Rotating objects can exhibit surprising and unexpected behaviors. For example, a common lecture demonstration makes use of a bicycle wheel with two handles along the axis. The wheel is spun, then handed to an unsuspecting student who is asked to turn the spinning wheel $\text{90}°$. Surprisingly, this is very hard to do. The reason is that the angular momentum is a vector, so the wheel’s rotation axis—the direction of $\vec{L}$—is highly resistant to change. If the wheel is spinning fast, a large torque is required to turn the wheel’s axis.

We’ll look at a related example: the precession of a gyroscope. A gyroscope—whether it’s a toy or a precision instrument used for navigation—is a rapidly spinning wheel or disk whose axis of rotation can assume any orientation. As it spins, it has angular momentum $\vec{L}=I\vec{\omega }$ along the rotation axis. A navigation gyroscope is mounted in gimbals that allow it to spin with virtually no torque from the environment. Once its axis is pointed north, conservation of angular momentum will ensure that the axis continues to point north no matter how the ship or plane moves.

We want to consider a horizontal gyroscope, with the disk spinning in a vertical plane, that is supported at only one end of its axle, as shown in FIGURE 12.57. You would expect it to simply fall over—but it doesn’t. Instead, the axle remains horizontal, parallel to the ground, while the entire gyroscope slowly rotates in a horizontal plane. This steady change in the orientation of the rotation axis is called precession, and we say that the gyroscope precesses about its point of support. The precession frequency $\text{Ω}$ (capital Greek omega) is much less than the disk’s rotation frequency $\omega$. Note that $\text{Ω}$, like $\omega$, is in rad/s.

FIGURE 12.57 A spinning gyroscope precesses in a horizontal plane.

You might object that angular momentum is not conserved during precession. This is true. The magnitude of $\vec{L}$ is constant, but its direction is changing. However, angular momentum is conserved only for an isolated system, one on which there is no net torque. The spinning gyroscope is not an isolated system because gravity is exerting a torque on it. Indeed, understanding the relationship between the gravitational torque and the angular momentum is the key to understanding why the gyroscope precesses.

FIGURE 12.58a shows a gyroscope that is not spinning. When released, it most definitely falls over by rotating about the point of support until the disk hits the table. Because the motion is rotation, rather than the translational motion of a gyroscope that is simply dropped, we can analyze it using the concepts of torque and angular momentum.

FIGURE 12.58 The gravitational torque on a nonspinning gyroscope causes it to fall over.

There are two forces acting on the gyroscope: gravity pulling downward at the disk’s center of mass (we’ll assume that the axle is massless) and the normal force of the support pushing upward. The normal force exerts no torque about the pivot point because it acts at the pivot point, so the net torque on the gyroscope is entirely a gravitational torque:

where $d=\left|\vec{r}\right|$ is the distance from the pivot to the center of the disk. To evaluate the cross product, we redrew ${\overrightarrow{F}}_{\text{G}}$ at the pivot and established a coordinate system with the z-axis along the axle. Vectors $\vec{r}$ and ${\overrightarrow{F}}_{\text{G}}$ are perpendicular $(\text{sin}\alpha =1)$, and by using the right-hand rule we see that $\vec{\tau }$ points along the x-axis.

We found in the previous section that a torque causes the angular momentum to change. In particular,

So in a small interval of time $dt$, the torque causes the gyroscope’s angular momentum about the point of support to change by $d\vec{L}=\vec{\tau }dt$.

FIGURE 12.58b shows graphically what happens. Initially, when the gyroscope is first released, $\vec{L}=\vec{0}$. After a small interval of time, the gyroscope acquires a small amount of angular momentum $d\vec{L}$ in the direction of $\vec{\tau }$, the $î$ direction. An angular momentum along the x-axis means that the gyroscope is rotating in the yz-plane—which is exactly what it does as it starts to fall. During the next interval of time, $\vec{L}$ increases a bit more in the $î$ direction, and then a bit more. This is what we expect as the falling gyroscope picks up speed, increasing its angular momentum.

Now the magnitude of $\vec{\tau }$ does not remain constant—the angle between $\vec{r}$ and $\vec{F}{}_{\text{G}}$ changes as the gyroscope falls, changing the cross product—so integrating Equation 12.57 symbolically is difficult. Nonetheless, the direction of $\vec{\tau }$ is always the $î$ direction, so we can see that the angular momentum keeps increasing in the $î$ direction as the gyroscope falls.

What’s different about a spinning gyroscope that causes it to precess rather than fall? In FIGURE 12.59a we’ve again just released the gyroscope, its axle is again along the z-axis, but now it’s spinning with angular velocity $\vec{\omega }=\omega \widehat{k}$. Consequently, the gyroscope has initial angular momentum $\vec{L}=I\vec{\omega }=I\omega \widehat{k}$ along the z-axis. The torque is exactly as we calculated above, and that torque again causes the angular momentum to change by $d\vec{L}=\vec{\tau }dt$. The only difference is that the gyroscope starts with initial angular momentum—but that makes all the difference.

FIGURE 12.59 For a spinning gyroscope, the gravitational torque changes the direction but not the magnitude of the angular momentum.

FIGURE 12.59b, looking down from above, shows the initial angular momentum $\vec{L}$. A very small time interval $dt$ after we release the gyroscope, its angular momentum will have changed to $\vec{L}+d\vec{L}$. The small change in angular momentum, $d\vec{L}$, is parallel to the torque and thus perpendicular to the spinning gyroscope’s angular momentum $\vec{L}$. Because we’re adding vectors, not scalars, the “new” angular momentum has rotated to a new position but not increased in magnitude. So during $dt$, the angular momentum—and thus the entire gyroscope—rotates through a small angle $d\varphi$ in the horizontal plane.

The torque vector is always perpendicular to the axle, and thus $d\vec{L}$ is always perpendicular to $\vec{L}$. With each subsequent time interval $dt$, the gyroscope rotates through another small angle $d\varphi$ while the magnitude of the angular momentum (and hence the disk’s angular velocity $\omega )$ is unchanged. The gyroscope is precessing in the horizontal plane!

You’ve encountered a similar situation previously. If a ball is initially at rest, pulling on it with a string causes the ball to accelerate (increasing $\vec{v})$ in the direction of the pull. That is, $\vec{v}$ increases in magnitude but doesn’t change direction. But if a ball on a string is in uniform circular motion, a force directed to the center—the string tension—has a very different effect. $d\vec{v}$ points toward the center, because that’s the direction of the centripetal acceleration, but now $d\vec{v}$ is perpendicular to $\vec{v}$. Adding them as vectors to get $\vec{v}+d\vec{v}$ changes the direction of the velocity vector but not its magnitude. Then, as now, having an initial vector ($\vec{v}$ or $\vec{L})$ leads to very different behavior than not having an initial vector.

The small horizontal rotation $d\varphi$ is a small piece of the precession. Because it occurs during the small time interval $dt$, the rate of horizontal rotation—the precession frequency—is

In Figure 12.59b the small length $dL$ is the arc length spanned by $d\varphi$, hence $d\varphi =dL/L$ and the precession frequency is

From Equations 12.56 and 12.57, $dL/dt=\tau =Mgd$. Further, the gyroscope’s angular momentum has magnitude $L=I\omega$, where I is the moment of inertia of the disk rotating about the axle. Thus the precession frequency of the gyroscope—in rad/s—is

Because the spin angular velocity $\omega$ is in the denominator, a very rapidly spinning gyroscope precesses very slowly. As the gyroscope runs down, due to any little bit of friction, it begins to precess faster and faster.

We have made one tacit assumption. As the gyroscope precesses, the precessional motion has its own angular momentum along the vertical axis. The gyroscope’s angular momentum $\vec{L}$ is not simply the angular momentum of the spinning disk, as we assumed, but the vector sum $\vec{L}{}_{\text{spin}}+\vec{L}{}_{\text{precess}}$. As long as the gyroscope precesses slowly, with $\text{Ω}\ll \omega$, the precessional angular momentum is very small compared to the spin angular momentum and our assumption is well justified. But toward the end of the gyroscope’s motion, as $\omega$ decreases and $\text{Ω}$ increases, our model of precession breaks down and the gyroscope’s motion becomes more complex.

CHAPTER 12 CHALLENGE EXAMPLE The ballistic pendulum revisited

A $2.0\text{kg}$ block hangs from the end of a $1.5\text{kg}$, $1.0\text{-m-long rod,}$ together forming a pendulum that swings from a frictionless pivot at the top end of the rod. A $10\text{g}$ bullet is fired horizontally into the block, where it sticks, causing the pendulum to swing out to a $30\text{°}$ angle. What was the speed of the bullet?

MODEL Model the rod as a uniform rod that can rotate around one end, and assume the block is small enough to model as a particle. There are no external torques on the bullet $+$ block $+$ rod system, so angular momentum is conserved in the inelastic collision. Further, the mechanical energy of the system is conserved after (but not during) the collision as the pendulum swings outward.

VISUALIZE FIGURE 12.60 is a pictorial representation. This is a two-part problem, so we’ve separated the collision’s before-and-after from the pendulum swing’s before-and-after. The end of the collision is the beginning of the swing.

FIGURE 12.60 Pictorial representation of the bullet hitting the pendulum.

FIGURE 12.60 Full Alternative Text

SOLVE This is a ballistic pendulum. 《 EXAMPLE 11.5 considered a simpler ballistic pendulum with a mass on a string, rather than on a rod, and a review of that example is highly recommended. The key to both is that a different conservation law applies to each part of the problem.

Angular momentum is conserved in the collision, thus $L_1=L_0$. Before the collision, the angular momentum—which we’ll measure about the pendulum’s pivot point—is entirely that of the bullet. The angular momentum of a particle is $L=mrv\text{sin}\beta$. An instant before the collision, just as the bullet reaches the block, $r=d$ and, because $\vec{v}$ is perpendicular to $\vec{r}$ at that instant, $\beta =90\text{°}$. Thus $L_0=m_{\text{b}}d{v}_{0\text{b}}$. (This is the magnitude of the angular momentum; from the right-hand rule, the angular momentum vector points out of the page.)

An instant after the collision, but before the pendulum has had time to move, the rod has angular velocity ${\omega }_1$ and the block, with the embedded bullet, is moving in a circle with speed $v_1={\omega }_{1}r={\omega }_{1}d$. The angular momentum of the $\text{block}+\text{bullet system}$ is that of a particle, still with $\beta =90\text{°}$, while that of the rod—an object rotating on a fixed axle—is $I_{\text{rod}}{\omega }_1$. Thus the post-collision angular momentum is

$$L_1=(m_{\text{B}}+m_{\text{b}})v_{1}r+I_{\text{rod}}{\omega }_1=(m_{\text{B}}+m_{\text{b}}){}_{}{d}^2{\omega }_1+\frac{1}{3}{m}_{\text{R}}{d}^2{\omega }_1$$

The moment of inertia of the rod was taken from Table 12.2.

Equating the before-and-after angular momenta, then solving for $v_{0\text{b}}$, gives

$$\begin{array}{l}{m}_{\text{b}}d{v}_{0\text{b}}=(m_{\text{B}}+m_{\text{b}})d^2{\omega }_1+\frac{1}{3}{m}_{\text{R}}{d}^2{\omega }_1\\ v_{0\text{b}}={\displaystyle \frac{m_{\text{B}}+m_{\text{b}}+\frac{1}{3}{m}_{\text{R}}}{m_{\text{b}}}}d{\omega }_1=251d{\omega }_1\end{array}$$

Once we know ${\omega }_1$, which we’ll find from energy conservation in the swing, we’ll be able to compute the bullet’s speed.

Mechanical energy is conserved during the swing, but you must be careful to include all the energies. The kinetic energy has two components: the translational kinetic energy of the $\text{block}+\text{bullet}$ system and the rotational kinetic energy of the rod. The gravitational potential energy also has two components: the potential energy of the $\text{block}+\mathrm{bullet system}$ and the potential energy of the rod. The latter changes because the center of mass moves upward as the rod swings. Thus the energy-conservation statement is

$$\frac{1}{2}(m_{\text{B}}+m_{\text{b}})v_2{}^2+\frac{1}{2}{I}_{\text{rod}}{\omega }_2{}^2+(m_{\text{B}}+m_{\text{b}})g{y}_2+m_{\text{R}}g{y}_{\text{cm}2}=\frac{1}{2}(m_{\text{B}}+m_{\text{b}})v_1{}^2+\frac{1}{2}{I}_{\text{rod}}{\omega }_1{}^2+(m_{\text{B}}+m_{\text{b}})g{y}_1+m_{\text{R}}g{y}_{\text{cm}1}$$

Although this looks very complicated, you should convince yourself that we’ve done nothing more than add up two kinetic energies and two potential energies before and after the swing.

We know that $v_2=0$ and ${\omega }_2=0$ at the end of the swing, and that $v_1=d{\omega }_1$ at the beginning. We also know the moment of inertia of a rod pivoted at one end. Combining the potential energy terms and using $\text{Δ}y=y_{\text{f}}-y_{\text{i}}$, we thus have

$$\frac{1}{2}(m_{\text{B}}+m_{\text{b}}+\frac{1}{3}{m}_{\text{R}})d^2{\omega }_1{}^2=(m_{\text{B}}+m_{\text{b}})g\text{Δ}y+m_{\text{R}}g\text{Δ}{y}_{\text{cm}}$$

We see from Figure 12.60 that the block, at its highest point, is distance $d\text{cos}\theta$ below the pivot. It started distance $d$ below the pivot, so the bullet$+$block system gained height $\text{Δ}y=d-d\text{cos}\theta =$ $d(1-\text{cos}\theta )$. The rod’s center of mass started distance $d\text{/}2$ below the pivot and rises only half as much as the block, so $\text{Δ}{y}_{\text{cm}}=\frac{1}{2}d(1-\text{cos}\theta )$. With these, the energy equation becomes

$$\frac{1}{2}(m_{\text{B}}+m_{\text{b}}+\frac{1}{3}{m}_{\text{R}})d^2{\omega }_1{}^2=(m_{\text{B}}+m_{\text{b}}+\frac{1}{2}{m}_{\text{R}})gd(1-\text{cos }\theta )$$

We can now solve for ${\omega }_1$:

$${\omega }_1=\sqrt{{\displaystyle \frac{m_{\text{B}}+m_{\text{b}}+\frac{1}{2}{m}_{\text{R}}}{m_{\text{B}}+m_{\text{b}}+\frac{1}{3}{m}_{\text{R}}}}{\displaystyle \frac{2g(1-\text{cos}\theta )}{d}}}=1.70\text{rad}\text{/}\text{s}$$

and with that

$$v_{0\text{b}}=251d{\omega }_1=430\text{m}\text{/}\text{s}$$

REVIEW $430\text{m}\text{/}\text{s}$ seems a reasonable speed for a bullet. This was a challenging problem, but one that you can solve if you focus on the problem-solving strategies—drawing a careful pictorial representation, defining the system, and thinking about which conservation laws apply—rather than hunting for the “right” equation.