivative dv d = (ω × r) dt dt ⎛ dω ⎞ ⎛ dr⎞ = × r + ω × (4.24) ⎝⎜ dt ⎠⎟ ⎝⎜ dt ⎠⎟ = α × r + ω × v. - 67- On the second line of equation (4.24) we use the fact that the derivative of a product of functions, say, f and g , yields d df dg ( fg) = g + f . (4.25) dt dt dt The same result holds for the scalar or cross products between two vectors a and b . We can verify this as follows d d (a ⋅b) = axbx + ayby + azbz dt dt ( ) d d d = (axbx ) + ayby + azbz dt dt ( ) dt ( ) da db da db da db = x b + a x + y b + a y + z b + a z (4.26) dt x x dt dt y y dt dt z z dt da da da db db db = x b + y b + z b + a x + a y + a z dt x dt y dt z x dt y dt z dt da db = ⋅b + a ⋅ dt dt and if we consider the x component of a × b (see equation (4.10)) d d (a × b) = aybz − azby dt x dt ( ) d d = a b − a b dt ( y z ) dt ( z y ) ⎛ da db ⎞ ⎛ da db ⎞ = y b + a z − z b + a y (4.27) ⎝⎜ dt z y dt ⎠⎟ ⎝⎜ dt y z dt ⎠⎟ ⎛ da da ⎞ ⎛ db db ⎞ = y b − z b + a z − a y ⎝⎜ dt z dt y ⎠⎟ ⎝⎜ y dt z dt ⎠⎟ ⎛ da ⎞ ⎛ db⎞ = ⎜ × b⎟ + ⎜ a × ⎟ .