Rotation Torque Moment of Inertia
Lana Sheridan
De Anza College
Mar 9, 2020 Last time
• rotational quantities
• torque
• the cross product Overview
• net torque
• Newton’s second law for rotation
• calculating moments of inertia Quick review of Vector Expressions
#» #» #» Let a , b , and c be (non-null) vectors.
Could this possibly be a valid equation? #» #» #» a = b · c
(A) yes (B) no Quick review of Vector Expressions
#» #» #» Let a , b , and c be (non-null) vectors.
Could this possibly be a valid equation? #» #» #» a = b · c
(A) yes (B) no ← Quick review of Vector Expressions
#» #» #» Let a , b , and c be vectors. Let m and n be non-zero scalars.
Could this possibly be a valid equation? #» #» m = b × c
(A) yes (B) no Quick review of Vector Expressions
#» #» #» Let a , b , and c be vectors. Let m and n be non-zero scalars.
Could this possibly be a valid equation? #» #» m = b × c
(A) yes (B) no ← Quick review of Vector Expressions
#» #» #» Let a , b , and c be vectors. Let m and n be non-zero scalars.
Suppose #» • a 6= 0 and #» • b 6= 0 and #» #» • a 6= n b for any value of n. Could this possibly be a true equation? #» #» #» #» a × b = b × a
(A) yes (B) no Quick review of Vector Expressions
#» #» #» Let a , b , and c be vectors. Let m and n be non-zero scalars.
Suppose #» • a 6= 0 and #» • b 6= 0 and #» #» • a 6= n b for any value of n. Could this possibly be a true equation? #» #» #» #» a × b = b × a
(A) yes (B) no ← Torque
Torque is a measure of force-causing-rotation.
It is not a force, but it is related. It depends on a force vector and its point of application relative to an axis of rotation.
Torque is given by:
#» #» #» τ = r × F
That is: the cross product between #» • a vector r , the displacement of the point of application of the force from the axis of rotation, and #» • an the force vector F 300 Chapter 10 Rotation of a Rigid Object About a Fixed Axis
▸ 10.2 continued v 1.3 m/s 2 Do the same for the outer track: vf 5 5 22 5 22 rad/s 5 2.1 3 10 rev/min rf 5.8 3 10 m The CD player adjusts the angular speed v of the disc within this range so that information moves past the objective lens at a constant rate. (B) The maximum playing time of a standard music disc is 74 min and 33 s. How many revolutions does the disc make during that time?
SOL U TION Categorize From part (A), the angular speed decreases as the disc plays. Let us assume it decreases steadily, with a constant. We can then apply the rigid object under constant angular acceleration model to the disc.
Analyze If t 5 0 is the instant the disc begins rotating, with angular speed of 57 rad/s, the final value of the time t is (74 min)(60 s/min) 1 33 s 5 4 473 s. We are looking for the angular displacement Du during this time interval. 1 Use Equation 10.9 to find the angular displacement of Du 5 uf 2ui 5 2 vi 1vf t the disc at t 5 4 473 s: 1 5 5 2 57 rad/s11 22 rad2 /s 4 473 s 5 1.8 3 10 rad 1 1 rev21 2 Convert this angular displacement to revolutions: Du 5 1.8 3 105 rad 5 2.8 3 104 rev 2p rad 1 2 a b (C) What is the angular acceleration of the compact disc over the 4 473-s time interval?
SOL U TION Categorize We again model the disc as a rigid object under constant angular acceleration. In this case, Equation 10.6 gives the value of the constant angular acceleration. Another approach is to use Equation 10.4 to find the average angular acceleration. In this case, we are not assuming the angular acceleration is constant. The answer is the same from both equations; only the interpretation of the result is different.
Analyze Use Equation 10.6 to find the angular vf 2vi 22 rad/s 2 57 rad/s a5 5 5 27.6 3 1023 rad/s2 acceleration: t 4 473 s
Finalize The disc experiences a very gradual decrease in its rotation rate, as expected from the long time interval required for the angular speed to change from the initial value to the final value. In reality, the angular acceleration of the disc is not constant. Problem 90 allows you to explore the actual time behavior of the angular acceleration.
The component F sin f 10.4 Torque Torque tends to rotate the wrench about an axis through O. In our study of translational motion, after investigating the description of motion, we studied the cause of changes in motion: force. We follow the same plan here: F sin f S F What is the cause of changes in rotational motion? Imagine trying to rotate a door by applying a force of magnitude F perpendic- r f ular to the door surface near the hinges and then at various distances from the hinges. You will achieve a more rapid rate of rotation for the door by applying the O f F cos f Sr force near the doorknob than by applying it near the hinges. d Line of When a force is exerted on a rigid object pivoted about an axis, the object tends action to rotate about that axis. The tendency of a force to rotate an object about some axis is measured by a quantity called torque St(Greek letter tau). Torque is a vector, S but we will consider only its magnitude here; we will explore its vector nature in Figure 10.7 The force F has a Chapter 11. greater rotating#» #»tendency about an axis throughτ = r ×O asF F= increasesrF sin φ andnˆ Consider the wrench in Figure 10.7 that we wish to rotate around an axis that is as the moment arm d increases. perpendicular to the page and passes through the center of the bolt. The applied #» #» where φ is the angle between r and F, and nˆ is the unit vector #» #» perpendicular to r and F, as determined by the right-hand rule. Question
B
A
A torque is supplied by applying a force at point A. To produce the same torque, the force applied at point B must be: (A) greater (B) less (C) the same 1Image from Harbor Freight Tools, www.harborfreight.com Question
B
A
A torque is supplied by applying a force at point A. To produce the same torque, the force applied at point B must be: (A) greater ← (B) less (C) the same 1Image from Harbor Freight Tools, www.harborfreight.com 268 Chapter 9 Linear Momentum and Collisions
is applied at the center of mass, the system moves in the direction of the force with- The system rotates clockwise 268when a force is appliedChapter 9 Linear outMomentum rotating and (see Collisions Fig. 9.13c). The center of mass of an object can be located with above the center of mass. this procedure. 268 Chapter 9 Linear is Momentum appliedThe center at and the Collisionsof center mass of themass, pair the of system particles moves described in the direction in Figure of 9.14 the is force located with on- The system rotates clockwise the x axis and lies somewhere between the particles. Its x coordinate is given by when a force is applied out rotating (see Fig. 9.13c). The center of mass of an object can be located with m x 1 m x CMabove the center of mass. isthis applied procedure. at the center of mass, the system1 moves1 in2 2the direction of the force with- Torque The system rotates clockwise xCM ; (9.28) out The rotating center (see of massFig. 9.13c).of the pairThe ofcenter particles ofm 1mass 1describedm of2 an objectin Figure can 9.14 be locatedis located with on Torque: when a force is applied the x axis and lies somewhere between the particles. Its x coordinate is given by above the center of mass. thisFor procedure.example, if x 5 0, x 5 d, and m 5 2m , we find that x 5 2d. That is, the a 1 2 2 1 1 CM 3 CM centerThe centerof mass of lies mass closer of the to pairthe moreof particles mmassive1x1 describedm particle.2x2 in If Figure the two 9.14 masses is located are equal, on xCM ; (9.28) thethe x center axis and of masslies somewhere lies midway between between the mthe particles.1 1 particles.m2 Its x coordinate is given by The system rotates counter- 1 2 For We example, can extend if x this5 0, concept x 5 d, toand a system mm 15x1 2ofm manym, 2wex2 findparticles that xwith5 massesd. That mi inis, threethe clockwiseaCM when a force is applied 1 2 x ; 2 1 CM 3 (9.28) dimensions. The x coordinate ofCM the center of mass of n particles is defined to be below the center of mass. center of mass lies closer to the more massivem1 1 mparticle.2 If the two masses are equal, Forthe example,center of massif x 5lies 0, midway x 5 d ,between and m 5the 2 mparticles., we find that x 5 2d. That is, the Thea system rotates counter- 1 2 2 1 mixi CMmixi3 We can extendm1x1 1 thism2 xconcept2 1 m3x to3 1 a systemc1 m ofn xmanyn a particlesi withai masses1 m in three clockwise when a force is applied center x of ;mass lies closer to the more massive particle.5 If the5 two masses5 arei equal,m x Torque: thedimensions. centerCM of Themassm x lies 1coordinatem midway1 m betweenof1 thec 1center them particles. of mass of n particlesM is definedM a to ibei belowCM the center of mass. 1 2 3 n m i The system rotates counter- We can extend this concept to a system of many particlesa i with masses m in three clockwise when a force is applied mi ixi mixi i (9.29) dimensions.m The1x1 1x coordinatem2x2 1 m3x of3 1 thec center1 mn xofn massai of n particlesai is defined1 to be below the center of mass. xCM ; 5 5 5 mixi b where xi is the mx coordinate1 m 1 m of1 thec ith1 particlem and the total massM is MM ;a oi mi where CM 1 2 3 n mmx m x i the sum runs over1 all n1 particles.1 cThe1 y and z coordinatesaa i ii aof thei i center of mass are m1x1 m2x2 m3x3 m nxn i i i 1 (9.29) similarlyxCM ; defined by the equations 5 5 5 mixi The system moves in the m 1 m 1 m 1 c1 m M M a bCM where x is the x coordinate1 2 of3 the ith particlen and them total mass is M ; oi m where direction of the force without i 1 a i1 i i the sum runs over all ny particles.; Them y y andand z coordinatesz i; ofm thez center of mass(9.29)(9.30) are rotating when a force is applied CM M a i i CM M a i i No torque: at the center of mass. similarly defined by the equationsi i b where xi is the x coordinate of the ith particle and the total mass is M ; oi mi where The system moves in the S direction of the force without the Thesum centerruns over of mass all n can particles. be1 located The yin and three z coordinates dimensions1 of by the its centerposition of vectormass are r CM . rotating when a force is applied similarly The components defined by of the thisyCM equations vector; arem ixyi ,and y , andz CM z; , definedmi zi in Equations 9.29(9.30) and M a CM CM CMM a Theat the system center moves of mass. in the 9.30. Therefore, i i directionCM of the force without 1 1 S The center of massy can be locatedm y inand three z dimensionsm byz its position vector(9.30) r CM. rotating when a force is applied S CM ; i i 1 CM ; 1 i i 1 The componentsr 5 ofx this ^i 1 vectory M j^a arei1 zx k,^ y5 , and mz x,M ^i defined1ai inm Equationsy j^ 1 9.29m z and k^ at the center of mass. CM CM CM CMCM CMM a CMi i M a i i M a i i c 9.30. Therefore, i i i S CM The center of mass can be located in three dimensions by its position vector r CM. 1 Figure 9.13 A force is applied The componentsS of this vector are SxCM, yCM1, and zCMS , defined1 in Equations1 9.29 and r 5 x ^i 1 y j^ 1 zr CM k^ ;5 mi xr i ^i 1 m y j^ 1 m z k^(9.31) to a system of two particles of 9.30. Therefore,CM CM CM CM M a i i i i i i cCM M aii M ai M ai unequal mass connected by a S 1 1 1 light, rigid rod. where r i isS the position^ vector^ of the^ ith particle, ^defined by ^ ^ Figure 9.13 A force is applied r CM 5 xCM i 1 yCM j 1 zSCM k 5 1 mSixi i 1 mi yi j 1 mi zi k r CM ; M ami r i M a M a (9.31) to ca system of two particles of S ^ ai ^ ^ i i r i ; xi iM1 yi i j 1 zi k unequal mass connected by a Figurey 9.13 A force is applied S S 1 S light, rigid rod. where Although r i is the locating position the vector center of ofrtheCM mass i;th particle,for anmi extended,r i defined bycontinuous object is(9.31) some- to a system of two particles of M a S i unequal massx connected by a what more cumbersome than locating ^ the center ^ ^of mass of a small number of par- CM S r i ; xi i 1 yi j 1 zi k light, rigid rod. whereticles, rthei is basicthe position ideas we vector have ofdiscussed the ith particle,still apply. defined Think by of an extended object as a y m m1 2 Although locating the center of mass for an extended, continuous object is some- system containing a large numberS of small^ mass^ elements^ such as the cube in Figure x r i ; xi i 1 yi j 1 zi k what more cumbersome than locating the center of mass of a small number of par- xCM CM 9.15. Because the separation between elements is very small, the object can be con- y x1 ticles, the basic ideas we have discussed still apply. Think of an extended object as a m Although locating the center of mass for an extended, continuous object is some- m1 2 sidered to have a continuous mass distribution. By dividing the object into elements x 2 system containing a large number of small mass elements such as the cube in Figure what more cumbersome than locating the center of mass of a small number of par- xCM x of mass Dmi with coordinates xi, yi, zi, we see that the x coordinate of the center of CM ticles,9.15. Because the basic the ideas separation we have betweendiscussed elements still apply. is veryThink small, of an the extended object canobject be ascon a - x m mass is approximately Figure1 9.14m1 The center of2 mass systemsidered containing to have a continuousa large number mass of distribution. small mass elementsBy dividing such the as object the cube into in elements Figure of two particles of unequal mass x x2 of mass Dm with coordinates x , y , z , we1 see that the x coordinate of the center of on the x axis is locatedCM at x , a 9.15. Becausei the separation betweeni xiCMi
S force F acts at an angle f to the horizontal. We define the magnitude of the torque S Pitfall Prevention 10.4 associated with the force F around the axis passing through O by the expression Torque Depends on Your Choice t ; rF sin f 5 Fd (10.14) of Axis There is no unique value of the torque on an object. Its S where r is the distance between the rotation axis and the point of application of F, value depends on your choice of and d is the perpendicular distance from the rotation axis to the line of action of rotation axis. S F. (The line of action of a force is an imaginary line extending out both ends of the S vector representing the force. The dashed lineNet extending Torque from the tail of F in Fig. S 10.7 is part of the line of action of F.) From the right triangle in Figure 10.7 that has the wrench as its hypotenuse, we see that d 5 r sin f. The quantity d is called S the moment arm (or lever arm) of F. Object that can rotate about Moment an arm axis at O: S In Figure 10.7, the only component of F that tends to cause rotation of the wrench around an axis through O is F sin f, the component perpendicular to a S line drawn from the rotation axis to the point of application of the force. The hori- F1 zontal component F cos f, because its line of action passes through O, has no ten- dency to produce rotation about an axis passing through O. From the definition of torque, the rotating tendency increases as F increases and as d increases, which d explains why it is easier to rotate a door if we push at the doorknob rather than at a 1 point close to the hinges. We also want to apply our push as closely perpendicular O to the door as we can so that f is close to 908. Pushing sideways on the doorknob d2 (f 5 0) will not cause the door to rotate. If two or more forces act on a rigid object as in Figure 10.8, each tends to pro- S duce rotation about the axis through O. In this example, F tends to rotate the S 2 object clockwise and F1 tends to rotate it counterclockwise. We use the convention S that the sign of the torque resulting from a force is positive if the turning tendency F2 of the force is counterclockwise and negative if the turning tendency is clockwise. S S Figure 10.8 The force F tends For Example, in Figure 10.8, the torque resulting from F , which has a moment arm 1 S 1 to rotate the object counterclock- #» d , is positive and equal to 1F d ; the torque from F2There is negative are and two equal forces to 2F d .acting, wise about but an axis the through two O, and torques produced, τ and 1 1 1 2 2 S 1 Hence, the net torque about an axis through O is #» F 2 tends to rotate it clockwise. τ 2 point in opposite directions. t 5 t 1 t 5 2 o 1 2 F1d1 F2d2 #» Torque should not be confused with force. Forcesτ can causewould a change produce in transla a- counterclockwise rotation tional motion as described by Newton’s second law. #»Forces1 can also cause a change in rotational motion, but the effectiveness of the τ forces2 would in causing produce this change a clockwise rotation depends on both the magnitudes of the forces and the moment arms of the forces, in the combination we call torque. Torque has units of force times length—newton meters (N ? m) in SI units—and should be reported in these units. Do not confuse torque and work, which have the same units but are very different concepts.
Q uick Quiz 10.4 (i) If you are trying to loosen a stubborn screw from a piece of wood with a screwdriver and fail, should you find a screwdriver for which the handle is (a) longer or (b) fatter? (ii) If you are trying to loosen a stubborn y bolt from a piece of metal with a wrench and fail, should you find a wrench for which the handle is (a) longer or (b) fatter? S T1
R 1 Example 10.3 The Net Torque on a Cylinder R 2 x A one-piece cylinder is shaped as shown in Figure 10.9, with a core section protrud- O ing from the larger drum. The cylinder is free to rotate about the central z axis shown in the drawing. A rope wrapped around the drum, which has radius R , z S 1 exerts a force T1 to the right on the cylinder. A rope wrapped around the core, S S T2 which has radius R2, exerts a force T2 downward on the cylinder. (A) What is the net torque acting on the cylinder about the rotation axis (which is Figure 10.9 (Example 10.3) A solid cylinder pivoted about the z axis the z axis in Fig. 10.9)? S through O. The moment arm of T1 is S continued R1, and the moment arm of T2 is R2. 10.4 Torque 301
S force F acts at an angle f to the horizontal. We define the magnitude of the torque S Pitfall Prevention 10.4 associated with the force F around the axis passing through O by the expression Torque Depends on Your Choice t ; rF sin f 5 Fd (10.14) of Axis There is no unique value of the torque on an object. Its S where r is the distance between the rotation axis and the point of application of F, value depends on your choice of and d is the perpendicular distance from the rotation axis to the line of action of rotation axis. S F. (The line of action of a force is an imaginary line extending out both ends of the S vector representing the force. The dashed line extending from the tail of F in Fig. S 10.7 is part of the line of action of F.) From the right triangle in Figure 10.7 that has the wrench as its hypotenuse, we see that d 5 r sin f. The quantity d is called S the moment arm (or lever arm) of F. Moment arm SNet Torque In Figure 10.7, the only component of F that tends to cause rotation of the wrench around an axis through O is F sin f, the component perpendicular to a S line drawn from the rotation axis to the point of application of the force. The hori- F1 zontal component F cos f, because its line of action passes through O, has no ten- dency to produce rotation about an axis passing through O. From the definition of torque, the rotating tendency increases as F increases and as d increases, which d explains why it is easier to rotate a door if we push at the doorknob rather than at a 1 point close to the hinges. We also want to apply our push as closely perpendicular O to the door as we can so that f is close to 908. Pushing sideways on the doorknob d2 (f 5 0) will not cause the door to rotate. If two or more forces act on a rigid object as in Figure 10.8, each tends to pro- S duce rotation about the axis through O. In this example, F tends to rotate the S 2 object clockwise and F1 tends to rotate it counterclockwise. We use the convention S that the sign of the torque resulting from a force is positive if the turning tendency F2 of the force is counterclockwise and negative if the turning tendency is clockwise. S S Figure 10.8 The force F tends For Example, in Figure 10.8, the torque resulting from F , which has a moment arm 1 SThe1 net torque is the sumto rotate of the the object torques counterclock- acting on the object: d , is positive and equal to 1F d ; the torque from F2 is negative and equal to 2F d . wise about an axis through O, and 1 1 1 2 2 S Hence, the net torque about an axis through O is F 2 tends to rotate it clockwise. t 5 t 1 t 5 2 o 1 2 F1d1 F2d2 #» #» Torque should not be confused with force. Forces can cause a change in transla- τ = τ tional motion as described by Newton’s second law. Forces can also cause a change net i in rotational motion, but the effectiveness of the forces in causing this change i depends on both the magnitudes of the forces and the moment arms of the forces, X in the combination we call torque. Torque has units of force times length—newton meters (N ? m) in SI units—and should be reportedIn in these this units. case, Do with not confusenˆ pointing out of the slide: torque and work, which have the same units but are very different concepts.#» #» #» Q uick Quiz 10.4 (i) If you are trying to loosen a stubborn screw from a pieceτ netof = τ 1 + τ 2 = (d1F1 − d2F2)nˆ wood with a screwdriver and fail, should you find a screwdriver for which the handle is (a) longer or (b) fatter? (ii) If you are trying to loosen a stubborn y bolt from a piece of metal with a wrench and fail, should you find a wrench for which the handle is (a) longer or (b) fatter? S T1
R 1 Example 10.3 The Net Torque on a Cylinder R 2 x A one-piece cylinder is shaped as shown in Figure 10.9, with a core section protrud- O ing from the larger drum. The cylinder is free to rotate about the central z axis shown in the drawing. A rope wrapped around the drum, which has radius R , z S 1 exerts a force T1 to the right on the cylinder. A rope wrapped around the core, S S T2 which has radius R2, exerts a force T2 downward on the cylinder. (A) What is the net torque acting on the cylinder about the rotation axis (which is Figure 10.9 (Example 10.3) A solid cylinder pivoted about the z axis the z axis in Fig. 10.9)? S through O. The moment arm of T1 is S continued R1, and the moment arm of T2 is R2. 10.4 Torque 301
S force F acts at an angle f to the horizontal. We define the magnitude of the torque S Pitfall Prevention 10.4 associated with the force F around the axis passing through O by the expression Torque Depends on Your Choice t ; rF sin f 5 Fd (10.14) of Axis There is no unique value of the torque on an object. Its S where r is the distance between the rotation axis and the point of application of F, value depends on your choice of and d is the perpendicular distance from the rotation axis to the line of action of rotation axis. S F. (The line of action of a force is an imaginary line extending out both ends of the S vector representing the force. The dashed line extending from the tail of F in Fig. S 10.7 is part of the line of action of F.) From the right triangle in Figure 10.7 that has the wrench as its hypotenuse, we see that d 5 r sin f. The quantity d is called S the moment arm (or lever arm) of F. Moment arm S In Figure 10.7, the only component of F that tends to cause rotation of the wrench around an axis through O is F sin f, the component perpendicular to a S line drawn from the rotation axis to the point of application of the force. The hori- F1 zontal component F cos f, because its line of action passes through O, has no ten- dency to produce rotation about an axis passing through O. From the definition of torque, the rotating tendency increases as F increases and as d increases, which d explains why it is easier to rotate a door if we push at the doorknob rather than at a 1 point close to the hinges. We also want to apply our push as closely perpendicular O to the door as we can so that f is close to 908. Pushing sideways on the doorknob d2 (f 5 0) will not cause the door to rotate. If two or more forces act on a rigid object as in Figure 10.8, each tends to pro- S duce rotation about the axis through O. In this example, F tends to rotate the S 2 object clockwise and F1 tends to rotate it counterclockwise. We use the convention S that the sign of the torque resulting from a force is positive if the turning tendency F2 of the force is counterclockwise and negative if the turning tendency is clockwise. S S Figure 10.8 The force F tends For Example, in Figure 10.8, the torque resulting from F , which has a moment arm 1 S 1 to rotate the object counterclock- d1, is positive and equal to 1F1d1; the torque from F2 is negative and equal to 2F2d2. wise about an axis through O, and Example 10.3 - NetS Torque on a Cylinder Hence, the net torque about an axis through O is F 2 tends to rotate it clockwise. t 5 t 1 t 5 2 A one-piece cylinder is shaped as shown, with a core section o 1 2 F1d1 F2d2 Torque should not be confused with force. Forces can causeprotruding a change fromin transla the- larger drum. The cylinder is free to rotate tional motion as described by Newton’s second law. Forcesabout can also the cause central a changez axis shown in the drawing. A rope wrapped in rotational motion, but the effectiveness of the forces in causing this change around the drum, which has radius R1, exerts a force T1 to the depends on both the magnitudes of the forces and the moment arms of the forces, in the combination we call torque. Torque has units of forceright times on length—newton the cylinder. A rope wrapped around the core, which has meters (N ? m) in SI units—and should be reported in theseradius units.R Do2, not exerts confuse a force T2 downward on the cylinder. torque and work, which have the same units but are very different concepts. What is the net torque acting on the cylinder about the rotation Q uick Quiz 10.4 (i) If you are trying to loosen a stubborn screw from a piece of wood with a screwdriver and fail, should you find a screwdriveraxis (which for which is the the z axis)? handle is (a) longer or (b) fatter? (ii) If you are trying to loosen a stubborn y bolt from a piece of metal with a wrench and fail, should you find a wrench for which the handle is (a) longer or (b) fatter? S T1
R 1 Example 10.3 The Net Torque on a Cylinder R 2 x A one-piece cylinder is shaped as shown in Figure 10.9, with a core section protrud- O ing from the larger drum. The cylinder is free to rotate about the central z axis shown in the drawing. A rope wrapped around the drum, which has radius R , z S 1 exerts a force T1 to the right on the cylinder. A rope wrapped around the core, S S T2 which has radius R2, exerts a force T2 downward on the cylinder. (A) What is the net torque acting on the cylinder about the rotation axis (which is Figure 10.9 (Example 10.3) A solid cylinder pivoted about the z axis the z axis in Fig. 10.9)? S through O. The moment arm of T1 is S continued R1, and the moment arm of T2 is R2. #» τ net = 2.5 Nm, counter-clockwise (or ˆk)
10.4 Torque 301
S force F acts at an angle f to the horizontal. We define the magnitude of the torque S Pitfall Prevention 10.4 associated with the force F around the axis passing through O by the expression Torque Depends on Your Choice t ; rF sin f 5 Fd (10.14) of Axis There is no unique value of the torque on an object. Its S where r is the distance between the rotation axis and the point of application of F, value depends on your choice of and d is the perpendicular distance from the rotation axis to the line of action of rotation axis. S F. (The line of action of a force is an imaginary line extending out both ends of the S vector representing the force. The dashed line extending from the tail of F in Fig. S 10.7 is part of the line of action of F.) From the right triangle in Figure 10.7 that has the wrench as its hypotenuse, we see that d 5 r sin f. The quantity d is called S the moment arm (or lever arm) of F. Moment arm S In Figure 10.7, the only component of F that tends to cause rotation of the wrench around an axis through O is F sin f, the component perpendicular to a S line drawn from the rotation axis to the point of application of the force. The hori- F1 zontal component F cos f, because its line of action passes through O, has no ten- dency to produce rotation about an axis passing through O. From the definition of torque, the rotating tendency increases as F increases and as d increases, which d explains why it is easier to rotate a door if we push at the doorknob rather than at a 1 point close to the hinges. We also want to apply our push as closely perpendicular O to the door as we can so that f is close to 908. Pushing sideways on the doorknob d2 (f 5 0) will not cause the door to rotate. If two or more forces act on a rigid object as in Figure 10.8, each tends to pro- S duce rotation about the axis through O. In this example, F tends to rotate the S 2 object clockwise and F1 tends to rotate it counterclockwise. We use the convention S that the sign of the torque resulting from a force is positive if the turning tendency F2 of the force is counterclockwise and negative if the turning tendency is clockwise. S S Figure 10.8 The force F tends For Example, in Figure 10.8, the torque resulting from F , which has a moment arm 1 S 1 to rotate the object counterclock- d , is positive and equal to 1F d ; the torque from F2 is negative and equal to 2F d . wise about an axis through O, and 1 1 1 2 2 S Hence, the net torque about an axis through O is F 2 tends to rotate it clockwise. t 5 t 1 t 5 2 o 1 2 F1d1 F2d2 Torque should not be confused with force. Forces can cause a change in transla- tional motion as described by Newton’s second law. Forces can also cause a change in rotational motion, but the effectiveness of the forces in causing this change depends on both the magnitudes of the forces and the moment arms of the forces, in the combination we call torque. Torque has units of force times length—newton meters (N ? m) in SI units—and should be reported in these units. Do not confuse torque and work, which have the same units but are very different concepts. Q uick Quiz 10.4 (i) If you are trying to loosen a stubbornExample screw from 10.3 a piece of - Net Torque on a Cylinder wood with a screwdriver and fail, should you find a screwdriver for which the handle is (a) longer or (b) fatter? (ii) If you are trying to loosen a stubborn y bolt from a piece of metal with a wrench and fail, should you find a wrench for which the handle is (a) longer or (b) fatter? S T1
R 1 Example 10.3 The Net Torque on a Cylinder R 2 x A one-piece cylinder is shaped as shown in Figure 10.9, with a core section protrud- O ing from the larger drum. The cylinder is free to rotate about the central z axis shown in the drawing. A rope wrapped around the drum, which has radius R , z S 1 exerts a force T1 to the right on the cylinder. A rope wrapped around the core, S S T2 which has radius R2, exerts a force T2 downward on the cylinder. (A) What is the net torque acting on the cylinder about the rotation axis (which is Figure 10.9 (Example 10.3) A solid cylinder pivoted about the z axis the z axis in Fig. 10.9)? First: Find an expression for the net torqueS acting on the cylinder through O. The moment arm of T1 is S about thecontinued rotation axis.R1, and the moment arm of T2 is R2.
Second: Let T1 = 5.0 N, R1 = 1.0 m, T2 = 15 N, and R2 = 0.50 m. What is the net torque? Which way is the rotation? 10.4 Torque 301
S force F acts at an angle f to the horizontal. We define the magnitude of the torque S Pitfall Prevention 10.4 associated with the force F around the axis passing through O by the expression Torque Depends on Your Choice t ; rF sin f 5 Fd (10.14) of Axis There is no unique value of the torque on an object. Its S where r is the distance between the rotation axis and the point of application of F, value depends on your choice of and d is the perpendicular distance from the rotation axis to the line of action of rotation axis. S F. (The line of action of a force is an imaginary line extending out both ends of the S vector representing the force. The dashed line extending from the tail of F in Fig. S 10.7 is part of the line of action of F.) From the right triangle in Figure 10.7 that has the wrench as its hypotenuse, we see that d 5 r sin f. The quantity d is called S the moment arm (or lever arm) of F. Moment arm S In Figure 10.7, the only component of F that tends to cause rotation of the wrench around an axis through O is F sin f, the component perpendicular to a S line drawn from the rotation axis to the point of application of the force. The hori- F1 zontal component F cos f, because its line of action passes through O, has no ten- dency to produce rotation about an axis passing through O. From the definition of torque, the rotating tendency increases as F increases and as d increases, which d explains why it is easier to rotate a door if we push at the doorknob rather than at a 1 point close to the hinges. We also want to apply our push as closely perpendicular O to the door as we can so that f is close to 908. Pushing sideways on the doorknob d2 (f 5 0) will not cause the door to rotate. If two or more forces act on a rigid object as in Figure 10.8, each tends to pro- S duce rotation about the axis through O. In this example, F tends to rotate the S 2 object clockwise and F1 tends to rotate it counterclockwise. We use the convention S that the sign of the torque resulting from a force is positive if the turning tendency F2 of the force is counterclockwise and negative if the turning tendency is clockwise. S S Figure 10.8 The force F tends For Example, in Figure 10.8, the torque resulting from F , which has a moment arm 1 S 1 to rotate the object counterclock- d , is positive and equal to 1F d ; the torque from F2 is negative and equal to 2F d . wise about an axis through O, and 1 1 1 2 2 S Hence, the net torque about an axis through O is F 2 tends to rotate it clockwise. t 5 t 1 t 5 2 o 1 2 F1d1 F2d2 Torque should not be confused with force. Forces can cause a change in transla- tional motion as described by Newton’s second law. Forces can also cause a change in rotational motion, but the effectiveness of the forces in causing this change depends on both the magnitudes of the forces and the moment arms of the forces, in the combination we call torque. Torque has units of force times length—newton meters (N ? m) in SI units—and should be reported in these units. Do not confuse torque and work, which have the same units but are very different concepts. Q uick Quiz 10.4 (i) If you are trying to loosen a stubbornExample screw from 10.3 a piece of - Net Torque on a Cylinder wood with a screwdriver and fail, should you find a screwdriver for which the handle is (a) longer or (b) fatter? (ii) If you are trying to loosen a stubborn y bolt from a piece of metal with a wrench and fail, should you find a wrench for which the handle is (a) longer or (b) fatter? S T1
R 1 Example 10.3 The Net Torque on a Cylinder R 2 x A one-piece cylinder is shaped as shown in Figure 10.9, with a core section protrud- O ing from the larger drum. The cylinder is free to rotate about the central z axis shown in the drawing. A rope wrapped around the drum, which has radius R , z S 1 exerts a force T1 to the right on the cylinder. A rope wrapped around the core, S S T2 which has radius R2, exerts a force T2 downward on the cylinder. (A) What is the net torque acting on the cylinder about the rotation axis (which is Figure 10.9 (Example 10.3) A solid cylinder pivoted about the z axis the z axis in Fig. 10.9)? First: Find an expression for the net torqueS acting on the cylinder through O. The moment arm of T1 is S about thecontinued rotation axis.R1, and the moment arm of T2 is R2.
Second: Let T1 = 5.0 N, R1 = 1.0 m, T2 = 15 N, and R2 = 0.50 m. What is the net torque? Which way is the rotation? #» τ net = 2.5 Nm, counter-clockwise (or ˆk) Rotational Version of Newton’s Second Law
Tangential components of forces give rise to torques.
They also cause tangential accelerations. Consider the tangential component of the net force, Fnet,t :
Fnet,t = m at
from Newton’s second law. #» #» #» τ net = r × F net = r Fnet,t nˆ
Now let’s specifically consider the case of a single particle, mass m, at a fixed radius r. 302 Chapter 10 Rotation of a Rigid Object About a Fixed Axis
▸ 10.3 continued
SOL U TION S Conceptualize Imagine that the cylinder in Figure 10.9 is a shaft in a machine. The force T1 could be applied by a S drive belt wrapped around the drum. The force T2 could be applied by a friction brake at the surface of the core. Categorize This example is a substitution problem in which we evaluate the net torque using Equation 10.14. S The torque due to T1 about the rotation axis is 2R T . (The sign is negative because the torque tends to produce S 1 1 clockwise rotation.) The torque due to T2 is 1R2T2. (The sign is positive because the torque tends to produce counter- clockwise rotation of the cylinder.) t 5 t 1 t 5 2 Evaluate the net torque about the rotation axis: o 1 2 R2T2 R1T1 As a quick check, notice that if the two forces are of equal magnitude, the net torque is negative because R . R . Start- 1 S2 ing from rest with both forces of equal magnitude acting on it, the cylinder would rotate clockwise because T1 would S be more effective at turning it than would T2.
(B) Suppose T1 5 5.0 N, R1 5 1.0 m, T2 5 15 N, and R2 5 0.50 m. What is the net torque about the rotation axis, and which way does the cylinder rotate starting from rest?
SOL U TION Substitute the given values: o t 5 (0.50 m)(15 N) 2 (1.0 m)(5.0 N) 5 2.5 N ? m Because this net torque is positive, the cylinder begins to rotate in the counterclockwise direction.
The tangential force on the 10.5 Analysis Model: Rigid Object Under a Net Torque Rotational Versionparticle of results Newton’s in a torque on Second the Law A single particle, massparticlem, at about a fixedan axis radiusthrough r. In Chapter 5, we learned that a net force on an object causes an acceleration of the the center of the circle. object and that the acceleration is proportional to the net force. These facts are the S basis of the particle under a net force model whose mathematical representation ⌺ Ft is Newton’s second law. In this section, we show the rotational analog of Newton’s second law: the angular acceleration of a rigid object rotating about a fixed axis is m proportional to the net torque acting about that axis. Before discussing the more
S complex case of rigid-object rotation, however, it is instructive first to discuss the ⌺ Fr case of a particle moving in a circular path about some fixed point under the influ- ence of an external force. r Consider a particle of mass m rotating in a circle of radius r under the influence S S of a tangential net force g Ft and a radial net force g Fr as shown in Figure 10.10. The radial net force causes the particle to move in the circular path with a centrip- S Figure 10.10 A particle rotating etal acceleration. The tangential force provides a tangential acceleration at , and in a circle under the influence of a For such a particle, Fnet,t = mat S tangential net force F . A radial Ft 5 mat S g t o net force#» g Fr also must be present S to maintainτ net the= circularr Fnet, motion.t nˆ The magnitude of the net torque due to g Ft on the particle about an axis perpen- dicular to the page through the center of the circle is = r m at nˆ #» t 5 5 = r m (αr) o o Ft r (mat)r #» = (mr 2) α Because the tangential acceleration is related to the angular acceleration through the relationship at 5 ra (Eq. 10.11), the net torque can be expressed as o t 5 (mra)r 5 (mr 2)a (10.15) Let us denote the quantity mr 2 with the symbol I for now. We will say more about this quantity below. Using this notation, Equation 10.15 can be written as o t 5 Ia (10.16) That is, the net torque acting on the particle is proportional to its angular accelera- tion. Notice that o t 5 Ia has the same mathematical form as Newton’s second law of motion, o F 5 ma. Scalar I is not impulse! This is just an unfortunate notation coincidence.
I is called the moment of inertia of this system, for this particular axis of rotation. #» Replacing the constant quantity in our expression for τ net: #» #» τ net = I α
Rotational Version of Newton’s Second Law
(mr 2) is just some constant for this particle and this axis of rotation.
Let this constant be (scalar) I = mr 2. I is called the moment of inertia of this system, for this particular axis of rotation. #» Replacing the constant quantity in our expression for τ net: #» #» τ net = I α
Rotational Version of Newton’s Second Law
(mr 2) is just some constant for this particle and this axis of rotation.
Let this constant be (scalar) I = mr 2.
Scalar I is not impulse! This is just an unfortunate notation coincidence. #» Replacing the constant quantity in our expression for τ net: #» #» τ net = I α
Rotational Version of Newton’s Second Law
(mr 2) is just some constant for this particle and this axis of rotation.
Let this constant be (scalar) I = mr 2.
Scalar I is not impulse! This is just an unfortunate notation coincidence.
I is called the moment of inertia of this system, for this particular axis of rotation. Rotational Version of Newton’s Second Law
(mr 2) is just some constant for this particle and this axis of rotation.
Let this constant be (scalar) I = mr 2.
Scalar I is not impulse! This is just an unfortunate notation coincidence.
I is called the moment of inertia of this system, for this particular axis of rotation. #» Replacing the constant quantity in our expression for τ net: #» #» τ net = I α The moment of inertia measures the rotational inertia of an object (how hard is it to rotate an object?), just as mass is a measure of inertia.
Rotational Version of Newton’s Second Law
Compare! #» #» τ net = I α #» #» F net = m a
Now the moment of inertia, I, stands in for the inertial mass, m. Rotational Version of Newton’s Second Law
Compare! #» #» τ net = I α #» #» F net = m a
Now the moment of inertia, I, stands in for the inertial mass, m.
The moment of inertia measures the rotational inertia of an object (how hard is it to rotate an object?), just as mass is a measure of inertia. And we sum over these torques to get the net torque.
# » 2 #» τnet = mi ri α i X All particle in the object rotate together, so for an object made of a collection of particles, masses mi at radiuses ri : 2 I = mi ri i X
Moment of Inertia We just found that for a single particle, mass m, radius r, I = mr 2
For an extended object with mass distributed over varying distances from the rotational axis, each particle of mass mi experiences a torque: #» 2 #» τ i = mi ri α Moment of Inertia We just found that for a single particle, mass m, radius r, I = mr 2
For an extended object with mass distributed over varying distances from the rotational axis, each particle of mass mi experiences a torque: #» 2 #» τ i = mi ri α
And we sum over these torques to get the net torque.
# » 2 #» τnet = mi ri α i X All particle in the object rotate together, so for an object made of a collection of particles, masses mi at radiuses ri : 2 I = mi ri i X Moment of Inertia
If the object’s mass is far from the point of rotation, more torque is needed to rotate the object (with some angular acceleration).
The barbell on the right has a greater moment of inertia.
1Diagram from Dr. Hunter’s page at http://biomech.byu.edu (by Hewitt?) Moment of Inertia
And we sum over these torques to get the net torque. So, for a collection of particles, masses mi at radiuses ri :
2 I = mi ri i X
For a continuous distribution of mass, we must integrate over each small mass ∆m:
I = r 2 dm Z Moment of Inertia
For a continuous distribution of mass, we must integrate over each small mass ∆m:
I = r 2 dm Z
For an object of density ρ:
I = r 2 ρ dV Z If ρ varies with position, it must stay inside the integral. Also notice that these integrals and sums are similar to the expression for the center-of-mass, but for I we have r 2 and we do not divide by the total mass.
Units: kg m2
Moment of Inertia
Important caveat: Moment of inertia depends on the object’s mass, shape, and the axis of rotation.
A single object will have different moments of inertia for different axes of rotation. Moment of Inertia
Important caveat: Moment of inertia depends on the object’s mass, shape, and the axis of rotation.
A single object will have different moments of inertia for different axes of rotation.
Also notice that these integrals and sums are similar to the expression for the center-of-mass, but for I we have r 2 and we do not divide by the total mass.
Units: kg m2 Example: Calculating Moment of Inertia
Calculate the moment of inertia of two equal point-like masses connected by a light rod, length `, rotating about the y-axis through the center of mass.
m ↑ y m
l m`2 = 2
Example: Calculating Moment of Inertia
m m
l
Moment of inertia:
2 I = mi xi i X ` 2 ` 2 = m − + m 2 2 Example: Calculating Moment of Inertia
m m
l
Moment of inertia:
2 I = mi xi i X ` 2 ` 2 = m − + m 2 2 m`2 = 2 Summary
• net torque • Newton’s second law for rotation • moments of inertia
3rd Assignment due Friday.
(Uncollected) Homework Serway & Jewett, • Ch 10, Probs: 27 (net torque), 29 (and ang. acc.), 40, 45(a) (moment of interia) • Look at examples 10.4, and 10.11