∝(19A.2)(19A.2)at=r∝ a_t=r \propto \label{19-1} A Rotating Rigid Body The characterization of the motion of a rotating rigid body has a lot in common with that of a particle traveling on a circle. In fact, every particle making up a rotating rigid body is undergoing circular motion. But different particles making up the rigid body move on circles of different radii and hence have speeds and accelerations that differ from each other. For instance, each time the object goes around once, every particle of the object goes all the way around its circle once, but a particle far from the axis of rotation goes all the way around circle that is bigger than the one that a particle that is close to the axis of rotation goes around. To do that, the particle far from the axis of rotation must be moving faster. But in one rotation of the object, the line from the center of the circle that any particle of the object is on, to the particle, turns through exactly one rotation. In fact, the angular motion variables that we have been using to characterize the motion of a line extending from the center of a circle to a particle that is moving on that circle can be used to characterize the motion of a spinning rigid body as a whole. There is only one spin rate for the whole object, the angular velocity ωω\omega, and if that spin rate is changing, there is only one rate of change of the spin rate, the angular acceleration ∝∝\propto. To specify the angular position of a rotating rigid body, we need to establish a reference line on the rigid body, extending away from a point on the axis of rotation in a direction perpendicular to the axis of rotation. This reference line rotates with the object. Its motion is the angular motion of the object. We also need a reference line segment that is fixed in space, extending from the same point on the axis, and away from the axis in a direction perpendicular to the axis. This one does not rotate with the object. Imagining the two lines to have at one time been collinear, the net angle through which the first line on the rigid body has turned relative to the fixed line is the angular position θθ\theta of the object. The Constant Angular Acceleration Equations While physically, there is a huge difference, mathematically, the rotational motion of a rigid body is identical to motion of a particle that only moves along a straight line. As in the case of linear motion, we have to define a positive direction. We are free to define the positive direction whichever way we want for a given problem, but we have to stick with that definition throughout the problem. Here, we establish a viewpoint some distance away from the rotating rigid body,but on the axis of rotation, and state that, from that viewpoint, counterclockwise is the positive sense of rotation, or alternatively, that clockwise is the positive sense of rotation. Whichever way we pick as positive, will be the positive sense of rotation for angular displacement (change in angular position), angular velocity, angular acceleration, and angular position relative to the reference line that is fixed in space. Next, we establish a zero for the time variable; we imagine a stopwatch to have been started at some instant that we define to be time zero. We call values of angular position and angular velocity, at that instant, the initial values of those quantities. Given these criteria, we have the following table of corresponding quantities. Note that a rotational motion quantity is in no way equal to its linear motion counterpart, it simply plays a role in rotational motion that is mathematically similar to the role played by its counterpart in linear motion. Linear Motion Quantity Corresponding Angular Motion Quantity xxx θθ\theta vvv ωω\omega aaa ∝∝\propto The one variable that the two different kinds of motion do have in common is the stopwatch reading ttt. Recall that, by definition, ω=dθdtω=dθdt \omega=\frac{d\theta}{dt} \nonumber and∝=dωdtand∝=dωdt\mbox{and} \quad \propto= \frac{d\omega}{dt} \nonumber While it is certainly possible for ∝∝\propto to be a variable, many cases arise in which ∝∝\propto is a constant. Such a case is a special case. The following set of constant angular acceleration equations apply in the special case of constant angular acceleration: (The derivation of these equations is mathematically equivalent to the derivation of the constant linear acceleration equations. Rather than derive them again, we simply present the results.) θ=θ0+ω0t+12∝t2(19A.3)(19A.3)θ=θ0+ω0t+12∝t2 \theta=\theta_0+\omega_0t+\frac{1}{2} \propto t^2 \label{19-2} θ=θ0+ω0+ω2t(19A.4)(19A.4)θ=θ0+ω0+ω2t \theta=\theta_0+\frac{\omega_0+\omega}{2} t \label{19-3} ω=ω0+∝t(19A.5)(19A.5)ω=ω0+∝t \omega=\omega_0+\propto t \label{19-4} ω2=ω20+2∝Δθ(19A.6)(19A.6)ω2=ω02+2∝Δθ \omega^2=\omega_0^2 +2\propto \Delta\theta \label{19-5} The rate at which a sprinkler head spins about a vertical axis increases steadily for the first 2.00 seconds of its operation such that, starting from rest, the sprinkler completes 15.0 revolutions clockwise (as viewed from above) during that first 2.00 seconds of operation. A nozzle, on the sprinkler head, at a distance of 11.0 cm from the axis of rotation of the sprinkler head, is initially due west of the axis of rotation. Find the direction and magnitude of the acceleration of the nozzle at the instant the sprinkler head completes its second (good to three significant figures) rotation. Solution We’re told that the sprinkler head spin rate increases steadily, meaning that we are dealing with a constant angular acceleration problem, so, we can use the constant angular acceleration equations. The fact that there is a non-zero angular acceleration means that the nozzle will have some tangential acceleration at→at→\vec{a_t}. Also, the sprinkler head is spinning at the instant in question so the nozzle will have some centripetal acceleration ac→ac→\vec{a_c}. We’ll have to find both at→at→\vec{a_t} and ac→ac→\vec{a_c} and add them like vectors to get the total acceleration of the nozzle. Let’s get started by finding the angular acceleration ∝∝\propto. We start with the first constant angular acceleration equation (equation 19A.319A.3\ref{19-2}): θ=0+0⋅t+12∝t2θ=0+0⋅t+12∝t2 \theta= 0+ 0\cdot t+\frac{1}{2} \propto t^2 \nonumber The initial angular velocity ω0ω0\omega_0 is given as zero. We have defined the initial angular position to be zero. This means that, at time t=2.00st=2.00st = 2.00 s, the angular position θθ\theta is 15.0rev=15.0 rev2π radrev=94.25rad15.0rev=15.0 rev2π radrev=94.25rad15.0 \, \mbox{rev}=15.0\space \mbox{rev}\frac{2\pi \space\mbox{rad}}{\mbox{rev}}=94.25 \mbox{rad}. Solving equation 19A.319A.3\ref{19-2} above for ∝∝\propto yields: ∝=2θt2∝=2θt2 \propto=\frac{2\theta}{t^2} \nonumber ∝=2(94.25rad)(2.00s)2∝=2(94.25rad)(2.00s)2 \propto=\frac{2(94.25 \mbox{rad})}{(2.00s)^2} \nonumber ∝=47.12rads2∝=47.12rads2 \propto=47.12 \frac{\mbox{rad}}{s^2} \nonumber Substituting this result into equation 19A.219A.2\ref{19-1}: at=r∝at=r∝a_t= r \propto \nonumber gives us at=(.110m)47.12rad/s2at=(.110m)47.12rad/s2 a_t=(.110m)47.12\mbox{rad}/s^2 \nonumber which evaluates to at=5.18ms2at=5.18ms2a_t=5.18 \frac{m}{s^2} \nonumber Now we need to find the angular velocity of the sprinkler head at the instant it completes 2.00 revolutions. The angular acceleration ∝∝\propto that we found is constant for the first fifteen revolutions, so the value we found is certainly good for the first two turns. We can use it in the fourth constant angular acceleration equation (equation 19A.619A.6\ref{19-5}): ω2=0+2∝Δθω2=0+2∝Δθ \omega^2= 0+2\propto \Delta\theta \nonumber where Δθ=2 rev=2.00 rev2π radrev=4.00π radΔθ=2 rev=2.00 rev2π radrev=4.00π rad\Delta\theta=2\space \mbox{rev}=2.00\space\mbox{rev}\frac{2\pi\space\mbox{rad}}{\mbox{rev}}=4.00\pi \space\mbox{rad} ω=2∝Δθ−−−−−−√ω=2∝Δθ\omega=\sqrt{2\propto\Delta\theta} \nonumber ω=2(94.25rad/s2)4.00πrad−−−−−−−−−−−−−−−−−−−√ω=2(94.25rad/s2)4.00πrad\omega=\sqrt{2(94.25 \mbox{rad}/s^2)4.00\pi \mbox{rad}} \nonumber \omega=48.67 \mbox{rad}/s\label{19-6}\omega=48.67 \mbox{rad}/s\label{19-6}\omega=48.67 \mbox{rad}/s\label{19-6} (at that instant when the sprinkler head completes its 2nd turn) Now that we have the angular velocity, to get the centripetal acceleration we can use equation ??????\ref{18-6}: ac=rω2ac=rω2 a_c=r\omega^2 \nonumber ac=.110m(48.67rad/s)2ac=.110m(48.67rad/s)2 a_c=.110m(48.67 \mbox{rad}/s)^2 \nonumber ac=260.6ms2ac=260.6ms2a_c=260.6 \frac{m}{s^2} \nonumber Given that the nozzle is initially at a point due west of the axis of rotation, at the end of 2.00 revolutions it will again be at that same point. Now we just have to add the tangential acceleration and the centripetal acceleration vectorially to get the total acceleration. This is one of the easier kinds of vector addition problems since the vectors to be added are at right angles to each other. From Pythagorean’s theorem we have a=a2c+a2t−−−−−−√a=ac2+at2 a=\sqrt{a_c^2+a_t^2} \nonumber a=(260.6m/s2)2+(5.18m/s2)2−−−−−−−−−−−−−−−−−−−−−−√a=(260.6m/s2)2+(5.18m/s2)2 a=\sqrt{(260.6m/s^2)^2+(5.18m/s^2)^2} \nonumber a=261m/s2a=261m/s2 a=261m/s^2 \nonumber From the definition of the tangent of an angle as the opposite over the adjacent: tanθ=atactanθ=atac tan\theta=\frac{a_t}{a_c} \nonumber θ=tan−15.18m/s2260.6m/s2θ=tan−15.18m/s2260.6m/s2 \theta=tan^{-1} \frac{5.18 m/s^2}{260.6m/s^2} \nonumber θ=1.14∘θ=1.14∘\theta=1.14^{\circ} \nonumber Thus, a=261m/s2at 1.14∘ North of Easta=261m/s2at 1.14∘ North of Easta=261m/s^2 \quad \mbox{at \(1.14^{\circ}\) North of East} \nonumber When the Angular Acceleration is not Constant The angular position of a rotating body undergoing constant angular acceleration is given, as a function of time, by our first constant angular acceleration equation, equation 19A.319A.3\ref{19-2}: θ=θ0+ω0t+12∝t2θ=θ0+ω0t+12∝t2\theta=\theta_0+\omega_0t+\frac{1}{2}\propto t^2 \nonumber If we take the 2nd derivative of this with respect to time, we get the constant ∝∝\propto. (Recall that the first derivative yields the angular velocity ωω\omega and that ∝=dωdt∝=dωdt\propto=\frac{d\omega}{dt}. ) The expression on the right side of θ=θ0+ω0t+12∝t2θ=θ0+ω0t+12∝t2\theta=\theta_0+\omega_0t+\frac{1}{2}\propto t^2 contains three terms: a constant, a term with ttt to the first power, and a term with ttt to the 2nd power. If you are given θθ\theta in terms of ttt, and it cannot be rearranged so that it appears as one of these terms or as a sum of two or all three such terms; then; ∝∝\propto is not a constant and you cannot use the constant angular acceleration equations. Indeed, if you are being asked to find the angular velocity at a particular instant in time, then you’ll want to take the derivative dθdtdθdt\frac{d\theta}{dt} and evaluate the result at the given stopwatch reading. Alternatively, if you are being asked to find the angular acceleration at a particular instant in time, then you’ll want to take the second derivative d2θdt2d2θdt2\frac{d^2\theta}{dt^2} and evaluate the result at the given stopwatch reading. Corresponding arguments can be made for the case of ωω\omega. If you are given ωω\omega as a function of ttt and the expression cannot be made to “look like” the constant angular acceleration equation ω=ω0+∝