✓TIp Specify the Axis
Torque depends on where the force is applied relative to some rotation axis. The same physical force results in different torques about different axes. Be sure the rotation axis is specified before you make a calculation involving torque.
GoT IT? 10.2 The forces in Figs. 10.5 and 10.6 all have the same magnitude. (1) Which of Figs. 10.5a, 10.5b, and 10.6b has the greatest torque? (2) Which of these has the least torque?
10.3 Rotational Inertia and the Analog of Newton’s Law
Torque and angular acceleration are the rotational analogs of force and linear accelera- tion. To develop a rotational analog of Newton’s law, we still need the rotational analog of mass.
The mass m in Newton’s law is a measure of a body’s inertia—of its resistance to changes in motion. So we want a quantity that describes resistance to changes in rotational motion. Figure 10.9 shows that it’s easier to set an object rotating when its mass is concen- trated near the rotation axis. Therefore, our rotational analog of inertia must depend not only on mass itself but also on the distribution of mass relative to the rotation axis.
Suppose the object in Fig. 10.9 consists of an essentially massless rod of length R with a ball of mass m on the end. We allow the object to rotate about an axis through the free end of the rod and apply a force FS to the ball, always at right angles to the rod (Fig. 10.10).
The ball undergoes a tangential acceleration given by Newton’s law: F = mat. (There’s also a tension force in the rod, but because it acts along the rod, it doesn’t contribute to the torque or angular acceleration.) We can use Equation 10.5 to express the tangen- tial acceleration in terms of the angular acceleration a and the distance R from the rota- tion axis: F = mat = maR. We can also express the force F in terms of its associated torque. Since the force is perpendicular to the rod, Equation 10.10 gives t = RF. Using our expression for F, we have
t = 1mR22a
Here we have Newton’s law, F = ma, written in terms of rotational quantities.
The torque—analogous to force—is the product of the angular acceleration and the quantity mR2, which must therefore be the rotational analog of mass. We call this quantity the rotational inertia or moment of inertia and give it the symbol I. Rotational inertia is measured in kg#m2 and accounts for both an object’s mass and the distribution of that mass. Like torque, the value of the rotational inertia depends on the location of the rotation axis. Given the rotational inertia I, our rotational analog of Newton’s law becomes
t = Ia 1rotational analog of Newton>s second law2 (10.11) Although we derived Equation 10.11 for a single, localized mass, it applies to extended objects if we interpret t as the net torque on the object and I as the sum of the rotational inertias of the individual mass elements making up the object.
Calculating the Rotational Inertia
When an object consists of a number of discrete mass points, its rotational inertia about an axis is the sum of the rotational inertias of the individual mass points:
I = a mi ri2 1rotational inertia2 (10.12) Here mi is the mass of the ith mass point, and ri is its distance from the rotation axis.
Figure 10.9 It’s easier to set an object rotating if the mass is concentrated near the axis.
Rotating the mass near the axis is easy.
Farther away, it’s harder to spin.
Rotation axis
Figure 10.10 A force applied perpendicular to the rod results in angular acceleration.
FS FS
FS
PheT: Torque (Torque)
PheT: Torque (Moment of Inertia)
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174 Chapter 10 Rotational Motion
assEss Make sense? Even though there are two masses, our answer is less than the rotational inertia mL2 of a single mass rotated about a rod of length L. That’s because distance from the rotation axis is squared, so it contributes more in determining rotational inertia than does mass. ■ A dumbbell-shaped object consists of two equal masses m = 0.64 kg
on the ends of a massless rod of length L = 85 cm. Calculate its rota- tional inertia about an axis one-fourth of the way from one end of the rod and perpendicular to it.
IntErprEt Here we have two discrete masses, so this problem is asking us to calculate the rotational inertia by summing over the in- dividual masses.
DEvElop Figure 10.11 is our sketch. We’ll use Equation 10.12, I = g mi ri2, to sum the two individual rotational inertias.
EvaluatE I = a mi ri2= m114L22+ m134L22= 58 mL2
= 5810.64 kg210.85 m22 = 0.29 kg#m2
Figure 10.11 Our sketch for Example 10.4, showing rotation about an axis perpendicular to the page.
GoT IT? 10.3 Would the rotational inertia of the two-mass dumbbell in Example 10.4 (a) increase, (b) decrease, or (c) stay the same (1) if the rotation axis were at the center of the rod? (2) If it were at one end?
With continuous distributions of matter, we consider a large number of very small mass elements dm throughout the object, and sum the individual rotational inertias r2 dm over the entire object (Fig. 10.12). In the limit of an arbitrarily large number of infinitesimally small mass elements, that sum becomes an integral:
I =
Lr2 dm arotational inertia,
continuous matterb (10.13)
where the limits of integration cover the entire object.
Figure 10.12 Rotational inertia can be found by integrating the rotational inertias r2 dm of the mass elements making up an object.
The mass element dm contributes rotational inertia r2 dm.
Rotation axis
r
dm
ExAmpLE 10.4 Rotational Inertia: A Sum
We’re almost done. But the integral in Equation 10.13 contains r, and we’ve related dm and dx. No problem: On the one-dimensional rod, distances from the rotation axis are just the coordinates x. So r becomes x in our integral, and we have
I =
Lr2 dm = L
L/2 -L/2
x2M L dx
We chose the limits to include the entire rod; with the origin at the center, it runs from -L/2 to L/2.
Find the rotational inertia of a uniform, narrow rod of mass M and length L about an axis through its center and perpendicular to the rod.
IntErprEt The rod is a continuous distribution of matter, so calcu- lating the rotational inertia is going to involve integration. We identify the rotation axis as being in the center of the rod.
DEvElop Figure 10.13 shows the rod and rotation axis; we added a coordinate system with x-axis along the rod and the origin at the rota- tion axis. With a continuous distribution, Equation 10.13, I = 1r2 dm, applies. To develop a solution plan, we need to set up the integral in Equation 10.13. That equation may seem confusing because the inte- gral contains both the geometric variable r and the mass element dm.
How are they related? At this point you might want to review Tactics 9.1 (page 148); we’ll follow its steps here. (1) We’re first supposed to find a suitable mass element; here, with a one-dimensional rod, that can be a short section of the rod. We marked a typical mass element in Fig. 10.13.
(2) This step is straightforward in this one-dimensional case; the length of the mass element is dx, signifying an infinitesimally short piece of the rod. (3) Now we form ratios to relate dx and the mass element dm. The total mass of the rod is M, and its total length is L. With the mass distrib- uted uniformly, that means dx is the same fraction of L that dm is of M, or dx/L = dm/M. (4) We solve for the mass element: dm = 1M/L2 dx.
Figure 10.13 Our sketch of the uniform rod of Example 10.5.
The mass element has mass dm and length dx.
ExAmpLE 10.5 Rotational Inertia by Integration: A Rod
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10.3 Rotational Inertia and the Analog of Newton’s Law 175
EvaluatE Because the sum of the mass elements over the ring is the total mass M, we find
I = MR2 1thin ring2 (10.15) assEss The rotational inertia of the ring is the same as if all the mass were concentrated in one place a distance R from the rotation axis;
the angular distribution of the mass about the axis doesn’t matter.
Notice, too, that it doesn’t matter whether the ring is narrow like a loop of wire or long like a section of hollow pipe, as long as it’s thin enough that all of it is essentially equidistant from the rotation axis (Fig. 10.15).
Find the rotational inertia of a thin ring of radius R and mass M about the ring’s axis.
IntErprEt This example is similar to Example 10.5, but the geom- etry has changed from a rod to a ring.
DEvElop Figure 10.14 shows the ring with a mass element dm. All the mass elements in the ring are the same distance R from the rotation axis, so r in Equation 10.13 is the constant R, and the equation becomes
I =
LR2 dm = R2 Ldm where the integration is over the ring.
Figure 10.14 Our sketch of a thin ring, showing one mass element dm.
Figure 10.15 The rotational inertia is MR2 for any thin ring, whether it’s narrow
like a wire loop or long like a pipe. ■
ExAmpLE 10.6 Rotational Inertia by Integration: A Ring
you probably realized that the rotational inertia would be 12 mL2 for ro- tation about the rod’s center. The total mass for that one was M = 2m, so in terms of total mass the rotational inertia about the center would be I = 14 ML2—a lot larger than what we’ve found for the continuous rod. That’s because much of the continuous rod’s mass is close to the rotation axis, so it contributes less to the rotational inertia. ■
EvaluatE The constants M and L come outside the integral, so we have I =
L
L/2 -L/2
x2M L dx = M
LL
L/2 -L/2
x2 dx = M L
x3 3 `L/2
-L/2 = 121 ML2 (10.14) assEss Make sense? In Example 10.4 we found I = 58 mL2 for a rod with two masses m on the ends. If you thought about GOT IT? 10.3,
A disk of radius R and mass M has uniform density. Find the rota- tional inertia of the disk about an axis through its center and perpen- dicular to the disk.
IntErprEt Again we need to find the rotational inertia for a piece of continuous matter, this time a disk.
DEvElop Because the disk is continuous, we need to integrate us- ing Equation 10.13, I = 1r2 dm. We’ll condense the strategy we applied in Example 10.5. The result of Example 10.6 suggests di- viding the disk into rings, as shown in Fig. 10.16a. Equation 10.15, with MSdm, shows that a ring of radius r and mass dm contributes r2 dm to the rotational inertia of the disk. Then the total inertia will be I = 10Rr2 dm, where we chose the limits to pick up contributions from all the mass elements on the disk. Again we need to relate r and dm. Think of “unwinding” the ring, as shown in Fig. 10.16b;
it becomes essentially a rectangle whose area dA is its circumfer- ence multiplied by its width: dA = 2pr dr. Next, we form ratios.
Figure 10.16 A disk may be divided into ring-shaped mass elements of mass dm, radius r, and width dr.
(a)
(b)
ExAmpLE 10.7 Rotational Inertia by Integration: A Disk
The ring area dA is to the total disk area pR2 as the ring mass dm is to the total mass M: 2pr dr/pR2 = dm/M. Solving for dm gives dm = 12Mr/R22 dr.
(continued)
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176 Chapter 10 Rotational Motion
The rotational inertias of other shapes about various axes are found by integration as in these examples. Table 10.2 lists results for some common shapes. Note that more than one rotational inertia is listed for some shapes, since the rotational inertia depends on the rotation axis.
If we know the rotational inertia Icm about an axis through the center of mass of a body, a useful relation called the parallel-axis theorem allows us to calculate the rotational in- ertia I through any parallel axis. The parallel-axis theorem states that
I = Icm + Md2 (10.17)
where d is the distance from the center-of-mass axis to the parallel axis and M is the total mass of the object. Figure 10.17 shows the meaning of the parallel-axis theorem, which you can prove in Problem 78.
GoT IT? 10.4 Explain why the rotational inertia of the solid sphere in Table 10.2 is less than that of the spherical shell with the same radius and the same mass.
✓TIp Constants and Variables
Note the different roles of R and r here. R represents a fixed quan- tity—the actual radius of the disk—and it’s a constant that can go outside the integral. In contrast, r is the variable of integration, and it changes as we range from the disk’s center to its edge, add- ing up all the infinitesimal mass elements. Because r is a variable over the region of integration, we can’t take it outside the integral.
■
EvaluatE We now evaluate the integral:
I = L
R 0
r2 dm= L
R 0
r2a2Mr R2 b dr
= 2M R2L
R 0
r3 dr = 2M R2
r4 4`
R
0 = 12 MR2 1disk2 (10.16) assEss Again, this result makes sense. In the disk, some of the mass is closer to the rotation axis, so the rotational inertia should be less than the value MR2 for the ring.
Table 10.2 Rotational Inertias
12 13
2 3 1
12
121
Thin rod about center I = ML2
Thin rod about end I = ML2
R
R
R R
Thin ring or hollow cylinder about its axis
I = MR2
Disk or solid cylinder about its axis I = MR2
Hollow spherical shell about diameter I = MR2
Solid sphere about diameter I = MR2
b
a
b
a Flat plate about perpendicular axis I = M1a2 + b22
Flat plate about central axis I = Ma2
L
L
2
5 1
12
Figure 10.17 Meaning of the parallel-axis theorem.
2 5
25 7
5
This axis is through the sphere’s center, so I = MR2.
This parallel axis is a distance d = R away from the original axis, so I = MR2 + Md2 = MR2.
(a) (b)
d = R
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10.3 Rotational Inertia and the Analog of Newton’s Law 177
Rotational Dynamics
Knowing a body’s rotational inertia, we can use the rotational analog of Newton’s second law (Equation 10.11) to determine its behavior, just as we used Newton’s law itself to ana- lyze linear motion. Like the force in Newton’s law, the torque in Equation 10.11 is the net external torque—the sum of all external torques acting on the body.
to the jets using Equation 10.10, t = rF sin u. (3) Use the rotational analog of Newton’s law—Equation 10.11, t= Ia—to find the angu- lar acceleration. (4) Use the change in angular speed to get the time.
EvaluatE Following our plan, (1) the rotational inertia from Table 10.2 is I = 12 MR2. (2) With the jets tangent to the satellite, sin u in Equation 10.10 is 1, so each jet contributes a torque of magnitude RF, where R is the satellite radius and F the jet thrust force. With two jets, the net torque then has magnitude t = 2RF. (3) Equation 10.11 gives a = t/I = 12RF2/112 MR22 = 4F/MR. (4) We want this torque to drop the angular speed from v0 = 10 rpm to zero, so the magni- tude of the speed change is
∆v =10 rev/min =110 rev/min212p rad/rev2/160 s/min2
= 1.05 rad/s
Since angular acceleration is a = ∆v/∆t, our final answer is
∆t = ∆v
a = MR ∆v 4F
= 1940 kg210.70 m211.05 rad/s2 142120 N2 =8.6 s
assEss Make sense? Yes: The thrust F appears in the denominator, showing that a larger force and hence torque will bring the satellite more rapidly to a halt. Larger M and R contribute to a larger rotational inertia, thus lengthening the stopping time—although a larger R also means a larger torque, an effect that reduces the R dependence from the R2 that appears in the expression for rotational inertia. ■ A cylindrical satellite is 1.4 m in diameter, with its 940-kg mass dis-
tributed uniformly. The satellite is spinning at 10 rpm but must be stopped so that astronauts can make repairs. Two small gas jets, each with 20-N thrust, are mounted on opposite sides of the satellite and fire tangent to the satellite’s rim. How long must the jets be fired in order to stop the satellite’s rotation?
IntErprEt This is ultimately a problem about angular acceleration, but we’re given the forces the jets exert. So it becomes a problem about calculating torque and then acceleration—that is, a problem in rotational dynamics using the rotational analog of Newton’s law.
DEvElop Figure 10.18 shows the situation. We’re asked about the time, which we can get from the angular acceleration and initial angu- lar speed. We can find the acceleration using the rotational analog of Newton’s law, Equation 10.11, if we know both torque and rotational inertia. So here’s our plan: (1) Find the satellite’s rotational inertia from Table 10.2, treating it as a solid cylinder. (2) Find the torque due
Figure 10.18 Torque from the jets stops the satellite’s rotation.
ExAmpLE 10.8 Rotational Dynamics: De-Spinning a Satellite
A single problem can involve both rotational and linear motion with more than one object.
The strategy for dealing with such problems is similar to the multiple-object strategy we de- veloped in Chapter 5, where we identified the objects whose motions we were interested in, drew a free-body diagram for each, and then applied Newton’s law separately to each ob- ject. We used the physical connections among the objects to relate quantities appearing in the separate Newton’s law equations. Here we do the same thing, except that when an object is rotating, we use Equation 10.11, the rotational analog of Newton’s law. Often the physical connection will entail relations between the force on an object in linear motion and the torque on a rotating object, as well as between the objects’ linear and rotational accelerations.
A solid cylinder of mass M and radius R is mounted on a frictionless horizontal axle over a well, as shown in Fig. 10.19. A rope of negli- gible mass is wrapped around the cylinder and supports a bucket of mass m. Find an expression for the bucket’s acceleration as it falls down the well shaft.
ExAmpLE 10.9 Rotational and Linear Dynamics: Into the Well
IntErprEt If it weren’t connected to the cylinder, the bucket would fall with acceleration g. But the rope exerts an upward tension force TS on the bucket, reducing its acceleration and at the same time exerting a torque on the cylinder. So we have a problem involving both rotational and linear dynamics. We identify the bucket and the cylinder as the (continued)
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178 Chapter 10 Rotational Motion
Figure 10.20 Our free-body diagrams for the bucket and cylinder.
Figure 10.19 Example 10.9.
Pulley mass M
m m
GoT IT? 10.5 The figure shows two identical masses m connected by a string that passes over a frictionless pulley whose mass M is not negligible. One mass rests on a frictionless table;
the other hangs vertically, as shown. Is the magnitude of the ten- sion force in the vertical section of the string (a) greater than, (b) equal to, or (c) less than that in the horizontal section? Explain.