Physics 53. Rotational Motion 3. Sir, I have found you an argument, but I am not obliged to find you an understanding.


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1 Physcs 53 Rotatonal Moton 3 Sr, I have found you an argument, but I am not oblged to fnd you an understandng. Samuel Johnson Angular momentum Wth respect to rotatonal moton of a body, moment of nerta plays the same role that mass plays n the translatonal moton of a partcle. It measures the ntrnsc reluctance of a body to have ts state of rotaton changed. Torque plays the role n rotaton that force plays n the translatonal moton of a partcle. It descrbes the external nfluence that causes changes n the state of rotaton. But what descrbes the state of rotaton tself? For a partcle, the state of translatonal moton s descrbed by the lnear momentum, p = mv. The correspondng quantty for rotatonal moton s the angular momentum. Lke torque, angular momentum has meanng only wth respect to some specfed reference pont. Also lke torque, ts magntude depends on the dstance from that pont. We begn wth the smplest system, a sngle partcle. Later we wll generalze to systems of partcles, wth specal nterest n rgd bodes. The defnton for a partcle s: Angular momentum of a partcle L = r p Here r s the poston vector of the partcle relatve to the reference pont, and p = mv s ts lnear momentum. Some propertes of L: L s a vector, perpendcular to the plane contanng r and p, and thus perpendcular to both r and p. L s zero f the partcle moves along the lne of r,.e., drectly toward or away from the reference pont. PHY 53 1 Rotatons 3
2 The magntude s gven by L = r p, where r s the moment arm, defned to be the perpendcular dstance from the reference pont to the lne along whch the partcle moves. Alternatvely, L = rp, where p s the component of p perpendcular to r. L s a maxmum, equal to rp, f p s perpendcular to r. Ths s the case f the partcle moves (at least momentarly) n a crcle about the reference pont. Lke lnear momentum and knetc energy, angular momentum s an mportant aspect of the state of moton of a partcle, especally of orbtal moton around some center of force. It s also an mportant property of the behavor of a system of partcles. Torque as a vector Here s the general defnton of the torque of a force about a gven reference pont: Torque τ = r F Here r specfes the locaton, relatve to the reference pont, of the pont at whch the force F s appled. Some propertes of τ: Torque s a vector, perpendcular to the plane contanng r and F, and thus perpendcular to both r and F. The torque s zero f F acts along the lne of r,.e., drectly toward or away from the reference pont. The magntude s gven by the two formulas ntroduced earler: τ = rf = r F. The magntude s a maxmum, equal to rf, f F s perpendcular to r. The total torque on a system of partcles s the sum of the torques of all forces that act on any of the partcles n the system. We wll see that the torques due to nternal forces cancel, so the total torque s actually the sum of the torques due only to external forces. PHY 53 2 Rotatons 3
3 Crcular moton revsted Now we apply these new defntons to a famlar problem, that of a partcle constraned to move n a crcle. Shown s a partcle, attached to a massless rod of length r whch s pvoted at the reference pont, so that the partcle moves n a crcle about that pont. We see that L = r p has magntude L = rmv and drecton parallel to the angular velocty ω. Snce v = rω, L = mr 2 ω = Iω. In vector form L = Iω. ω r v Ths smple relatonshp between L and ω also holds, as we wll see below, n many but not all mportant stuatons nvolvng systems of partcles. It s the rotatonal counterpart of p = mv. Now consder the acton of an external force appled to the partcle. Let the force be tangent to the crcle. (Other components would be counteracted by forces exerted by the rod.) If the force s parallel to the velocty, the partcle speeds up; the torque τ = r F s drected parallel to ω. Its magntude s τ = rf. Ths force and the torque t produces gve rse to a tangental acceleraton (F = ma t ) and to an angular acceleraton (snce a t = α r ). The magntudes obey F = mrα, so τ = mr 2 α = Iα. In vector form: τ = Iα. Ths smple relaton between τ and α also holds n many but not all mportant stuatons nvolvng systems of partcles. It s the rotatonal counterpart of F = ma. These relatvely smple propertes of a sngle partcle movng n a crcle carry over to the rotatonal moton of a rgd body provded t s a symmetrc body rotatng about ts symmetry axs. Ths wll be shown below. If L = Iω and τ = Iα (as n the case here) then because α = dω /dt (by defnton) and I s a constant, we see that τ = dl/dt. As we wll show below, ths relaton s true n all cases. It s the rotatonal counterpart of F = dp/dt. Angular momentum of a system Here are some general propertes of torques and angular momentum for any system of partcles. The proofs of these statements are gven at the end of ths secton. Total angular momentum L tot = r CM Mv CM + L(rel. to CM) PHY 53 3 Rotatons 3
4 Angular momentum lke other propertes of a system breaks up nto the sum of two terms: what the angular momentum would be f the system were a sngle mass pont at the CM, plus the angular momentum as measured n the CM frame. Angular momentum (fxed axs) L( to axs) = Iω What about the components of L perpendcular to the axs? In general they can behave n a qute complcated way, changng wth tme as the body rotates. These changes must be brought about by external torques, caused by forces exerted on the body by the fxed axle about whch t rotates. But f the body s symmetrc about the axs, these perpendcular components of L contrbuted by the mrror mage mass ponts on opposte sdes of the axs cancel each other exactly, and no external torques are requred. For a symmetrc body rotatng about ts symmetry axs, the angular momentum s entrely parallel to the axs and s equal to Iω. Most of the cases consdered n ths course nvolve such symmetrc bodes. An example of a body that s not symmetrc s an unbalanced wheel on a car. The frcton caused by the normal forces exerted by the axle can rapdly wear out the wheel bearngs. Ths s why one has the wheels dynamcally balanced to make them symmetrc bodes. Rotatonal 2 nd law τ ext tot = dl tot dt Ths relaton holds n any nertal frame. The reference ponts for the torque and for the angular momentum must be the same, of course. One can show that ths relaton also holds f the reference pont s chosen to be the CM, even n cases where the CM frame s nonnertal. The reason for ths s that nertal forces arsng n an acceleratng frame act at the CM and thus cause no torques about that pont. From the rotatonal 2 nd law mmedately follows an mportant conservaton law: Conservaton of angular momentum If the total external torque on a system s zero, the total angular momentum of the system s conserved. Some aspects of ths law: The reference pont for the torques and the angular momentum must be the same. PHY 53 4 Rotatons 3
5 Snce torque and angular momentum are vectors, ths law holds component by component. Ths means that f one component of the external torque s zero that component of the total angular momentum s conserved, even f other components are not conserved. Ths s the last of the three major conservaton laws of classcal mechancs: Conservaton of total lnear momentum. If the net external force on a system s zero, the total lnear momentum of the system s conserved. Conservaton of total mechancal energy. If only conservatve forces do work on a system, the total mechancal energy of the system s conserved. Conservaton of total angular momentum. If the net external torque on a system s zero about some pont, the total angular momentum of the system about that pont s conserved. These laws have nnumerable applcatons, and a workng knowledge of them gves powerful nsght nto many physcal stuatons. Newton's laws and these conservaton laws form the core prncples of ths course. As noted earler, n classcal physcs there s also a law of conservaton of mass, sayng that the total mass of a closed system remans constant. It was noted that ths law fals when relatvty s taken nto account, snce mass can be converted nto other forms of energy. Because of that, the law of conservaton of mechancal energy as gven above also requres modfcaton. But the other two conservaton laws retan ther valdty even when relatvty s taken nto account. PHY 53 5 Rotatons 3
6 Proofs of the Statements Gven Above: Proof of L tot = r CM Mv CM + L(rel. to CM). We use the lab and CM reference frames dscussed earler, as shown: The partcle's momentum n the lab frame s p = m v, where v = v CM + v. Its angular momentum about the lab frame orgn s therefore L = r p = r m (v CM + v ). r Lab Frame m r CM r' CM Frame To get the total angular momentum of the system we sum ths over all the partcles: Ths s not as complcated as t looks: L tot = L = m (r CM + r ) (v CM + v ) = m (r CM v CM ) + m r v + m r v CM + r CM m v In the mddle lne, the sum n the frst term s just the total mass, so that term s r CM Mv CM, whch s the angular momentum the system would have f t were just one partcle located at the CM. Ths s often called the angular momentum of the center of mass. The second term n that lne s the angular momentum as measured n the CM frame. In the last lne, the sum n the frst term s proportonal to the poston vector of the CM n the CM frame, whle the sum n the second term s the total momentum n that frame; both of these are zero. There are only two survvng terms, whch are the two n the stated property. There are many examples of ths twopart structure. One s the moton of the earth. Its annual orbtal moton around the sun gves the angular momentum of ts CM, and ts daly rotaton about ts axs gves the angular momentum relatve to ts CM. PHY 53 6 Rotatons 3
7 Proof of L(parallel to axs) = Iω. Consder agan a rgd body constraned to rotate about a fxed axs. We choose the axs to pass through the CM, whch therefore does not move. The total angular momentum s thus only that relatve to the CM. Let the body rotate wth angular velocty ω, whch s a vector along the axs. We consder one of the mass ponts of the body, m. At a gven nstant ts ω Axs O r r v m stuaton s as shown. Its angular momentum about the orgn (the CM) s L = r (m v ). Ths vector has components both parallel and perpendcular to the axs. The parallel component s (L ) = m r v = m r 2 ω. If we sum ths component over all the partcles to get the total component parallel to the axs, we fnd 2 L = m r ω. But the sum multplyng ω s the moment of nerta about the axs. Ths proves the clam. Proof of the rotatonal 2 nd law. The total torque actng on a system of partcles s the sum of all the torques actng on the ndvdual partcles. Those torques can arse from ether external or nternal forces. Earler, when addng all the forces on the partcles of a system to get the total force, we found that because of Newton's 3rd law the nternal forces cancel each other exactly. A smlar thng happens wth the torques, f (as s the case for ordnary nternal forces) the nteracton forces between pars of partcles act along the lne between the partcles. As a result, the total torque on a system s just the sum of the external torques. We showed earler, for a sngle partcle, that the torque about a gven reference pont s equal to the rate of change of the angular momentum about that reference pont. We wll now see that ths s a general law for systems of partcles. The total angular momentum of the system s the sum of the angular momentum of the partcles: PHY 53 7 Rotatons 3
8 L = r p. The tme dervatve of ths quantty s, by the product rule: dl dt = v p + r dp dt. The frst sum on the rght s zero because v s parallel to p for each partcle. In the second sum, dp /dt = F, the total force on the th partcle. The second sum s thus the total torque, whch (as argued above) s just the sum of the external torques. Ths proves the clam. Statc equlbrum of a rgd body A rgd body n statc equlbrum s completely at rest and remans so. The CM remans at rest, and there s no rotaton about the CM. These thngs can happen only f both of the followng are true: The total external force s zero. The total torque about the CM s zero. The general condtons for statc equlbrum are thus: Statc Equlbrum Translatonal: F ext tot = 0. Rotatonal: τ ext tot = 0. These are both condtons on vectors, so they must hold for all components. One can easly show that any pont can be used for the reference pont for the torques (not just the CM) as long as the total force s zero. In the cases treated n ths course, the forces usually all le n a plane, so the condton on the forces nvolves two component equatons. Any torque about a pont n that plane wll have only a component perpendcular to the plane, so the condton on the torques gves only one equaton. Only stuatons wth three or fewer unknowns can be completely determned by these condtons. Stress and stran We have usually treated forces as though they act at a sngle pont. Ths s at best an approxmaton. Generally forces are dstrbuted over parts of the object. PHY 53 8 Rotatons 3
9 Gravty, for example, acts ndvdually on every partcle n the body. The total gravtatonal force s the sum of these ndvdual forces. For a small object near the earth's surface, the force on each partcle s smply mg. If g s the same at the locaton of all the partcles, the total gravtatonal torque about the CM of the body s easly shown to be zero. For that reason we say that gravty acts effectvely at the CM of the body. Ths s not true of large bodes lke the earth tself, acted on by gravty from other bodes lke the sun and moon, snce g s not the same at the locaton of all the partcles. One consequence of ths s tdal forces and torques, whch wll be dscussed later. The normal force exerted by a surface on a body acts at all the atoms and molecules of the common nterface. Frcton smlarly acts at all ponts on the nterface. To dscuss the dstrbuton of a force over a surface, one ntroduces the concept of force per unt area, whch s called the stress. Stresses are dvded nto three categores: Tenson acts to pull partcles at the surface drectly away from the body. Compresson acts to push partcles at the surface drectly nto the body. Shear acts to move partcles at the surface parallel to that surface. Tensle and compressve stresses nvolve forces normal to the surface of the body, whle shear stresses nvolve forces along the surface. Stresses generally produce (at least momentary) deformatons of the body. These are called strans. If the deformaton s small and not permanent, then we have a stuaton lke Hooke's law for sprngs. The stran s (approxmately) lnearly proportonal to the stress, and the constant of proportonalty s called an elastc modulus. In the case of tenson or compresson the fractonal change n the dmenson of the object perpendcular to the nterface s related to the tensle or compressve stress by Young's modulus: Y = Stress Stran = F / A ΔL/L. Here F/A s the force per unt area normal to the nterface, and the stran s the fractonal change n the dmenson normal to the nterface. In the case of shear, the stran s defned to be the rato of the lateral deformaton (Δx) to the dmenson perpendcular to the nterface (h). The correspondng modulus s called the shear modulus: S = F / A Δx/h. In ths case, F/A s the rato of the force along the nterface to the area of the nterface. PHY 53 9 Rotatons 3
10 An object may be subject to forces actng at all ponts on ts surface. If these forces are everywhere normal to the surface, the effect s to compress (or expand) the volume occuped by the object. In ths case the force per unt area s called pressure: P = F / A. P s postve f the force on the body s nward,.e., compressve. If the pressure ncreases by ΔP, the volume V occuped by the body wll decrease by ΔV. The relatonshp s gven by the bulk modulus: B = ΔP ΔV /V. We wll return to pressure when we dscuss fluds. PHY Rotatons 3
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