Circular Motion
Wheels on Railroad Trains
Rotational Inertia
Torque
Center of Mass and Center of Gravity
Locating the Center of Gravity
Stability
Centripetal Force
Centrifugal Force
Centrifugal Force in a Rotating Reference Frame
Simulated Gravity
Angular Momentum
Conservation of Angular Momentum
This chapter extends the translational ideas of previous chapters to rotation. A space habitat theme is briefly treated which opens the door to the myriad of interesting physics applications in a rotating frame of reference.
It is often said that rotational motion is analogous to linear motion and therefore should not be difficult to learn. But consider the many distinctions between 1) linear, 2) rotational, 3) revolutional, 4) radial, 5) tangential, 6) angular, and all related to a) speed, b) velocity, and c) acceleration. And we have centripetal and centrifugal forces, real and fictitious, not to mention torques. So there is a myriad of ideas to describe and understand. Is it any wonder why students who do well in linear motion fall apart here? A study of rotational motion is considerably more complex than a study of linear motion. Caution your students to be patient with themselves if they don’t immediately catch on to what has taken centuries to master.
One of the most intriguing examples of v = rw is the beveled shape of railroad wheels. A train is able to round a corner in the same way a tapered glass rolls in a circle along a tabletop. Or the way a person with one leg shorter than the other tends to walk in a circle when lost in the woods. This is the feature of the box on Railroad Train Wheels. Fascination, especially when demonstrated with a pair of tapered cups taped at their wide ends, that roll along a pair of metersticks. At CCSF, Will Maynez built a beautiful “roller coaster” along which a set of tapered wheels faithfully follow the curved track. Most impressive!
Note that this chapter omits entirely the “right hand rule,” where fingers of the right hand represent the motion of a rotating body and the thumb represents the positive vector of motion. I have always felt that the reason for this rule in introductory physics courses was to provide the instructor the opportunity to easily make up tricky exam questions. Please spare undue emphasis on this material, which your students can get into in a follow-up course.
Martha Lietz at Niles West High School
A neat activity for demonstrating centripetal force I learned from physics teacher Howie Brand is the following: Have a small group of students around a small table (ideally circular) blow air through straws at a Ping-Pong ball so that it will move in a circular path. They will experience the fact that the ball must be blown radially inward.
The classic oldie but goodie PSSC film, “Frames of Reference” goes nicely with this chapter. The book “Sport Science” by Peter Brancazio (Simon & Shuster, 1984) is an excellent source of physics applications, with emphasis on the human body.
Note that this chapter has more figures than any in the book—55 of them.
Centripetal force is nicely extended in the Practicing Physics book, and in the student supplement, Problem Solving in Conceptual Physics, a bit of trigonometry is employed.
This chapter can be skipped or skimmed if a short treatment of mechanics is desired.
There are 5 OHTs for this chapter: Figures 8.6, 8.7, 8.14, 8.18-8.19, and 8.53.
In the Practicing Physics book:
• Torques • Banked Airplanes
• Torques and Rotation • Banked Track
• Acceleration and Circular Motion • Leaning On
• The Flying Pig • Simulated Gravity and Frames of Reference
In the Next-Time Questions book:
• Falling Metersticks • Woman on the Plank
• Trucks on a Hill • Tether Ball
• Rotating Disk • Kagan Roll
• Wrench Pull • Two Spheres
• Rolling Cans • Can Spurt
• Berry Shake • Normal Forces
• Broom Balance
The activity Torque Feeler in the Laboratory Manual is especially worthwhile—even in a lecture setting. There are two other activities and two experiments to complement this chapter.
The suggested lecture should take two or three class periods.
SUGGESTED LECTURE PRESENTATION
Cite the difference between rotational and linear speed—the examples of riding at different radial positions on a merry-go-round, or the different speeds of different parts of a phonograph record. A couple of coins on a record player, one close to the axis and the other near the edge, dramatically show the greater speed of the outer one. Cite the motion of “tail-end Charlie” at the skating rink.
Circular Motion: Only for a rigid rotating system such as a solid turntable or a stiff spoke does the equation v = rw apply—the greater the distance from the axis of rotation, the greater the linear speed. Don’t be surprised to find students applying this relationship to a nonrigid system, such as a system of planets. They are confused about Mercury, which is relatively fast in its orbit about the Sun, and Pluto, which is very slow. Horses running around a circular track obey v = rw only if they are constrained, like joined by a giant nonflexible spoke.
CHECK QUESTION: Briefly describing what a phonograph record is. Two music pieces of equal times are on a record, one at the beginning of the record (outside), and the other at the end of the record (inside). Which piece is wider in the radial direction on the record? [Both are the same width, since w is constant at all parts of the record. The outer cut covers more length, but at a faster linear speed in the same time.]
Railroad Train Wheels: A fascinating application of v = rw is presented in the box on railroad train wheels. Be sure to use OHTs T-17 and T-18. OHT-17 shows tracks banked for turns and a level track. By shifting the wheels OHT-18 to-and-fro over the level-track of OHT-17, you can show how a train sways when self-correcting. Fasten a pair of cups with wide ends connected, and with small ends connected, and roll them along a pair of metersticks. Very impressive!
Side point: Today’s toilet tissue rolls are smaller in diameter than rolls of toilet tissue years ago. Since more tissue makes a complete circle on the outer part of the roll, decreasing the diameter only slightly means appreciably less tissue per roll.
While we’re on the subject of circles, you might ask why manhole covers are round (asked in a Check Yourself question). The answer is so some moron type doesn’t drop them accidentally into the manhole. If they were square, they could be tipped up on edge and dropped through the hole on the diagonal. Similarly with ovals. But a circular hole will defy the most determined efforts. Of course there is a lip around the inside of the manhole that cover rests on, making the diameter of the hole somewhat less than the diameter of the cover.
Rotational Inertia: Compare the idea of inertia and its role in linear motion to rotational inertia (moment of inertia) in rotational motion. The difference between the two involves the role of distance from a rotational axis. The greater the distance of mass concentration, the greater the resistance to rotation. Discuss the role of the pole for the tightrope walker in Figure 8.9. A novice tightrope walker might begin with the ends of the pole in supporting slots, similar to the training wheels on a beginner’s bicycle. If the pole has adequate rotational inertia, the slots mainly provide psychological comfort as well as actual safety. Just as the training wheels could be safely removed without the rider’s knowledge, the slots could be safely removed without the walker’s knowledge.
Show how a longer pendulum has a greater period and relate this to the different strides of long and short legged people. Imitate these strides yourself—or at least with your fingers walking across the desk.
DEMONSTRATION: This is a good one. Have two 1 meter pipes, one with two lead plugs in the center, the other with plugs in each end. They appear identical. Weigh both to show the same weight. Give one to a student (with plugs in ends) and ask her to rotate it about its center (like in Figure 8.8). Have another student do the same with the pipe that has the plugs in the middle. Then have them switch. Good fun. Then ask for speculations as to why one was noticeably more difficult to rotate than the other.
DEMONSTRATION: As in Check Yourself question in the chapter, have students try to balance on a finger a long stick with a massive lead weight at one end. Try it first with the weight at the fingertip, then with the weight at the top. Or you can use a broom, or long-handled hammer. Relate this to the ease with which a circus performer balances a pole full of people doing acrobatics, and cite how much more difficult it would be for the performer to balance an empty pole!
Also relate this demonstration and the continued adjustments you have to execute to keep the object balanced to the similar adjustments that must be made in keeping a rocket vertical when it is first fired. Amazing! As Lillian demonstrates in Figure 8.35 on page 143, the Segway Transporter employs the same physics. The Segway behave as we do—when it leans forward it increases its speed to keep its CG above a point of support.
DEMONSTRATION: Fasten a mass to the end of a meterstick. A blob of clay works fine. Set it on end on your lecture table, along with another meterstick with no attached mass. When you let go of the sticks, they’ll topple to the tabletop. Ask which stick will reach the tabletop first. [The plain stick wins due to the greater rotational inertia of the clay-top stick. There is more to this than simply greater rotational inertia, for torque is increased as well. If the clay is located at the middle of the stick, the effects of greater torque and greater rotational inertia balance each other and both sticks fall together.]
Discuss the variety of rotational inertias shown in Figure 8.14. Stress the formulas are for comparison, and point out why the same formula applies to the pendulum and the hoop (all the mass of each is at the same distance from the rotational axis). State how reasonable the smaller value for a solid disk is, given that much of its mass is close to the rotational axis.
DEMONSTRATION: Place a hoop and disk at the top of an incline and ask which will have the greater acceleration down the incline. Do not release the hoop and disk until students have discussed this with their neighbors. Try other shapes after your class makes reasoned estimates.
Center of Mass and Torque: Depart a bit from the order of the chapter and begin a discussion of center of mass before treating torque. Do this by tossing a small metal ball across the room, stating it follows a smooth curved path—a parabola. Then pick up an irregularly shaped piece of wood, perhaps an L-shape, and state that if this were thrown across the room it would not follow a smooth path, that it would wobble all over the place—but a special place, the place being discussed today—the center of mass, or center of gravity. Illustrate your definition with figures of different shapes, first those where the center of mass lies within the object and then to shapes where the center of mass lies outside the objects.
CHECK QUESTION: Where is the center of mass of a donut?
Consider the motion of a basketball tossed across the room when a heavy weight is attached to one side. The wobble is evident. Likewise for suns with planet, presently hot news to astronomy types.
Ask your students if they “have” a CG. Acknowledge that the CG in men is generally higher than in women (1%-2%), mainly because women tend to be proportionally smaller in the upper body, and heavier in the pelvis. On the average it lies about 6 inches above the crotch, a bit below the bellybutton. Interestingly enough, the reason for the bellybutton being where it is relates to CG. A baseball bat makes this point. Standing erect with heavy side down simulates an average woman. Standing with heavy side up, simulates an average man. Interestingly, a fetus turning in its mother’s womb would rotate about its CG, the likely place for its umbilical cord. When we bend over, of course, the CG extends beyond the physical body.
Pass around a meterstick with a weight that can be suspended at different places. This is “Torque Feeler,” as shown by Mary Beth Monroe in the photo chapter opener. It is an important activity that can be done in your classroom. Students hold the stick horizontally and note that different torques when the weight’s distance is varied. The difference between force and torque is felt!
Place an L-shaped body on the table and show how it topples—because its center of mass lies outside a point of support. Sketch this on the board. Then stand against a wall and ask if it is possible for one to bend over and touch their toes without toppling forward. Attempt to do so. Sketch this next to the L-shape as shown. By now your chalkboard looks like the following.
Discuss a remedy for such toppling, like longer shoes or the wearing of snowshoes or skis. Sketch a pair of skis on the feet of the person in your drawing. Seem to change the subject and ask why a pregnant woman often gets back pains. Sketch a woman before and after getting pregnant, showing how the CG shifts forward—beyond a point of support for the same posture. (This whole idea goes over much better in lecture than as reading material, so is not found in this edition. So now you can introduce it as a fresh idea in class.) Make a third sketch showing how a woman can adjust her posture so that her CG is above the support base bounded by her feet, sketching lastly, the “marks of pain.” Ask the class how she could prevent these pains, and if someone in class doesn’t volunteer the idea of wearing skis, do so yourself and sketch skis on her feet in the second drawing.
Lead your class into an alternate solution, that of carrying a pole on her shoulder, near the end of which is a load. Erase the skis and sketch in the pole and load as shown. Acknowledge the objection that she would have to increase the mass of the load as the months go by, and ask what else can be done. Someone should volunteer that she need only move the load closer to the end, which in effect shifts the overall CG in a favorable direction. This routine is effective and sparks much class interest. However, you must be very careful that you don’t offend your students, particularly your female students. Whenever you single out any “minority”(?) you run the risk of offending members of that minority group or those sensitive to the feelings of members of that group. We instructors, whether male or female ourselves, are for the most part conscious of this and therefore make our examples as general as possible—mixing “shes” and “hes” whenever these pronouns come up. But in the case of a person becoming pregnant, it’s a definite “she.” Any classroom laughter that your presentation elicits should be, after all, directed to the situation and not particularly toward the woman. In any event, we are in sad shape when we cannot laugh at ourselves occasionally.
CHECK QUESTION: Why does a hiker with a heavy backpack lean forward when standing or walking?
Return to your chalkboard sketches of L-shaped objects and relate their tipping to the torques that exist. Point out the lever arms in the sketches.CHECK QUESTION: An L-shaped object with CG marked by the X rests on a hill as shown. Draw this on your paper and mark it appropriately to determine whether the object will topple or not.
Comment: Be prepared for some students to sketch in the “vertical” line through the CG perpendicular to the slope as shown.
A simple example of this is to balance a pipe (smoking kind) on your hand when held at an angle.
Cite examples involving the CG in animals and people—how the long tails of monkeys enable them to lean forward without losing balance—and how people lean backwards when carrying a heavy load at their chests, and how the coolie method with the load distributed in two parts suspended at the ends of a pole supported in the middle is a better way.
Ask why a ball rolls down a hill. State that “because of gravity” is an incomplete answer. Gravity would have it slide down the hill. The fact it rolls, or rotates, is evidence of an unbalanced torque. Sketch this on your chalkboard.
DEMONSTRATION: Show how a “loaded disk” rolls up an inclined plane. After class speculation, show how the disk remains at rest on the incline. Modify your chalkboard sketch to show how both the CG with respect to the support point is altered, and the absence of a lever arm and therefore the absence of a torque.
On rolling: Cliff birds lay eggs that are somewhat pear-shaped. This shape assures that the eggs roll in circles, and don’t easily roll off precarious nesting places.
Discuss wrenches and clarify lever arm distances. Cite how a steering wheel is simply a modified wrench, and why trucks and heavy vehicles without power steering use large-diameter steering wheels.DEMONSTRATION: Attempt to stand from a seated position without putting your feet under the chair. Explain with center of gravity and torques.
DEMONSTRATION: Do as Michael Bimmerle does and stick a piece of masking tape on an easy-to-move door. Place the tape near the middle and when you pull the door, the tape becomes unstuck. Progressively move the tape closer toward the edge away from the hinges and the tape sticks better. Near the edge the tape will stick and open the door without pulling off. More torque for less force.
Interestingly enough, with regard to Figure 8.29, engineers recently stacked 400 tons of lead ingots on the base of the Leaning Tower
Seesaws: Extend rotation to seesaws, like in Figures 8.18 and 8.19.
DEMONSTRATION: Make a candle seesaw by trimming the wick so both ends are exposed, and balance the candle by a needle through the center. Rest the ends of the needle on a pair of drinking glasses. Light both ends of the candle. As the wax drips, the CG shifts, causing the candle to oscillate.
Explain how participants on a seesaw can vary the net torque by not only sliding back and forth, but by leaning. In this way the location of their CGs and hence the lever arm distance is changed. Discuss the boy playing by himself in the park (Exercise 30), and how he is able to rotate up and down by leaning toward and away from the fulcrum.
CHECK QUESTION: To balance a horizontal meterstick on one finger, you’d place your finger at the 50-cm mark. Suppose you suspend an identical meterstick vertically from one end, say the 0-cm end. Where would you place your finger to balance the horizontal stick? [At the 25-cm mark, where equal weights would each be 25 cm distant.]
DEMONSTRATION: Place a heavy plank on your lecture table so that it overhangs. Walk
out on the overhanging part and ask why you don’t topple. Relate this to the solitary seesaw example. (Note the version of this in the Next-Time Questions.)
Here’s a neat application of CG that is not in the text, but is another NTQ. If you gently shake a basket of berries, the larger berries will make their way to the top. In so doing the CG is lowered by the more compact smaller berries settling to the bottom. You can demonstrate this with a Ping-Pong ball at the bottom of a container of dried beans, peas, or smaller objects. When the container is shaken, the Ping-Pong ball surfaces, lowering the CG of the system. This idea can be extended to the Ping-Pong ball in a glass of water. The CG of the system is lowest when the Ping-Pong ball floats. Push it under the surface and the CG is raised. If you do the same with something more dense than water, the CG is lowest when it is at the bottom.
Centripetal Force: Whirl an object tied to the end of a string overhead and ask if there is an outward or an inward force exerted on the whirling object. Explain how no outward or centrifugal force acts on the whirling object (the only outward directed force is the reaction force on the string, but not on the object). Emphasize also that centripetal force is not a force in its own right, like gravity, but is the name for any force that pulls an object into a curved path.DEMONSTRATION: Swing a bucket of water in a vertical circle and show that the water doesn’t spill (when centripetal force is at least equal to the weight of the water). All your students have heard of this demonstration, but only a few have actually seen it done. Why doesn’t the water fall at the top of the path? The answer is intriguing—it does! You have to pull the bucket down as fast as the water falls. Similarly, a space shuttle above falls—just as much as the round Earth curves! Nothing holds the water up; nothing holds the satellite up. Both are falling—nice physics!
The water doesn’t spill at the top when the centripetal force is at least equal to the weight of the water. But before invoking this explanation, ask why the water doesn’t fall. After some thought state that your question is misleading, for the water indeed does fall! The “trick” is to pull the bucket down as fast as the water falls so both fall the same vertical distance in the same time. Too slow a swing produces a wet teacher. Interestingly enough, the water in the swinging bucket is analogous to the orbiting of a satellite. Both the swinging water and a satellite such as the orbiting space shuttle are falling. Because of their tangential velocities, they fall in a curve; just the right speed for the water in the bucket, and just the right greater speed for the space shuttle. Tying these related ideas together is good teaching!
CHECK QUESTION: A motorcycle runs on the inside of a bowl-shaped track (see sketch in Exercise 49). Is the force that holds the motorcycle in a circular path an inward- or outward-directed force? [It is an inward-directed force—a centripetal force. An outward-directed force acts on the inner wall, which may bulge as a result, but no outward-directed force acts on the motorcycle.]
centrifugal Force in a Rotating Frame: The concept of centrifugal force is useful when viewed from a rotating frame of reference. Then it seems as real as gravity to an occupant—like inside a rotating space habitat. State how it differs from a real force in that there is no agent such as mass. The magnetic force on a magnet, for example, is caused by the presence of another magnet; the force on a charge is caused by the presence of another charge. Whereas a real force is an interaction between one body and another, there is no reaction counterpart to centrifugal force. Distinguish centrifugal force from the action-reaction pairs of forces at the feet of an astronaut in a rotating habitat.
Discuss rotating space habitats. Show how g varies with the radial distance from the hub, and with the rotational rate of the structure. The Earth has been the cradle of humankind; but humans do not live in the cradle forever. We will likely leave our cradle and inhabit structures of our own building; structures that will serve as lifeboats for the planet Earth. Their prospect is exciting.
Angular momentum: Just as inertia and rotational inertia differed by a radial distance, and just as force and torque also differed by a radial distance, so momentum and angular momentum differ by a radial distance. Relate linear momentum to angular momentum for the case of a small mass at a relatively large radial distance—the object you swung overhead.
For the more general case, angular momentum is simply the product of rotational inertia I and angular velocity w. This is indicated in Figure 8.53.
DEMONSTRATION: With weights in your hand, rotate on a platform as shown in Figure 8.53. Simulate the slowing down of the Earth when ice caps melt and spread out.
DEMONSTRATION: Show the operation of a gyroscope—either a model or a rotating bicycle wheel as my late son James demonstrates in Figure 8.51.
Regarding Exercises 52 through 56: When answering these, demonstrate again on the rotating platform, holding the weights over your head to simulate Earth washing toward the equator, melting ice caps spreading toward the equator by lowering your hands in an outstretched position to simulate Earth and water flowing toward the equator. To simulate the effects of skyscraper construction, hold the weights short of fully stretched, then extend your arms full-length.
Going Further with Rolling: Rolling things have two kinds of kinetic energy: That due to linear motion, and that due to rotational motion. So an object rolling down an incline will lag behind a freely sliding object because a part of a rolling object’s kinetic energy is in rotation. If this is clear, the following question is in order for your better students.
NEXT-TIME QUESTION: Which will roll with the greater acceleration down an incline, a can of water or a frozen can of ice? Double credit for a good explanation of what is seen. [The can of liquid will undergo appreciably more acceleration because the liquid is not made to rotate with the rotating can. It in effect “slides” rather than rolls down the incline, so practically all the KE at the bottom is in linear speed with next-to none in rotation. Fine, one might say, then if the liquid doesn’t rotate, the can ought to behave as an empty can, with the larger rotational inertia of a “hoop” and lag behind. This brings up an interesting point: The issue is not which can has the greater rotational inertia, but which has the greater rotational inertia compared to its mass (note the qualifier in the legend of Figure 8.13). The liquid content has appreciably more mass than the can that contains it; hence the nonrolling liquid serves to increase the mass of the can without contributing to its rotational inertia. It gives the can of liquid a relatively small rotational inertia compared to its mass.]
You can follow through by asking which can will be first in rolling to a stop once they meet a horizontal surface. The can hardest to “get going” is also the can hardest to stop—so given enough horizontal distance, the slowest can down the incline rolls farther and wins the race!
CHALLENGE: At the bottom of an incline are two balls of equal mass—a solid one and a thin-walled hollow one. Each is given the same initial speed. Which rolls higher up the incline before coming to a stop? The answer is the hollow ball. In terms of rotational inertia, whether a ball is hollow or solid makes a big difference. A thin-walled hollow ball, having much of its mass along its radius, has a relatively large rotational inertia (2/3 MR2). A solid ball, having much of its mass near its center, has less rotational inertia (2/5 MR2). The ball with the greater rotational inertia out-rolls a lower-inertia ball. Hence the hollow ball rolls farther up the incline before it comes to a stop.
Another way to look at it is in terms of energy. The balls begin their upward travel with kinetic energy of two kinds—translational and rotational. Although their initial translational KEs are the same, the hollow ball begins with more rotational KE due to its greater rotational inertia. So the hollow ball has more total KE at the base of the incline, which means it must have more PE at the top. The hollow ball indeed goes higher.
Does mass make a difference? No. As with the mass of a pendulum bob, or the mass of a freely-falling object, mass makes no difference. Sent rolling up an incline with equal speeds, any hollow ball will out-roll any solid ball. That’s right—a tennis ball will roll higher than a bowling ball or a marble.
If you instead release both balls from a rest position at the top of an incline, the hollow ball “out-rests” the solid ball, is slower to gain speed, and lags behind the solid ball. The solid ball gets to the bottom first. Inertia is a resistance to change.
Solutions to Chapter 8 Exercises
1. Tape moves faster when r is greater, in accord with v = rw. So the reel with the most-filled reel corresponds to the tape with the greatest linear speed v.
2. The tangential speeds are equal, for they have the same speed as the belt. The smaller wheel rotates twice as fast because for the same tangential speed, and r half, w must be twice. v(big wheel) = rw; v(small wheel) = (r/2 ´ 2 w).
3. Large diameter tires mean you travel farther with each revolution of the tire. So you’ll be moving faster than your speedometer indicates. (A speedometer actually measures the RPM of the wheels and displays this as mi/h or km/h. The conversion from RPM to the mi/h or km/h reading assumes the wheels are a certain size.) Oversize wheels give too low a reading because they really travel farther per revolution than the speedometer indicates, and undersize wheels give too high a reading because the wheels do not go as far per revolution.
4. Sue’s tires have a greater rotational speed for they have to turn more times to cover the same distance.
5. The amount of taper is related to the amount of curve the railroad tracks take. On a curve where the outermost track is say 10% longer than the inner track, the wide part of the wheel will also have to be at least 10% wider than the narrow part. If it’s less than this, the outer wheel will rely on the rim to stay on the track, and scraping will occur as the train makes the curve. The “sharper” the curve, the more the taper needs to be on the wheels.
6. For the same twisting speed w, the greater distance r means a much greater speed v.
7. The CM is directly above the bird’s foot.
8. Two conditions are necessary for mechanical equilibrium, SF = 0 and ST = 0.
9. Rotational inertia and torque are most predominantly illustrated here, and the conservation of angular momentum also plays a role. The long distance to the front wheels increases the rotational inertia of the vehicle relative to the back wheels and also increases the lever arm of the front wheels without appreciably adding to the vehicle’s weight. As the back wheels are driven clockwise, the chassis tends to rotate counterclockwise (conservation of angular momentum) and thereby lift the front wheels off the ground. The greater rotational inertia and the increased clockwise torque of the more distant front wheels counter this effect.
Before leaving the cliff, front and back wheels provide the support base to support the car’s weight. The car’s center of mass is well within this support base. But when the car drives off the cliff, the front wheels are the first to leave the surface. This reduces the support base to the region between the rear wheels, so the car tips forward. In terms of torques, before driving off the cliff, the torques are balanced about the center of mass produced by the support forces at front and back wheels. But when the support force of the front wheels is absent, torque due to the support force of the rear wheels rotates the car forward about its center of mass making it nose forward as shown.
11. Friction by the road on the tires produces a torque about the car’s CM. When the car accelerates forward, the friction force points forward and rotates the car upward. When braking, the direction of friction is rearward, and the torque rotates the car in the opposite direction so the rear end rotates upward (and the nose downward).
12. The bowling ball wins. A solid sphere of any mass and size beats both a solid cylinder and a hollow ball of any mass and size. That’s because a solid sphere has less rotational inertia per mass than the other shapes. A solid sphere has the bulk of its mass nearer the rotational axis that extends through its center of mass, whereas a cylinder or hollow ball has more of its mass farther from the axis. The object with the least rotational inertia per mass is the “least lazy” and will win races.
13. The ball to reach the bottom first is the one with the least rotational inertia compared with its mass—that’s the hollow basketball.
14. If you roll them down an incline, the solid ball will roll faster. (The hollow ball has more rotational inertia compared with its weight.)
15. Don’t say the same, for the water slides inside the can while the ice is made to roll along with the can. When the water inside slides, it contributes weight rather than rotational inertia to the can. So the can of water will roll faster. (It will even beat a hollow can.)
16. Lightweight tires have less rotational inertia, and easier to get up to speed.
17. Advise the youngster to use wheels with the least rotational inertia—lightweight solid ones without spokes.
18. The lever arm is the same whether a person stands, sits, or hangs from the end of the seesaw, and certainly the person’s weight is the same. So the net torque is the same also.
19. In the horizontal position the lever arm equals the length of the sprocket arm, but in the vertical position, the lever arm is zero because the line of action of forces passes right through the axis of rotation. (With cycling cleats, a cyclist pedals in a circle, which means they push their feet over the top of the spoke and pull around the bottom and even pull up on the recovery. This allows torque to be applied over a greater portion of the revolution.)
20. No, for by definition, a torque requires both force and a lever arm.
21. No, because there is zero lever arm about the CM. Zero lever arm means zero torque.
22. In all three cases the spool moves to the right. In (a) there is a torque about the point of contact with the table that rotates the spool clockwise, so the spool rolls to the right. In (b) the pull’s line of action extends through (not about) the point of table contact, yielding no lever arm and therefore no torque; but with a force component to the right; hence the spool slides to the right without rolling. In (c) the torque produces clockwise rotation so the spool rolls to the right.
23. Friction between the ball and the lane provides a torque, which rotates the ball.
24. A rocking bus partially rotates about its center of mass, which is near its middle. The farther one sits from the center of mass, the greater is the up and down motion—as on a seesaw. Likewise for motion of a ship in choppy water or an airplane in turbulent air.
25. With your legs straight out, your CG is farther away and you exert more torque sitting up. So sit-ups are more difficult with legs straight out.
26. The long drooping pole lowers the CG of the balanced system—the tightrope walker and the pole. The rotational inertia of the pole contributes to the stability of the system also.
27. The wobbly motion of a star is an indication that it is revolving about a center of mass that is not at its geometric center, implying that there is some other mass nearby to pull the center of mass away from the star’s center. This is the way in which astronomers have discovered that planets exist around stars other than our own.
28. You bend forward when carrying a heavy load on your back to shift the CG of you and your load above the area bounded by your feet—otherwise you topple backward.
29. Two buckets are easier because you may stand upright while carrying a bucket in each hand. With two buckets, the CG will be in the center of the support base provided by your feet, so there is no need to lean. (The same can be accomplished by carrying a single bucket on your head.)
30. The weight of the boy is counterbalanced by the weight of the board, which can be considered to be concentrated at its CG on the opposite side of the fulcrum. He is in balance when his weight multiplied by his distance from the fulcrum is equal to the weight of the entire board multiplied by the distance between the fulcrum and the midpoint (CG) of the board. (How do the relative weight of boy and board relate to the relative lever arms?)
31. The CG of a ball is not above a point of support when the ball is on an incline. The weight of the ball therefore acts at some distance from the point of support which behaves like a fulcrum. A torque is produced and the ball rotates. This is why a ball rolls down a hill.
32. The top brick would overhang 3/4 of a brick length as shown. This is best explained by considering the top brick and moving downward; i.e., the CG of the top brick is at its midpoint; the CG of the top two bricks is midway between their combined length. Inspection will show that this is 1/4 of a brick length, the overhang of the middle brick. (Interestingly, with a few more bricks, the overhang can be greater than a brick length, and with a limitless number of bricks, the overhang can be made as large as you like.)
33. The Earth’s atmosphere is a nearly spherical shell, which like a basketball, has its center of mass at its center, i.e., at the center of the Earth.
34. It is dangerous to pull open the upper drawers of a fully-loaded file cabinet that is not secured to the floor because the CG of the cabinet can easily be shifted beyond the support base of the cabinet. When this happens, the torque that is produced causes the cabinet to topple over.
35. An object is stable when its PE must be raised in order to tip it over, or equivalently, when its PE must be increased before it can topple. By inspection, the first cylinder undergoes the least change in PE compared to its weight in tipping. This is because of its narrow base.
36. The CG of Truck 1 is not above its support base; the CGs of Trucks 2 and 3 are above their support bases. Therefore, only Truck 1 will tip.
37. The track will remain in equilibrium as the balls roll outward. This is because the CG of the system remains over the fulcrum. For example, suppose the billiard ball has twice the mass of the golf ball. By conservation of momentum, the twice-as-massive ball will roll outward at half the speed of the lighter ball, and at any time be half as far from the starting point as the lighter ball. So there is no CG change in the system of the two balls. We can see also that the torques produced by the weights of the balls multiplied by their relative distances from the fulcrum are equal at all points—because at any time the less massive ball has a correspondingly larger lever arm.
38. The equator has a greater tangential speed than latitudes north or south. When a projectile is launched from any latitude, the tangential speed of the Earth is imparted to the projectile, and unless corrections are made, the projectile will miss a target that travels with the Earth at a different tangential speed. For example, if the rocket is fired south from the Canadian border toward the Mexican border, its Canadian component of speed due to the Earth’s turning is smaller than Earth’s tangential speed further south. The Mexican border is moving faster and the rocket falls behind. Since the Earth turns toward the east, the rocket lands west of its intended longitude. (On a merry-go-round, try tossing a ball back and forth with your friends. The name for this alteration due to rotation is the Coriolis effect.)
39. In accord with the equation for centripetal force, twice the speed corresponds to four times the force.
40. No—in accord with Newton
41. Newton Newton Newton
42. Centripetal force is supplied by the component of the normal force that lies parallel to the radial direction.
43. Yes, in accord with Fc = mv2/r. Force is directly proportional to the square of speed.
44. There is no component of force parallel to the direction of motion, which work requires.
45. Centripetal force will be adequate when only the radial component of the normal force equals the mass of the car times the square of its speed divided by an appropriate radial distance from the center of the curve.
46. In accord with Newton Newton
. The resultant is a centripetal force.
49. (a) Except for the vertical force of friction, no other vertical force except the weight of the motorcycle + rider exists. Since there is no change of motion in the vertical direction, the force of friction must be equal and opposite to the weight of motorcycle + rider.
(b) The horizontal vector indeed represents the normal force. Since it is the only force acting in the radial direction, horizontally, it is also the centripetal force. So it’s both.
51. The rotational inertia of you and the rotating turntable is least when you are at the rotational axis. As you crawl outward, the rotational inertia of the system increases (like the masses held outward in Figure 8.53). In accord with the conservation of angular momentum, as you crawl toward the outer rim, you increase the rotational inertia of the spinning system and decrease the angular speed. You can also see that if you don’t slip as you crawl out, you exert a friction force on the turntable opposite to its direction of rotation, thereby slowing it down.
52. Soil that washed down the river is being deposited at a greater distance from the Earth’s rotational axis. Just as the man on the turntable slows down when one of the masses is extended, the Earth slows down in its rotational motion, extending the length of the day. The amount of slowing, of course, is exceedingly small. (Interestingly, the construction of many dams in the Northern Hemisphere has the opposite effect; shortening our days!)
53. In accord with the conservation of angular momentum, as the radial distance of mass increases, the angular speed decreases. The mass of material used to construct skyscrapers is lifted, slightly increasing the radial distance from the Earth’s spin axis. This would tend to slightly decrease the Earth’s rate of rotation, which in turn tends to make the days a bit longer. The opposite effect occurs for falling leaves, as their radial distance from the Earth’s axis decreases. As a practical matter, these effects are entirely negligible!
54. Rotational inertia would increase. By angular momentum conservation, the rotation of the Earth would decrease (as a skater spins slower with arms outstretched), making a longer day.
55. In accord with the conservation of angular momentum, if mass moves closer to the axis of rotation, rotational speed increases. So the day would be ever so slightly shorter.
56. In accord with the conservation of angular momentum, if mass moves farther from the axis of rotation, rotational speed decreases. So the Earth would slow in its daily rotation.
57. The angular momentum of the wheel-train system will not change in the absence of an external torque. So when the train moves clockwise, the wheel moves counterclockwise with an equal and opposite angular momentum. When the train stops, the wheel stops. When the train backs up, the wheel moves clockwise. If masses of the train and wheel are equal, they will move with equal speeds (since the mass of the wheel is as far from the axis as the mass of the train—equal masses at equal radial distances having equal rotational inertias). If the train is more massive than the wheel, the wheel will “recoil” with more speed than the train, and vice versa. (This is a favorite demonstration of Paul Robinson, whose children David and Kristen are shown in the photo.)
58. Without the small rotor on its tail, the helicopter and the main rotor would rotate in opposite directions. The small rotor provides a torque to offset the rotational motion that the helicopter would otherwise have.
59. Gravitational force acting on every particle by every other particle causes the cloud to condense. The decreased radius of the cloud is then accompanied by an increased angular speed because of angular momentum conservation. The increased speed results in many stars being thrown out into a dish-like shape.
60. In accord with Newton
Chapter 8 Problem Solutions
1. Since the bicycle moves 2 m with each turn of the wheel, and the wheel turns once each second, the linear speed of the bicycle is 2 m/s.
2. The linear speed, more correctly, tangential speed distance/time, will be the circumference of the Ferris wheel divided by the time for one revolution, 30 s. From geometry, the circumference = 2πr = 2(3.14)(10 m) = 62.8 m. So the linear speed v = (62.8 m)/(30 s) = 2.1 m/s.
3. The center of mass of the two weights is where a fulcrum would balance both—where the torques about the fulcrum would balance to zero. Call the distance (lever arm) from the 1-kg weight to the fulcrum x. Then the distance (lever arm) from the fulcrum to the 3-kg weight is (100 – x). Equating torques:
1x = (100 – x)3
x = 300 – 3x
x = 75.
So the center of mass of the system is just below the 75-cm mark. Then the three-times-as-massive weight is one-third as far from the fulcrum.
4. 7500 N at the left and 2500 N at the right. Since the distances from vehicle to supporting ends take the ratio 3/1, the support forces will have the inverse ratio; 1/3. That is, the left side will support 3 times as much as the farther right side. (Since the bridge is not set into rotation by the shared weight of the vehicle, the torques supplied by these forces must be equal. That is, Fd = fD, where the torque at the left is F times distance d from the vehicle, and the torque at the right is f times longer distance D from the vehicle. We consider torques about the CG of the vehicle, although torques about any point will produce the same result.) Note that F + f = 10,000 N, so f = (10,000 N - F), and distance D = 3d. Then Fd = (10,000 - F)3d, where F = 7500 N. This means f = 2500 N.
5. The mass of the stick is 1 kg. (This is a “freebie”; see Check Yourself question and answer in the chapter!)
6. (a) Torque = force ´lever arm = (80 N)(0.25 m) = 20 N m.
(b) Force = 200 N. Then (200 N)(0.10 m) = 20 N m.
(c) Yes. These answers assume that you are pushing perpendicular to the wrench handle. Otherwise, you would need to exert more force to get the same torque.
7. Centripetal force (and “weight” and “g” in the rotating habitat) is directly proportional to radial distance from the hub. At half the radial distance, the g force will be half that at his feet. The man will literally be “light-headed.” (Gravitational variations of greater than 10% head-to-toe are uncomfortable for most people.)
8. This follows the preceding answer. If one’s feet at full radius R undergo a centripetal acceleration of g, then the fractions of g closer to the hub will be in direct proportion to the fractions of R. At a distance from the hub of R/2, we get g/2, at 3R/4 from the hub, we get 3g/4, and so forth. Some thought will show that the fraction by which we decrease R is matched by an equal decrease of g. So for a decrease of 1/100 for g, we must move 1/100 closer to the hub. This means the hub would have to be 1.01 times as far from the feet as from the head. The radius of the habitat would have to be 100 times a person’s height.
9. The artist will rotate 3 times per second. By the conservation of angular momentum, the artist will increase rotation rate by 3. That is
Iwbefore = Iwafter
Iwbefore = [(I/3)I(3w)]after
10. From data listed on the inside back cover, by ratio;
=
=
where speed V of Earth about Sun is its circumference divided by the time for one orbit, 2πR/12 months, and speed v of Moon about the Earth is 2πr/1 month. Mass M of Earth is 6 ´1024 kg, and mass m of Moon is 7.4 ´1022 kg. Earth-Sun distance R is 1.5 ´1011 m, and Earth-Moon distance r is 3.8 ´108 m. Then
≈ ≈ 106.
So the Earth has a million times as much angular momentum about the Sun as the Moon has about the Earth.
