LAB # 4 - Torque. Introducton Through the use of Newton's three laws of moton, t s possble (n prncple, f not n fact) to predct the moton of any set of partcles. That s, n order to descrbe the moton of an electron or proton n an atom one smply wrtes Newton's second law of moton and ntegrates to get the poston as a functon of tme. We have done the same to descrbe the moton of baseballs and flyng hammers and so forth. One ssue that we have swept under the rug, however, s that baseballs and hammers (and even protons!) are not pont partcles, but are composed of consttuent partcles. In fact, to descrbe the moton of a large body, one would have to wrte Newton's second law for every partcle and solve each of these equatons. Consder that a baseball "shall wegh not less than fve nor more than 5 1/4 ounces avordupos" (from major league baseball offcal regulatons). In real unts, ths s about 14-149 grams. Thus, there are roughly (0.145 kg / 1.6 10-7 kg/nucleon) = 10 +6 nucleons n the baseball and a lke number of electrons. Assumng three quarks per nucleon, that s a total of ~10 +7 partcles n just one baseball. So, strctly speakng, we have to wrte 10 +7 equatons and solve them all to descrbe the moton of a baseball. So how dd we get away wth not dong t so far? Because a baseball s a member of a class of objects called rgd bodes. That s, t s an object composed of consttuent partcles that may move as a whole, but do not move relatve to each other. For the baseball example, even as the ball fles through the ar, the cork or rubber stays n the mddle, the horse- or cowhde stays on the outsde and the yarn stays between them. How does ths help? Well, let's do as I proposed above - wrte Newton's second law for every one of the partcles n a baseball: d r F, F, j = ma = m dt = F, EXT + j where the ndex counts the partcles from 1 to 10 +7. The total force, F,, s shown dvded nto two parts: a force on partcle due to some external agent, F,EXT, (n the baseball, t mght be the force of the bat or the force of gravty) and a force on partcle due to all the other partcles, ΣF,j, (labeled j, whch also counts all the partcles [except partcle ]). At ths pont, we have ~10 +7 equatons to solve - a hopeless cause. But suppose we add all 10 +7 of these equatons together. That s, sum equaton (1) over all values of : F = + = =, F, EXT F, j ma mr () all all all j all dt all d (1)
There are a couple of thngs to notce about ths equaton that make lfe easer. Consder the sum over and j of F,j. The F,j are all acton-reacton forces. That s, for every force F,j orgnatng from partcle j and actng on partcle, there s another term F j, orgnatng from and actng on j. But these are acton-reacton pars, and Newton's Thrd Law says that acton-reacton pars are equal n magntude and opposte n drecton. Therefore, these forces all add to zero. (That s, for each carbon atom pullng on ts neghbor, the neghbor pulls back just as hard). The sum over all contrbutons of the external forces smply adds to the total external force. (That s, f I add together the weght of each partcle n a baseball, the total wll smply be the total weght of the baseball). Fnally, the last summaton (after the second dervatve wth respect to tme) s defned as the product of the total mass of the object tmes the poston of the object's center of mass. That s, there s a specal locaton called the center of mass, defned as: mr mr all all r = (3) m M all What ths defnton allows us to do s to wrte Eq. as: d F EXT = = dt ( M r ) M a (4) Ths means that the object as a whole behaves exactly as a pont partcle wth mass M located at the pont r subject only to a force F EXT. Much easer (and faster!) than solvng 10 +7 equatons smultaneously. Ths s one of the reasons that we want to defne a center of mass accordng to Eq. 3. And note that we haven't made any assumptons about how each of those partcles behaves relatve to all the others. That s, Eq. 4 works as well for a chunk of gas (n whch the consttuent partcles move a great deal relatve to one another) as a chunk of baseball (n whch they don't). But we don't have the whole story. If we throw a baseball or a hammer, the center of mass may travel just lke a pont partcle, but the parts of the hammer or baseball may also move around the center of mass. In other words, the object may rotate as well as travel (f we gve the hammer a flp when we throw t, or f the chunk of gas s swrlng as well as movng, or f I'm throwng a curve ball nstead of a knuckle ball). To see how we deal wth ths mathematcally, magne that you are openng a door. If you are lke most people, you push on the door at or near the handle. If you try to open t by pushng near the hnge, you fnd that the force necessary to open the door s much larger. Therefore, t seems that when you want to rotate thngs around a pvot, t matters not only how bg a force you use, but where that force s appled relatve to the pvot (and further from the pvot requres a smaller force). In fact, as you wll hopefully demonstrate n ths lab, t s the product of the force and the dstance that gves an equvalent effect. (A bg force at a small dstance gves the same openng-power on the door as a small force at a large dstance - accordng to Pappus of Alexandra, ths realzaton led Archmedes to clam that gven a place to stand, he could move the Earth). But also, consder usng a gven force on the door handle. If you push on the door perpendcular to the drecton to the hnge, you get much more effect than f you push parallel to the hnge drecton. And every angle n between gves an effect n between.
Fgure 1 summarzes ths whole ssue wth a force of fxed magntude actng at varous ponts n varous drectons: Ths can be summarzed by wrtng the "force ntended to open the door" as " Door openng force" = rf sn Θ (5) where F s the magntude of the force, r s the dstance from the pont of rotaton to the pont at whch the force acts, and Θ s the angle between two lnes: the lne along whch the force acts and the lne between the pont of rotaton and pont at whch the force acts. Look at Fgure to see ths: Ths "door-openng force" or a force that tends to cause a rotaton s referred to as a torque, and s typcally symbolzed by the Greek letter τ (and rarely by the letter N). (Later n the semester, we wll wrte ths torque n terms of a vector cross product: τ = r x F). If we consder the torque produced by gravty (or any other central force), we can see another nterestng property of the center of mass. Consder the torque produced by gravty on all the dfferent parts of an rregularly shaped object: Θ m and note carefully that we are measurng the dstance, r, FROM the Θ center of mass of the object. Then we can wrte the torque produced about the center of mass due to the weght of the tny partcle as: m g τ = r F sn Θ = m g( r sn Θ ) = m gx (6) and the total torque due to all the partcles as the sum of Eq. 6: = m gx = τ g m x (7) NO dooropenng ablty Less dooropenng abltyopenng Good door- ablty Lne along whch force acts Less dooropenng ablty Lne between rotaton pont and pont of acton of force Note that we have taken g out of the summaton. Ths amounts to assumng that the acceleraton of gravty does not change apprecably over the sze of the object. For hammers and baseballs, ths s completely reasonable. For suffcently large objects, t Θ x = r snθ r
becomes less reasonable. (ASIDE: The fact that the acceleraton of gravty vares sgnfcantly across ts sze s the reason the moon only shows one face to the Earth). Returnng to Eq. 7, notce that the last summaton s the x-coordnate of the locaton of the center of mass (see Eq. 3). But snce we are measurng dstances FROM the center of mass, ths s zero. Therefore, gravty exerts no torque about the center of mass. (Ths result has great sgnfcance for the orbts of planets). We dscover another sgnfcant result f we consder that we dd NOT measure dstances about the center of mass (.e., the orgn was not at but elsewhere). All the mathematcs would be the same and the torque due to gravty would be: = τ τ = m gx = g m x = gm x = ( M g) r sn Θ (8) That s, the torque due to gravty about ANY pont s smply the total force of gravty (M g) tmes the dstance from the pont to the, tmes the sne of the angle between the drecton of the force (down) and the lne of r. In other words, Eqs. 4 and 8 say that we are completely justfed n pretendng that the force of gravty on an extended object acts only at the (subject to all the assumptons we made n gettng those equatons). The last pont to consder before we get on wth the lab s what happens f I push on the handle of a door that says "pull to open". I'm exertng a perfectly good torque (the force I exert s perpendcular to the hnge drecton, so the snθ s 1 and the torque s not zero). But I know from experence that the door wll not open. Why not? Because the door jamb exerts a torque n the opposte drecton that s just as bg. As I try to cause a rotaton so as to OPEN the door, the jamb exerts a torque so as to keep t closed. Therefore, the door remans at rest (closed). Ths s called equlbrum (the fact that the door s sttng stll as opposed to movng at constant velocty s why t s called STATIC equlbrum). So there are bascally two condtons n equlbrum: that the net force be zero (the force of the hnges, the door jamb and my hand and whatever other forces on the door add to zero) and the net torque be zero (all the torques due to all the forces add to zero). In ths lab, you wll consder a partcularly smple case of statc equlbrum and study the torques due to the varous forces appled. The Lab There are bascally three phases to the lab. In the frst, you wll consder a smplfed verson of Eq. 6. You wll be drected to balance a yardstck (you wll be told that ths s the same as fndng the center of mass - why s ths true?). Afterward, you wll suspend equal weghts at equal dstances (~5 cm or so) from the pont of support, so that t stll balances. (From what we sad, ths means that the torques add to zero - thnk carefully about what ths means about the sgns of the varous torques and what THAT means about the angles Θ). Then add weght to one sde and move the weght on the other sde untl the two sdes agan balance. Record the nformaton necessary to calculate the torque on each sde. Repeat ths untl one set of weghts s approachng the end of the yardstck. Plot the torque on one sde as a functon of the torque on the other sde. What does Eq. 6 say the slope and ntercept of ths plot SHOULD be? What s t? Dscuss the dfferences.
The next phase wll nvolve shftng the pont of support AWAY from the center of mass of the yardstck. Wthout a counterweght, the yardstck wll feel a net torque and wll fall over (does ths contradct our statement about no torques due to gravty about the? Why or why not?). Suspend a counterweght to restore balance. Use the fact that the net torque s zero (how do you know ths?) to fnd the weght of the yardstck. Move the pont of support agan and restore balance (by movng the counterweght and/or changng the weght). Repeat the calculaton. Do ths several more tmes and get a mean and standard devaton for the yardstck weght. Compare ths wth what the magc weghng box at the front of the class tells you. Use your data and what you know about error propagaton to determne whch of your measurements are more or less relable, f any. The fnal phase requres that you restore the pont of support to the of the yardstck so that t balances by tself. Then suspend an unknown weght on one sde and balance t wth a known weght at a known poston. Use ths nformaton to establsh the weght of the unknown. Move the locaton of both the known and unknown weghts and change the known weght untl balance s restored. Repeat the calculaton. Do ths several tmes, and get a mean and standard devaton for the unknown weght. Use your data and your knowledge of error propagaton to argue whch of your measurements s more or less relable. Compare your result wth what the magc weghng box at the front of the class tells you. Concluson Your report from ths lab should address the varous assumptons made n the course of the development of the varous equatons used (partcularly Eqs. 6 and 8, the workhorses of ths lab). In partcular, you should dscuss the effect these assumptons may have had on your results and you should defend any clams that there s no effect. Also, you should address the questons that were asked at varous ponts n the prevous development. Extra To test and enhance your understandng, consder the followng experment. Hold your fngers about 70 cm apart and have someone place a yardstck across them. Slowly and carefully move your fngers together. Use what you know about torques and frctonal forces to explan that when your fngers come together at the center, they are below the center of mass. Also, explan how a beam balance works and what t has to do wth today's lab.