Rotational Equilibrium and Rotational Dynamics

Transcription

1 8 Rotational Equilibrium and Rotational Dynamics Description: Reasoning with rotational inertia. Question CLICKER QUESTIONS Question E.0 The rotational inertia o the dumbbell (see igure) about axis A is twice the rotational inertia about axis B. The unknown mass is: 0 cm 0 cm 0 cm kg? A B. 4 7 kg. kg 3. 4 kg 4. 5 kg 5. 7 kg 6. 8 kg 7. 0 kg 8. None o the above 9. Cannot be determined 0. The rotational inertia cannot be dierent about dierent axes. Commentary Purpose: To practice problem solving with rotational motion ideas. Discussion: Rotational inertia is not an intrinsic property o an object; its value depends on the axis about which it is calculated. In this case, the object s rotational inertia is twice as large about one axis as another. Let s avoid numerical computations, even though we are given speciic values or all known quantities. Let d 0 cm, m kg, and M the unknown mass. We ll solve or M in terms o the given quantities. The rotational inertia about axis A is I m d + A M ( d ) m + 4 M d. The rotational inertia about axis B is I m d Md 4 m M d. (We are treating each dumbbell as a point mass.) + ( + ) B 369

2 370 Chapter 8 We are told that the rotational inertia about axis A is twice as large as that about axis B (I A I B ), so m + 4M (4m + M ). (Note that d has cancelled out, and does not aect the answer.) Thereore, M 7 m. Since m kg, M 7 kg. Key Points: An object's rotational inertia ( moment o inertia ) depends on the axis you choose to calculate it about. The rotational o inertia o a point-like object about an axis is md, where m is the object s mass and d is its distance rom the axis. The rotational o inertia o an object composed o multiple subobjects is just the sum o their individual rotational inertias (about the same axis). Avoid putting numbers into calculations until the very end. Instead, deine variables and work with those. (Oten, some will cancel out, simpliying your calculations.) For Instructors Only Students can get bogged down in unnecessary calculations and computations. This problem presents a good opportunity to discuss problem solving procedures. Many students will translate the given relationship between rotational inertias incorrectly, and will interpret The rotational inertia... about axis A is twice the rotational inertia about axis B as I A I B. Those who do this will get M 4 7 kg. (This is a well-documented error in translating verbal to algebraic representations.) Students should be encouraged to decide on qualitative grounds which mass is larger, so that they can check their answers or reasonability. Question E.0a Description: Integrating linear and rotational dynamics ideas. Question A disk, with radius 0.5 m and mass 4 kg, lies lat on a smooth horizontal tabletop. A string wound about the disk is pulled with a orce o 8 N. What is the acceleration o the disk? R F M ms 3. ms 4. ms 5. 4 ms 6. None o the above 7. Cannot be determined

3 Rotational Equilibrium and Rotational Dynamics 37 Commentary Purpose: To elicit and conront the common misconception that Newton s second law is somehow invalid or objects that rotate as well as translate, or or orces that are are also torques. Discussion: In the horizontal plane, the only orce acting on the disk is the tension F rom the string. (The vertical orces, gravity and the normal orce o the table, balance.) Thus, according to Newton s second law, a F m m s. Yes, it s really that easy. Newton s second law is always true, whether or not rotation occurs as well as translation. I the orce F were applied at the center o the disk, or anywhere else, the acceleration would still be ms. Where the orce is applied will aect how the disk rotates, but not how its center o mass accelerates. You may have diiculties with this idea. Intuitively, it may seem to you that part o the orce is going towards making the object rotate, so that not as much is available to cause acceleration. This kind o thinking is more appropriate or quantities such as work and energy, impulse and momentum that have conservation principles. Forces don t get used up. It can be helpul to compare this situation to one in which all given values are the same, but the string is attached to the center o the disk rather than wound around the edge. Both disks experience the same net orce, so both will have the same acceleration. However, the disk with the string wrapped around it will also have a nonzero angular acceleration, whereas the disk with the string attached to the center will not. I this seems to be getting something or nothing, consider that more work will be done in the irst case, explaining the act that it ends up with kinetic energy due to both translation and rotation. Key Points: The acceleration o a body as a whole depends on the net orce acting on the body, period. It does not depend on where on the body the orce acts or whether the body spins in addition to accelerating. Newton s second law is true or bodies that spin or rotate as well as those that don t. F ma and τ Iα are both true, always. For Instructors Only This is the irst in a two-question set exploring linear and angular acceleration rom a orce that exerts a torque. The most prevalent misunderstanding to conront here is students belie that somehow τ I α replaces F ma, rather than augmenting it. Students may assume that the question asks or angular acceleration. They should be cautioned against jumping to conclusions based on the supericial eatures o a question!

4 37 Chapter 8 Question E.0b Description: Integrating linear and rotational dynamics ideas. Question A disk, with radius 0.5 m and mass 4 kg, lies lat on a smooth horizontal tabletop. A string wound about the disk is pulled with a orce o 8 N. What is the angular acceleration o the disk? R F M rad s 3. 8 rad s 4. 4 rad s 5. rad s 6. None o the above 7. Cannot be determined Commentary Purpose: To explore the angular acceleration o an object experiencing both angular and linear acceleration due to a orce that also exerts a torque. Discussion: A orce exerted a distance R rom an axis o rotation causes a torque τ RF sin θ, where θ is the angle between the direction o the orce and the vector rom the axis to the point o application. In this case, the angle is 90, so the torque exerted about the center o mass is N m. The rotational equivalent o Newton s second law, τ I α, relates a body s angular acceleration to the net torque it experiences. The moment o inertia I o the disk is MR 05. kg m. Thereore, the angular acceleration o the disk about its center o mass is α 8 Nkgm 8 rads. (This is true even i the disk translates.) A common mistake is to calculate the angular acceleration rom the linear acceleration (ound in the previous problem) via a rα. This relationship between linear and angular acceleration is not generally true; it describes a physical ( geometric ) constraint that only applies in special circumstances, such as when a round object is rolling without slipping, or when it is rotating about a ixed axis through its center and a reers to the acceleration o a point on its rim. Key Points: τ I α describes the relationship between torque and angular acceleration the way Newton s second law describes that between orce and (translational) acceleration, and is always valid. The relationship a rα between translational and rotational acceleration is only valid in special cases. (The same is true o v rω and s rθ.) For Instructors Only This is the second in a two-question set exploring linear and angular acceleration rom a orce that exerts a torque.

5 Rotational Equilibrium and Rotational Dynamics 373 Some students will get the correct answer by misunderstanding the problem and thinking that the center o the disk is ixed in place. The question Is there riction at the pivot? indicates such a misunderstanding. Other students will give an answer less than 8 rad s, thinking that the translational motion somehow reduces the torque or its eect. Question E.03a Description: Linking orce and torque ideas in the context o mechanical advantage. Question A 00 kg crate is attached to a rope wrapped around the inner disk as shown. A person pulls on another rope wrapped around the outer disk with orce F to lit the crate. F 00 kg What orce F is needed to lit the crate m?. about 0 N. about 50 N 3. about 00 N 4. about 00 N 5. about 500 N 6. about 000 N 7. about 000 N 8. about N 9. Impossible to determine without knowing the radii 0. Impossible to determine or some other reason(s) Commentary Purpose: To link orce and torque ideas in the context o mechanical advantage. Discussion: Consider a static situation in which the crate is held motionless in the air. To support the crate, the tension in the rope attached to it must be 000 N (using g 0 N kg). The tension in the other rope is equal to F. I the disk arrangement is stationary, the torques exerted on it by the two ropes must balance. We do not know the radii o the pulleys, so let s use r or the smaller disk and R or the larger. The rope supporting the crate is tangential to its disk, so the torque exerted by this rope is 000r clockwise. The other rope is also tangential to its disk, so it exerts a torque o FR counter-clockwise. The orce o the pivot holding up the disks exerts zero torque, and we will assume the axle is rictionless.

6 374 Chapter 8 r R. In other words, the For the torques to balance each other, FR ( 000 N ) r, or F 000 N orce exerted by the person is a raction o the weight o the crate, and the raction depends on the ratio o the disk radii. We do not know the exact ratio r R, but we can estimate it rom the diagram. It looks to be about 5, so the orce needed is about 00 N. That s the orce required to hold the crate stationary in the air, or to lit or lower it with constant speed (no acceleration). In order to start the crate moving rom rest, a slightly larger orce is necessary, but it can be ininitesimally larger. (I there were riction in the pivot, the orce to get it moving would have to be enough larger to support the weight o the crate and overcome static riction.) The angle at which F is applied does not matter, as long as it acts tangentially to the disk. (It would, however, aect the orce exerted on and by the pivot axle.) Key Points: Two dierent orces can exert the same torque on an object. Mechanical advantage is gained by having the applied orce act arther rom the pivot point than the orce acting on the object to be move. The closer to the pivot point a orce acts, the larger it must be to balance other torque-causing orces. For Instructors Only This is the irst o two questions exploring the concepts o orce, torque, work, and energy in this mechanical advantage situation. Many students may say that F is impossible to determine, either because they are not given the radii and don t assume the diagram is to scale or they are not told i riction can be neglected. These are deensible responses. Other students may say that the answer is impossible to determine or another reason, such as not knowing the speed o the crate or the angle o the rope. These are not valid reasons. Some students will ignore units and treat 00 kg as the weight, and thereore say that F is about 0. Some will invert the ratio, thinking the applied orce is N, even though this does not agree with experience. Some students, not understanding mechanical advantage, will think that a orce o 000 N must be applied to lit the crate no matter what the radii are. Question E.03b Description: Linking orce, torque, work, and energy ideas in the context o mechanical advantage. Question A 00-kg crate is attached to a rope wrapped around the inner disk as shown. A person pulls on another rope wrapped around the outer disk with orce F to lit the crate.

7 Rotational Equilibrium and Rotational Dynamics 375 F 00 kg How much work is done by the person to lit the crate m?. about 400 J. slightly less than 000 J 3. exactly 000 J 4. slightly more than 000 J 5. much more than 000 J 6. Impossible to determine without knowing F 7. Impossible to determine without knowing the radii 8. Impossible to determine without knowing the mass o the pulley 9. Impossible to determine or two or more o the reasons given in 6, 7, and 8 above 0. Impossible to determine or some other reason(s) Commentary Purpose: To link orce, torque, work, and energy ideas in the context o mechanical advantage. Discussion: The only ambiguity here is whether or not to ignore rictional eects at the axle where the two disks are attached. I riction can be ignored, then energy conservation demands that the work done by the person is exactly equal to the work done on the crate, which is mg y 000 J. I riction is not ignored, then the work done by the person must be larger than the work done on the crate. Note that even though the orce F is much smaller than the weight o the crate, it acts through a much larger displacement than the crate travels. I, or example, the ratio o the disk diameters is to 5, then the orce F would be about 00 N and the displacement o the end o the rope would be about 0 m, even though the crate only moves up by m. Key Points: The presence o mechanical advantage in a system does not invalidate the work energy theorem. A smaller orce can do as much work as a larger one i it the smaller one acts through a longer distance. For Instructors Only This is the second o two questions exploring the concepts o orce, torque, work, and energy in this mechanical advantage situation. Some students will say that the work done is impossible to determine, because they are not told i riction can be neglected. This is a deensible response.

8 376 Chapter 8 Other students will say that it is impossible to determine or another reason, such as not knowing the speed o the crate, the angle o the rope, the orce F, the radii o the disks, or whether the diagram is to scale. These are not valid responses, since none o that inormation is required to answer the question. Some students will use g 9.8 N kg, and get a value slightly less than 000 J. Students who are including riction are likely to choose slightly more than 000 J, perhaps assuming that riction is small. I the coeicient o riction is large enough, a valid response would be much more than 000 J, though students might choose this response or other reasons as well. Question E.04a Description: Reasoning and problem solving with linear and rotational orms o Newtons laws in the context o rolling without slipping. Question A spool has string wrapped around its center axle and is sitting on a horizontal surace. I the string is pulled in the horizontal direction when tangent to the top o the axle, the spool will: F. Roll to the right. Not roll, only slide to the right 3. Spin and slip, without moving let or right 4. Roll to the let 5. None o the above 6. The motion cannot be determined. Commentary Purpose: To reason about a rotational system using the linear and rotational orms o Newton s second law. Discussion: There are our orces on the spool: () gravitation, down; () normal, up; (3) tension, right; and (4) riction, let or right. Gravitation is balanced by the normal orce, and their torques balance about any origin. We don t know yet which direction riction will point. First, imagine that the surace is rictionless. The net orce is then due exclusively to the tension, producing an acceleration to the right. The net torque about the spool s center is also due exclusively to the tension, producing a clockwise angular acceleration. The spool will start to move to the right and also rotate clockwise. It will be rolling to the right, and perhaps slipping at the same time. (An object only rolls without slipping when its rate o rotation and translation are just right so the contact point has zero velocity.)

9 Rotational Equilibrium and Rotational Dynamics 377 Now, add riction back in. A little bit o static riction will prevent the spool rom slipping and cause it to roll only; this is answer (). (I the contact point starts to slip, the riction orce will oppose that, exerting a countering torque.) I F is very large, the spool will not be able to roll without slipping, since the (static) riction orce has a maximum possible value. In that case, the spool will slide to the right while rotating slightly: not an available answer. Answer () is impossible, since the net torque on the spool cannot be zero. (I the spool slips at all, the torque rom the riction orce will supplement, not counteract, the torque rom the string.) Answer (3) is impossible, since the net orce on the spool cannot be zero. I the net orce were zero, the riction orce would have to point to the let and have the same magnitude as the tension. But i the spool spins, the bottom surace slides to the let and the riction orce must point to the right. Answer (4) is impossible, since it would require a net orce to the let and a counterclockwise net torque, which cannot both exist. (What direction would the riction orce point?) Key Points: When two orces are balanced (same strength, opposite directions) and colinear (having the same line o action), their torques about any origin balance. For rotational problems, τ I α and F net ma are both useul. In other words, the rotational orm o Newton s second law does not replace or supersede its linear orm. You can reason your way to answers by making assumptions and then checking or contradictions. For Instructors Only This the irst o three questions using this situation. The goal o the set is to sensitize students to the dependence o torque on the choice o origin, and to the tactic o choosing the origin so that an unknown orce exerts no torque. A powerul pattern with all three o the questions is to have students predict what they believe will happen, explain why, then observe a demonstration. This question exists largely to set up the subsequent ones. Most students will intuitively predict the correct motion. I your class generally agrees on answer (), we recommend moving on to the next question, where intuition is not so useul. Ater presenting and discussing the rest o the set, you may want to return to this one and show how easily it can be analyzed by choosing an origin at the contact point.

10 378 Chapter 8 Question E.04b Description: Reasoning and problem solving with linear and rotational orms o Newton s laws in the context o rolling without slipping. Question A spool has string wrapped around its center axle and is sitting on a horizontal surace. I the string is pulled in the horizontal direction when tangent to the bottom o the axle, the spool will: F. Roll to the right. Not roll, only slide to the right 3. Spin and slip, without moving let or right 4. Roll to the let 5. None o the above 6. The motion cannot be determined. Commentary Purpose: To explore the choice o origin and its eect on the torque. Discussion: Intuitively, it may not be obvious what will happen here. Pulling on the string seems like it might cause the spool to unwind, thus rotating counterclockwise and perhaps rolling to the let. On the other hand, the string pulls to the right, so perhaps it will cause the spool to roll to the right (clockwise) along the surace. A more careul analysis is required. There are our orces on the spool: () gravitation, down; () normal, up; (3) tension, right; and (4) riction, let or right. Gravitation is balanced by the normal orce. Because they are balanced and colinear, their torques also balance about any origin. I the spool rolls to the let without slipping, the spool s center o mass accelerates to the let. Since the tension acts to the right, the static riction orce must act to the let and must have a larger magnitude so that the net orce acts to the let. However, i that were true the net torque about the spool s center would be clockwise, causing the spool to rotate to the right. Contradiction! Since we don t know what direction the riction orce points, let s choose an origin about which the riction orce exerts no torque: the point o contact between the spool and surace. For this origin, the only orce exerting a nonzero torque is the tension orce, so the net torque is clockwise and the spool rotates to the right relative to the contact point. This means it rolls to the right. There is no reason the spool must necessarily slip or slide. I we pull gently enough, there will be enough static riction so that the spool rolls without slipping. Why doesn t the spool unroll to the let? Because although the string applies a torque in the counterclockwise direction, the static riction orce exerts a larger torque in the clockwise direction. (I we yank hard enough on the string, the spool will overcome static riction and slide to the right as it spins counterclockwise. This is not the intent, so it is not any o the answers provided.)

11 Rotational Equilibrium and Rotational Dynamics 379 Key Points: By choosing your origin careully, you can avoid dealing with torques due to an unknown orce. The torque exerted by a orce depends on the origin you calculate it about. You do not need to choose the center o the object as the origin. τ I α is true, valid, and useul in addition to, not instead o, F ma. In other words, Newton s second law in its rotational orm does not replace or supersede Newton s second law in its linear orm. Both are valid and useul, and oten both are needed to analyze a situation. For Instructors Only This the second o three questions using this situation. The goal o the set is to sensitize students to the dependence o torque on the choice o origin, and also to the strategic choice o origin to resolve conlict or inconsistency. A powerul pattern with all three o the questions is to have students predict what they believe will happen, explain why, then observe a demonstration (the predict and observe instructional tactic). As a whole, the set also demonstrates the compare and contrast tactic by varying only one eature rom question to question. This is a counterintuitive situation. Most students will have a strong intuitive belie that the spool should unroll to the let, and the demonstration will intrigue and motivate them. It is important, however, to collect explanations without comment, since most will be aulty in some way, and then show the demonstration. You can explain the result ater you have shown the demonstration. Students will have great diiculty believing you i you discuss the situation beore doing the demonstration. During the demonstration, only a gentle tug on the string is needed to cause the spool to roll to the right. Students might think that you will pull hard, which means it will move to the right while unwinding counterclockwise. This is not the intent, so they will need to answer None o the above. Most students automatically use the center o the spool as the deault origin. They might not realize that they can place the origin anywhere they want. Students oten abandon the linear orm o Newton s second law when they start learning its rotational orm. They oten think that the rotational orm supersedes and or replaces the linear orm. They are oten not given suicient opportunities to see that sometimes both are needed, and that the two relationships are completely separate and independent. Students also oten do not realize or ully appreciate that the rotational orm is actually an ininite number o relationships, one or every possible origin. (In practice, there are a limited number o independent relationships.) Even ater observing the demonstration, many students will not understand how the net orce can act to right at the same time as the net torque about the center is clockwise. The static riction orce must be only slightly smaller than the tension orce so that it can exert the larger torque about the center o the spool.

12 380 Chapter 8 Question E.04c Description: Reasoning and problem solving with linear and rotational orms o Newton s laws in the context o rolling without slipping. Question A spool has string wrapped around its center axle and is sitting on a horizontal surace. I the string is pulled at an angle to the horizontal when drawn rom the bottom o the axle, the spool will: q F. Roll to the right. Not roll, only slide to the right 3. Spin and slip, without moving let or right 4. Roll to the let 5. None o the above 6. It is impossible to determine the motion. Commentary Purpose: To explore the choice o origin and its eect on the torque. Discussion: There are our orces on the spool: () gravitation, down; () normal, up; (3) tension, as shown; and (4) riction. Gravitation and the normal orce exert zero torques about any origin along a vertical line through the center o the spool. (They no longer exert balancing torques, because they are no longer equal to each other in strength. The normal orce is smaller than the weight, because the tension orce has a component up. ) Thus, whether we choose the center o the spool or the contact point, the torques are zero. Tension exerts a counter-clockwise torque about the center o the spool. It is hard to predict what direction the riction orce will point, and thereore what direction its torque about the center o the spool will be. So, rom this analysis it is not obvious what will happen. As with the previous problem, we can choose our origin to be at the point o contact between the spool and surace, so that riction exerts zero torque. About this origin, the net torque is exerted exclusively by the tension in the string. But in what direction is that torque? To igure that out, we need to know the line o action o the tension orce. The line o action is a straight line having the same orientation as the orce and passing through the point o application o the orce, as shown. The torque due to F can be ound by treating it as though it is applied anywhere along the line o action. Thus, the way the given diagram is drawn, the torque is clockwise. This means the spool will roll to the right.

13 Rotational Equilibrium and Rotational Dynamics 38 F Line o action Origin q However, the angle θ is let unspeciied. I we don t assume the drawing is to scale, the angle could be anything. There is an angle or which the line o action passes through the chosen origin, in which case the net torque on the spool is zero, the net orce is to the right, and the spool will slide to the right without spinning. I the angle is even larger than this, so that the line o action passes to the right o the origin contact point, the net torque is counter-clockwise. This would cause the spool to unroll to the let. I the tension is vertical, the spool will also unroll to the let. Demonstrations can conirm all o these outcomes. Thus, the motion o the spool depends upon the angle θ. The angle drawn in the igure will cause the spool to roll to the right. Key Points: The line o action is a useul concept or determining the direction o the torque exerted by a orce. Each orce has its own line o action. We can treat the torque as though the orce is applied anywhere along the line o action. A clever choice o origin can make torque problems much easier to analyze. For Instructors Only This is the last o three questions using this situation. This question drives home the idea that the choice o origin should be done with some strategic thinking and goal. It also shows the utility o the line o action concept. A physical demonstration is extremely valuable here, so that students can see that the angle o the tension orce critically aects the motion o the spool. A amily o diagrams showing the line o action or dierent angles o the tension orce can help students understand the analysis and come to a better understanding o torques and lines o action.

14 38 Chapter 8 Question E.06a Description: Reasoning with orce, energy, and torque ideas in the context o mechanical advantage. Question Two blocks hang rom strings wound around dierent parts o a double pulley as shown. Assuming the system is not in equilibrium, what happens to the system s potential energy when it is released rom rest? 500 g 00 g. It remains the same.. It decreases. 3. It increases. 4. It is impossible to determine without knowing the radii o the two pulleys 5. It is impossible to determine without knowing the ratio o the radii o the two pulleys 6. It is impossible to determine or some other reason Commentary Purpose: To develop qualitative reasoning and problem-solving skills by applying energy ideas in a rotational dynamics context. Discussion: We do not know the radii o the two pulleys, or even their ratio, so we cannot predict which block will all and which will rise when the system is released rom rest. To answer the question, however, we do not need to know any o these eatures. We do know that the system is not in equilibrium, which means that one block will start to all and the other will start to rise, and the double pulley will begin to rotate. Thus, the kinetic energy o the system will rise. Where will this energy come rom? There are no external orces doing work on the system, so it can only come rom the potential energy o the gravitational interaction between the blocks and the Earth. I the kinetic energy is increasing, the potential energy must decrease so that total mechanical energy will be conserved. Note that we are treating the Earth as part o the system. Properly speaking, gravitational potential energy is not a property o an object such as a block, but rather o the interaction between two objects in this case, between each block and the Earth. (There is also gravitational potential energy between the two blocks and between each and the double pulley, but these are truly miniscule.) I we did not treat the Earth as part o the system, we would not talk about gravitational potential energy in this question. Instead, we would talk about the work done by external orces: the gravitational orce o the Earth on each o the blocks. This would just be a dierent way o describing the situation, and (i we were calculating numbers) would produce the same results.

15 Rotational Equilibrium and Rotational Dynamics 383 Key Points: Many questions can be answered through qualitative reasoning rom general principles, without numerical calculations or solving or anything. I the kinetic energy o a system increases, then either external orces are doing positive work on the system, or the potential energy o an interaction among parts o the system is decreasing. To be precise, we talk about the potential energy o an interaction between objects, not the potential energy o an object. For Instructors Only This is the irst o two questions on this situation. This one explores qualitative reasoning with energy ideas; the next is similar but applies orce ideas. Although the topics o these two questions are conservation o energy and orces, they are useul as broader integrating questions that teach students to use their inventory o basic physics principles or reasoning about various situations. The most likely stumbling block or students with this problem is that they will want to solve or the motion and will not be able to. That makes it a good context or teaching the value o qualitative, principlebased reasoning. Treating the diagram as a scale drawing will not help students determine which way the pulley rotates, since the ratio o the radii is about :5, the same as the ratio o the block masses. Question E.06b Description: Reasoning with orce, energy, and torque ideas in the context o mechanical advantage. Question Two blocks hang rom strings would around dierent parts o a -kg double pulley as shown. The pivot exerts a normal orce F N supporting the double pulley. Assuming the system is not in equilibrium, which statement about F N is true ater the system is released rom rest? (Use g 0 N kg.) 500 g 00 g. F N 0 N. 0 N < F < 7 N 3. F N 7 N 4. F N > 7 N 5. It is impossible to predict what the normal orce on the double pulley will be.

16 384 Chapter 8 Commentary Purpose: To develop qualitative reasoning and problem-solving skills by applying orce ideas in a rotational dynamics context. Discussion: We do not know the radii o the two pulleys, or even their ratio, so we cannot predict which block will all and which will rise when the system is released rom rest. To answer the question, however, we do not need to know any o these eatures. We do know that the center o mass o the system is alling, and in act is accelerating downward. This means that a net orce in the downward direction must be acting on the system. The only (external) orces are gravitation and the normal orce exerted by the pivot, which means the normal orce must be smaller than the total weight o the system (7 N). Further, the double pulley is not accelerating, so the net orce on it must be zero. There is tension in both strings pulling down, so the normal orce must be larger than the weight o the double pulley (0 N). Note that the tensions in the strings are not N and 5 N, the weights o the blocks. For the alling block (whichever that turns out to be), the tension will be slightly smaller than the weight; or the rising block, the tension will be slightly larger than the weight. This is needed to satisy Newton s second law applied to each hanging mass. In the previous question, we considered the Earth to be part o the system we were analyzing. In this question, it is more convenient not to, but rather to treat the gravitational orce as an external orce acting upon a system comprised o the double pulley, ropes, and two blocks. Reasoning about the center o mass motion o the system, i the Earth were included in that system, would be diicult! Key Points: Many questions can be answered through qualitative reasoning rom general principles, without numerical calculations or solving or anything. I the center o mass o a body or system is accelerating even i part o it is held in place there must be a nonzero net external orce acting on one or more components o system. I the center o mass o a body or system is not accelerating, all external orces on that body or system must sum to zero. For Instructors Only This is the second o two questions on this situation. This one explores qualitative reasoning with orce ideas; the previous was similar but applied energy ideas. Although the topics o these two questions are conservation o energy and orces, they are useul as broader integrating questions that teach students to use their inventory o basic physics principles or reasoning about various situations. Revisiting old ideas in new contexts is valuable: it enriches the new context and helps students cross-link new and old ideas. Students may have diiculty ocusing on the pivot and the orces it exerts. They are not accustomed to applying Newton s second law (linear) to situations involving pulleys and torque. Some students will say that the orce supporting the pulley is equal to the pulley s weight, 0 N, ignoring the tensions pulling down. Others will say that the tensions are N and 5 N, so the orce supporting the system is 7 N. This set o two questions presents an excellent opportunity to hold a higher-level discussion about choosing a system as part o strategic problem solving: or example, why one would decide to include the Earth as part o the system sometimes but not others.

17 Rotational Equilibrium and Rotational Dynamics 385 Note that treating the diagram as a scale drawing will not help students determine which mass alls and which rises, since the ratio o the radii is the same as the ratio o the hanging masses (:5). Question E.07 Description: Developing problem solving skills by choosing an origin or statics problems. Question A uniorm rod o length L, mass M, is suspended by two thin strings. Which o the ollowing statements is true regarding the tensions in the strings? T T. T T. T.5 T 3. T 0.6 T 4. T 0.8 T 5. None o the above 6. Not enough inormation to determine Commentary Purpose: To practice determining torques in static situations, once again making the point that a good choice o pivot point can make a problem easy. Discussion: A consideration o orces tells us that the two tensions must add up to the bar s weight, in order to have zero net orce in the y-direction. Thus, we must turn to torques to answer this. There must be zero net torque about any point on the bar i the bar is to remain static. The question is, what choice o pivot point will make the problem easiest? Since the question doesn t require us to know the weight, choosing the center o mass as the pivot point is advantageous: gravity exerts no torque about that point. Then, our equation that states the sum o the torques equals zero will relate the two tensions, providing the answer we seek. We need to use the markings on the rod to determine where the center o mass is and how ar each string is rom it. We don t know the units how long each segment is but it doesn t matter, since we re looking or a ratio between T and T. Counting segments, we see that T acts 5 units rom the center and T acts 3 units away, so T must be 3 5 o T. Thus, answer (3) is appropriate. Key Points: Look or the most convenient origin about which to calculate torques. Use all the inormation provided in a question, including the diagram. I you think you need a quantity that is not given, deine a variable or it and proceed. The variable will oten cancel out.

18 386 Chapter 8 For Instructors Only I any students choose Not enough inormation, we suggest asking them what it is they would need to know to make the question answerable, and why they need it. It s likely they want to know physical dimensions or the locations o the string. They may not realize they can count segments to ind relative distances, or they may not be comortable working with ratios rather than actual distances. Question E.08 Description: Developing problem-solving skills by working with orces and torques in a nontrivial statics situation. Question A uniorm rod is hinged to a wall and held at a 30 angle by a thin string that is attached to the ceiling and makes a 90 angle to rod. Which statement(s) must be true? (At least one o them is true and at least one is alse.) 30. The hinge orce is purely vertical.. The hinge orce is purely horizontal. 3. The string tension is equal to the hinge orce. 4. The string tension is smaller than the rod s weight. 5. and 3 are true. 6. and 3 are true. 7. and 4 are true. 8. and 4 are true and 4 are true. 0. Three o the statements are true. Commentary Purpose: To help you learn to reason using orces and torques. Discussion: The rod is at rest, so the net orce on it must be zero, and the net torque about any origin must also be zero. This yields many possible relationships, all o which are valid, but only some o which bring out relevant eatures o this situation. In other words, we do not need to write down every valid equation or relationship to answer this question. Rather, thoughtul choices about how to proceed will yield eicient results. It is useul to assume nothing about the hinge orce and to think o it as having a vertical and a horizontal component. These components can be treated as independent orces. (We oten separate one orce into two separate component orces, or example when treating the contact orce between two suraces as a normal orce and a riction orce.) Let s ocus on each statement and determine its truth or alsehood.

19 Rotational Equilibrium and Rotational Dynamics 387 The hinge orce is purely vertical. This statement is alse, because the net orce in the horizontal direction must be zero. The tension orce has a component pulling to the right, so the hinge must pull to the let with an equal orce. The hinge orce is purely horizontal. This statement is alse, because the net torque about the center o mass must be zero. Both the tension orce and the horizontal component o the hinge orce exert counterclockwise torques about the center o mass o the rod. Thereore, the hinge must have a vertical component to provide a balancing clockwise torque. The string tension is equal to the hinge orce. This statement is alse, since the two orces have dierent directions, with the hinge pulling up and to the let and the tension pulling up and to the right. (It turns out that the two orces do have the same magnitude.) The string tension is smaller than the rod s weight. This statement is true, since the net torque about the let end o the rod must be zero. The torque exerted by the hinge is zero (about this point), so the tension must balance the weight. The moment arm is larger or the tension, so the orce must be smaller. Key Points: For an object at rest, all components o the net orce and the net torque about any origin are zero. Strategic choices o relationships and origins can make analysis and reasoning particularly eicient. To isolate an unknown, choose an origin such that the torque due to the other unknown(s) is zero. You can consider the situation using several dierent origins i you want; τ I α must be true or all o them. For Instructors Only This question provides an excellent opportunity to explore problem-solving approaches and strategies. The equations are simple; inding the most eicient use o those equations is more diicult. Students can be either unaware o or overwhelmed by the decision making needed to solve statics problems. It all looks so easy when the instructor does it, yet when students are doing homework or exams, it becomes impossible, largely because they have not practiced the skill o strategic thinking. The key to evaluating statements and is to ocus on the other component to determine its validity. That is, to determine i the hinge orce is purely vertical, one must ind out whether the horizontal component is zero. (Students sometimes think that hinges only exert vertical orces.) Some good students might igure out that the hinge orce and the tension orce have the same magnitude, since their horizontal components balance and their vertical components each support hal the weight o the rod. They might not realize, however, that statement 3 is about vector equality, requiring magnitudes and directions to be the same. (This alls into the category o students giving the right answer to the wrong question.)

20 388 Chapter 8 Question E.09 Description: Problem solving with orces and torques in a statics context. Question A uniorm rod o length 4L, mass M, is suspended by two thin strings, lengths L and L as shown. What is the tension in the string at the let end o the rod? L L 4L. Mg. Mg 3. Mg 3 4. Mg 4 5. None o the above Commentary Purpose: To help you to understand the deinition o torque and its application to a static situation. Description: This is a statics problem: the object has zero linear acceleration and zero angular acceleration, so the net orce and the net torque on it must both be zero. Note that a torque is always determined about some origin; the net torque on the object about any origin, anywhere in space, must be zero. I the net orce must be zero, the sum o the tensions in the two strings must be Mg, so that the net orce on the rod is zero. So i the strings have equal tensions, the tension in each must be Mg. But are the tensions equal? Since the rod is uniorm, its center o mass is at its middle. This is a particularly convenient choice o origin, since the torque due to gravitation is zero about this point. However, we are ree choose any point in space as the origin to answer this question. Let s choose the middle o the rod as the origin. Even though the rod is not perectly horizontal, the moment arms or the two tension orces are equal. They are equal to the horizontal distance rom the center o the rod to the point o attachment o the string (less than L), which is the same or each string. Since the net torque about any origin must be zero or an object at rest, the two tensions must exert balancing torques. Since the moment arms are the same, the tensions must be the same also. In other words, each string supports hal the weight o the rod. The lengths o the strings does not matter. Another way to solve the problem is to choose the origin to be at the right end o the rod. This is also a strategic choice, because then the torque due to the tension in the right string is zero. Two orces on the rod exert nonzero torques: the tension in the let string, which is the desired unknown, and gravitation. The moment arm or the tension is twice as large as the moment arm or gravitation. (Remember, gravitation acts as though the orce is exerted at the center o mass.) As beore, since the rod is at rest, the two torques must balance each other. Thereore, since the moment arm or the tension is twice as large, the tension must be hal as large as the weight o the rod.

21 Rotational Equilibrium and Rotational Dynamics 389 Key Points: For a body in static equilibrium (i.e., one that is stationary), the net orce on the body and the net torque on the body about any point must both be zero. Any origin may be chosen to analyze and solve a problem, but thoughtul, strategic choices o origin can make the analysis much simpler. The moment arm o a orce about an origin, used to determine the torque it exerts, is the shortest distance rom the origin to the imaginary line you get by extending the line along which the orce acts to ininity in both directions. It is not necessarily the distance rom the origin to the point at which the orce acts on the object. Gravitation acts as though the orce is exerted at the center o mass. For Instructors Only This is one o the simplest static situations we can create, yet students oten cannot sort out the relevant eatures because o the angle o the rod. Many conclude intuitively that the let string exerts the larger orce, and thoughts o torque, center o mass, and moment arms are oten neglected. Students oten have diiculty applying the concept o moment arm to a situation in which orces are exerted at an angle relative to the rod or object. A diagram may help many sort this out. Students can be lustered by having to choose which point is the origin. They requently get hung up on making the right choice, not realizing that all choices are correct but some are easier to work with than others. It is useul to give students opportunities to think about strategic choices o origin, and also to have students solve the problem two or more times with dierent origins, as this will encourage comparison o approaches. I students only see one choice or any given situation, they will (reasonably but incorrectly) conclude that there is one best (and thereore right ) choice to any problem. Question E.0a Description: Developing problem-solving skills by choosing an origin or torque problems. Question A uniorm disk with mass M and radius R sits at rest on an incline 30 to the horizontal. A string is wound around the disk and attached to the top o the incline as shown. The string is parallel to incline. What is the tension in the string?. Mg. Mg 3. Mg 5 4. Mg 4 5. None o the above 6. Cannot be determined 30

22 390 Chapter 8 Commentary Purpose: To develop your problem-solving skills by considering multiple approaches to a question, and explore the signiicance o where you choose your origin or torque calculations. Discussion: There are oten multiple ways to solve a problem, and part o learning to do physics well is learning to select the easiest approach to a given problem. This statics question is conceptually straightorward to answer, but the algebra can be relatively simple or complicated depending on the coordinate system you choose. The unknown orces acting on the disk are the tension orce o the string that we seek, and the normal and riction orces o the plane. We know the direction o each o these and the location at which it acts on the disk, but not its magnitude. Gravity, whose magnitude and direction we know, also acts on the disk. So, we have three unknowns: the magnitudes o the tension, normal, and riction orces. We thereore need three independent equations in order to solve or them. Since the disk is stationary, we know it is not accelerating in the x- or y-directions, and that it is not rotating about any pivot point we choose to consider. This means that the net orce in the x-direction is zero, the net orce in the y direction is zero, and the net torque about any point is zero. By writing each o these statements in terms o the actual orces (and trigonometric unctions o the incline angle), we get three equations, and all we have to do is eliminate the variables we don t care about and solve or the tension. This is a classic statics problem. The question is, in order to ind the tension, what choice o coordinate system and pivot point is best? Any choice will work, but some will lead to rather ugly algebra. How to choose? I we choose the pivot point or our torque equation to be at the point o contact between the disk and incline, then neither the normal nor riction orces exert any torque (since both act at the pivot point itsel ). So, the only two torques are due to gravity and the string tension; the only unknown is the magnitude o the tension, and we can solve this one equation or the tension. Simple! We don t need to use the two orce equations or solve or the riction or normal orces at all. The tension exerts a torque o RT out o the page. The weight acts at the center o the disk, and the component o the weight perpendicular to the vector rom pivot point to the center o the disk is Mgsinθ (the component parallel to the plane), so the weight exerts a torque o MgRsinθ into the page. Thus, RT MgR sinθ, and T Mg sinθ. The sine o 30 is, so answer (4) is correct. Had we chosen a dierent pivot point, the system o equations we d have to solve would be signiicantly more complex. Key Points: For a body in static equilibrium, the orces on it must add to zero (vector sum, i.e., along all axes), and the torques must add to zero about any pivot point you choose. Choosing the orientation o your coordinate axes and the pivot point or your torque equation can make the algebra o a problem easier or harder. It s generally wise to choose a pivot point where as many orces as possible, especially unknown ones you aren t interested in solving or, exert no torque: that is, the orces act at that point, or along a line that passes through the point. For Instructors Only In this question, the issue is not so much which answer is right as what approach provides the easiest path to it. Many students will place their pivot point at the center o the disk or perhaps where the string attaches to the disk, and then dive into two (or three) equations in two (or three) unknowns (depending on their