# Moments of Inertia. and reminds us of the analogous equation for linear momentum p= mv, which is of the form. The kinetic energy of the body is.

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1 Moments of Inerta Suppose a body s movng on a crcular path wth constant speed Let s consder two quanttes: the body s angular momentum L about the center of the crcle, and ts knetc energy T How are these quanttes related to ts angular velocty ω? L P, ω P The angular momentum about the center of the crcle has magntude L = mvr, and the velocty has magntude v = r ω The frst relaton we seek s therefore Ths s of the form r P L= mr 2 ω angular momentum = constant H angular velocty, v P and remnds us of the analogous equaton for lnear momentum p= mv, whch s of the form lnear momentum = mass H lnear velocty The knetc energy of the body s T = 1 2 mv2, and s of the form knetc energy = (1/2) mass H lnear velocty 2 Substtutng n for the velocty, we fnd T= 1 2 mr2 ω 2 whch s of the form knetc energy = (1/2) constant H angular velocty 2 The constant n these equatons s the rotatonal analog of mass It s called the moment of nerta of the body about the pont at the center of the crcle, and s symbolzed by I:,

2 L= I ω and T = 1 2 I ω 2 2 I = 3 m r In ths partcular case, the value of the moment of nerta s I = mr 2 Now let s suppose that we have a system consstng of several bodes held rgdly together n a sngle plane, and that ths composte body s rotatng about a pont n the plane of the body: Pω P r 2 Then each pece of the whole thng goes around n a crcle, and they all have the same angular velocty The angular momentum of the system s just the sum of the angular momenta of the parts: L= ä å 3 m r 2 ë ì ω, ã so the moment of nerta of the system s the sum of the ndvdual moments of nerta: P r 1 P v P v 2 1 In the case of a contnuous body, the sum becomes an ntegral We wll see some cases of ths below Example: moment of nerta of a rng about ts center Suppose that nstead of a sngle body movng n a crcular path, we have a thn rng spnnng around on ts axs lke ths: Pω The mass of the rng s m and ts radus s What s the moment of nerta of the rng about ts center? We magne the rng splt up nto tny peces All are at the same dstance from the center of the crcle The expresson for the moment of nerta smplfes, becomng I = ä ã å 3 m ë ì 2

3 The sum of the masses of the peces s just m, so we get the result I = m 2 Ths s the same as the moment of nerta of a sngle body at dstance from the center Example: moment of nerta of a dsc about ts center Now let s soup up our example even more Suppose we have a crcular dsc spnnng around on ts axs: Pω Let s let the mass of the dsc be m, and ts radus be We ll also suppose that the mass s unformly dstrbuted over the dsc What s the moment of nerta of the dsc about ts center? Well, we can thnk of the dsc as beng made up of a bunch of thn rngs We can add up the moments of nerta of all the rngs usng calculus, and the result wll be the moment of nerta of the dsc Let s see how ths works Consder a typcal rng, of radus r and (nfntesmal) thckness dr: (The thckness has been exaggerated n the fgure) What s the mass of ths rng? Well, the mass wll be the mass of the whole dsc, tmes the rato of the area of the rng to the area of the dsc The area of the rng s just ts length tmes ts thckness, or 2 π r dr So the mass s 2 π r dr m = 2 m r dr π 2 2 The moment of nerta of the rng s ts mass tmes the square of ts radus r, and we want to add these up for all rad from 0 to Ths means we have to do an ntegral I = I 0 r 2 ä å 2 m ã The result s 2 r drë r ì = 2 m 2 I 0 dr r 3 dr = 2 m

4 I = 1 2 m 2 Ths s an nterestng result If all the mass were concentrated at the edge (that s, f the plate were actually a rng), then the moment of nerta would be m 2 The moment of nerta of the dsc s actually only half as bg as ths, because the rngs nearer to the center contrbute less than they would f they were rght at the edge Example llustratng dependence of moment of nerta on the pont of rotaton Suppose we return to our thn rng of mass m and radus, and we constran t to rotate about a pont on the rng Here s a top vew: multply ther mass tmes the square of ther dstance from the pont of rotaton We then add these up over the whole rng Warnng! Ths s the knd of calculaton that looks very easy when done by someone else, but can appear nearly mpossble when you try t yourself! Don t be fooled! Practce on bodes of other shapes yourself One of the bggest challenges when dong a calculaton of ths knd s the very frst step: the choce of sutable coordnates Here s a good choce for the present case: calculatng moment of nerta about ths pont θ/2 r θ/2 θ dθ dθ rotates about ths pont What s the moment of nerta now? Well, the prncple s no dfferent from before Only the detals of ts applcaton change We just take all the lttle peces of the rng and The mass of the arc subtended by the angle dθ, shown n blue, s the mass of the rng tmes the rato of the length of the arc to the crcumference of the rng, or dθ 2 π m = m 2 π d θ

5 The dstance of the lttle arc to the pont about whch we are calculatng the moment of nerta s r = 2 cos θ 2 Multplyng the mass tmes the square of the dstance, and ntegratng over the whole rng, gves for the moment of nerta I = 2 π I 0 ä å 4 2 cos 2 θ ë ì ä å m ã 2 ã 2 π d θ ë ì = 2 m 2 π 2 π I ( 1 + cos θ ) d θ The value of the ntegral s π, so the result s I = 2 m 2 Notce that ths s the same body for whch we earler calculated the moment of nerta to be half as large! That s because the two moments of nerta are taken about dfferent ponts The moment of nerta s not an ntrnsc property of the body, but rather depends on the choce of the pont around whch the body rotates Exercse: moment of nerta of a wagon wheel about ts center Consder a wagon wheel made up of a thck rm and four thn spokes The rm s outer radus s a and nner radus s b, and ts mass M s unformly dstrbuted The spokes have length b and each has mass m, agan unformly dstrbuted Show that the moment of nerta of the wheel about ts axle s I = 1 2 M ( a 2 + b 2 ) m b 2 Hnt: ths exercse s desgned to be rather complcated You have to break up the wheel nto separate parts, calculate ther moments of nerta ndvdually, and add them up n the end Example of use of energy of rotaton: body rollng down slope Here s a famous example Consder a crcular body whch rolls wthout slppng down an nclned plane Let s not make any assumptons about the nature of the body, except that ts mass s m and ts moment of nerta about ts central axs s I What s the acceleraton along the slope? a b

6 v ω x θ x sn θ You could tackle ths by tryng to fgure out all the forces nvolved, but by far the easest way s to use the method of energy Let the dstance travelled along the slope be x Then the vertcal dstance travelled n the downward drecton s x sn θ The total energy s E = 1 2 mv I ω 2 mgx sn θ The frst term s the knetc energy due to the moton of the center of mass and the second s the knetc energy due to rotaton about the center of mass (We used a result of an earler secton to splt up the knetc energy n ths way) The thrd term s the gravtatonal potental energy A lttle reflecton wll convnce you that f no slppng occurs, then the angular frequency of rotaton s related to the speed along the plane by ω = v Substtutng, we fnd E = 1 2 m å ä 1 + ã I ë m 2 ì v 2 mgx sn θ We know that ths s constant n tme, so we may set ts tme dervatve to zero: 0 = mv ä å I ë 1 + ì a mgv sn θ ã m 2 Solvng for the acceleraton, we fnd a = g sn θ I 1 + m 2 Notce that ths s less than the acceleraton of a body sldng down a frctonless plane at the same angle Why s ths? Well, n the present case the force of frcton pushes uphll along the slope, causng the body to rotate as t moves Ths extra frctonal force slows down the body, compared wth a body sldng down a frctonless plane Exercse - practce wth forces Derve the above result by consderng the forces nvolved

7 Exercse - how to measure the moment of nerta Thnk of a way n whch you could use the above result to measure the moment of nerta of a body wth crcular cross-secton The nerta tensor Up untl now, we have been consderng only the rotaton of two-dmensonal planar bodes about axes perpendcular to the body But you can easly see that ths s only a very small subset of all possble rotatons of a rgd body So now let s move on to the more general case of the rotaton of a three-dmensonal rgd body about an arbtrary axs We wll assume only that the drecton of the axs of rotaton does not change Let s let the angular velocty vector of the body be ωp Ths s the same for every pece of the body As we learned n our secton on rotatonal moton, the velocty of any partcular pece s v P = ω P H P r Let s ask the queston: what s the relaton between the total angular momentum of the body and the angular velocty vector? We begn by wrtng down the expresson for the angular momentum and substtutng for the velocty: L P = 3 P r H m v P = 3 m P r H ( ω P H P r ) We then make use of the standard relaton for the trple cross product A P H ( B P H C P ) = ( A P A C P ) B P ( A P A B P ) C P We fnd the result L P = 3 m ä ã r 2 ω P ( P r A ω P ) P r ë Ths s a nce compact form of the result, but there s another form that gves more physcal nsght nto the relaton between the angular momentum and the angular velocty Ths other form s what we get when we re-wrte the above equaton n matrx notaton Let s begn by lookng at the components of the last equaton We wrte the poston vector as P r = x ˆ x + y ˆ y + z ˆ z, and obtan for the x-component of the angular momentum L x = 3 m ä ã r 2 ω x ( x ω x + y ω y + z ω z ) x ë, wth smlar equatons for the other components We can re-wrte ths as a lnear combnaton of the three components of the angular velocty:

8 L x = ä ã å 3 m x 2 ) ë ì ω x ä å 3 ã m x y ë ì ω y ä å 3 ã m x z ë ì ω z We recognze the rght-hand sde as the top component of the product of a matrx and the angular velocty vector: L x = ä å 3 m x 2 ) ë ì ã ä å 3 ã m x y ë ì ä å 3 ã m x z ë ì (For clarty, the other components have been temporarly suppressed) Ths matrx s called the nerta tensor We wll symbolze t by Wrtten out n full, t s = ä å 3 m x 2 ) ë ì ã ä å 3 m y x ë ì ã ä å 3 m z x ë ì ã ä å 3 m x y ë ì ã ä å 3 m y 2 ) ë ì ã ä å 3 m z y ë ì ã ä å 3 m x z ë ì ã ä å 3 m y z ë ì ã ä å 3 m z 2 ) ë ì ã and the relaton between the angular momentum and the angular velocty s wrtten compactly as ω x ω y ω z L P = ω P

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