Torque, Angular Momentum and Rotational Kinetic Energy
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1 Toque, Angula Moentu and Rotational Kinetic Enegy In ou peious exaples that inoled a wheel, like fo exaple a pulley we wee always caeful to specify that fo the puposes of the poble it would be teated as assless. ass M This was because the act of giing a wheel that has ass an (angula) acceleation equies a fo of foce and enegy tansfe into what is called otational kinetic enegy. We wanted to aoid the additional coplication until we fixed soe of the ideas in ind using puely linea systes. We now coect this deficiency to include otating objects that hae ass.
2 Rotational Kineatics Conside the otation of a solid body about an axis. We defined angula paaetes that let us wite equations haing the sae fo as the equations fo linea tanslation that we aleady know (labo saing, less to eoize). θ θ x Conside otation of the disk aboe about the axis shown. The yellow spot poides a ak that specifies the angula position of the disk θ and θ at two positions of the otation. By conention, angula position, θ, is easued w..t. the +x axis and is positie in the diection shown (counte clockwise).
3 The angula displaceent, θ is siply the change in the angula position, θ = θ θ Note that eey point of the body otates though the sae angula displaceent. θ θ x If the disk otates so that it sweeps out angle θ in tie t, its aeage angula elocity is θ θ θ ω ae = = t t t
4 Taking the liit as t 0 yields the instantaneous angula elocity, ω= li t 0 θ t θ θ x The angula elocity can also change so that at tie t the angula elocity is ω and at tie t the angula elocity is ω. The aeage angula acceleation is then, ω ω ω α ae = = t t t Again taking the liit as t 0 yields the instantaneous angula acceleation: α= li t 0 ω t
5 In Chapte 5 we saw that fo constant angula acceleation we had 4 kineatic equations fo the angula aiables that ioed those fo constant linea acceleation: Kineatic equations Rotation Tanslation ω=ω +αt o θ=θ +ω t + αt o o = o + at x = x + t + at o o ω =ω o + α( θ θo) θ=θ o + ω o +ω ( )t = + a(x x ) o x = x + ( + )t o o o With elations between the two sets of equations: s a = θ = ω = α (θ in adians) (ω in adians/s) (α in adians/s )
6 To go futhe we conside soething called the dubbell oto. This consists of two equal asses of sall diaete attached to the ends of a ey light (effectiely assless) od, with the syste fee to otate in a hoizontal plane about an axis though the iddle of the od). The ac length, s, of the cicula path swept out by a ass as it otates is elated to the angle swept out by the adius ia, θ s s = θ (θ in adians) Top iew
7 If the dubbell is otating then the linea elocity of the asses is elated to thei angula elocity by, = ω (ω in adians/s) And if the linea elocity of the asses is changing they ae acceleating in a diection tangent to the cicle, which is elated to the angula acceleation by, Top iew at = α (α in adians/s ) This tangential linea acceleation should not be confused with the adial (centipetal) acceleation needed to keep the asses on thei cicula tajectoy.
8 To eind you The oto has tangential linea acceleation, at = α only when its angula elocity changes (i.e. while its ate of otation is changing). It has a adial acceleation, a, een when its angula elocity ω is constant (i.e. α = 0) Top iew gien by, a = adial (centipetal) acceleation (inwads) a a t a a t Using the elation between linea and angula elocity, can also be witten a ω = = = ω = ω, this
9 We conside now the case whee at = α= 0 & the oto otates with constant angula elocity ω. What is the kinetic enegy of the syste in this case? Each ass has linea elocity so we hae K = + But = ω = ω so this can be witten K = ω + ω
10 K = ω + ω ( ) K = + ω K ( ) = ω Copaing this to the expession fo linea kinetic enegy, K = K we see that it has the sae fo if the te in backets takes on the otational equialent of ass. Defining fo the dubbell oto the otational inetia, I = We hae that its otational kinetic enegy can be expessed as, = Iω
11 Rotational Inetia Fo the dubbell oto its otational inetia was gien by the su of the indiidual asses ties thei distance to the otation axis squaed, I = + This genealizes to n, distinct asses i at distances i fo the axis of otation so we can wite that I n = i= i i Fo exaple, fo the igid oto shown hee whee the as ae igidly welded to each othe (otation about sae axis) n I= = + + i= i i 3 3
12 The otational kinetic enegy of this oto is still gien by, But hee with, K = Iω I= Solid objects like the industial flywheel shown hee (used fo enegy stoage as otational kinetic enegy) also hae a otational inetia. These ae soewhat oe difficult to calculate but ae tabulated fo specific coon shapes aound paticula axes.
13 Fo exaple fo GRR table 8. we hae the otational inetia fo a thin cylindical shell (ass M, adius R) o hoop about the axis shown, fo which I = I = While fo a solid cylinde o disk otating about the siila axis
14 Fo the sae oute shape if thei asses ae the sae the shell has twice the otational inetia of the solid cylinde. I = MR This is because all the ass of the shell is out at the one adius R fo the axis. Recall that fo discete asses, shell I n = i= i i So the oe ass futhe fo the axis the lage the alue of I. I = solid MR
15 Exaple A solid cylinde of ass 0 kg and 0.30 adius olls along flat gound without slipping at 3.0 /s. It encountes a ap angled at 0 o. How high does the cylinde oll? θ If the cylinde wee sliding without fiction its initial tanslational kinetic enegy (befoe hitting the ap) would be, K T = Since the cylinde is olling, howee, it is additionally otating about its cental axis so it will hae an additional kinetic enegy of otation.
16 Fo the point of iew of the oing axis, the oad at the botto edge of the cylinde is oing backwads with linea speed. Hence the linea speed of the i of the cylinde is so that its angula elocity is: = ω ω= θ The otational kinetic enegy is, K R = Iω with I = Then K R = = 4
17 The total (initial) kinetic enegy is then: 3 K = KTans + K Rot = + = 4 4 The cylinde will clib the ap coneting kinetic to potential enegy until K f = 0 and U f = gh (h = 0 at botto of ap) and h is the axiu height eached. Mechanical enegy is conseed (no wok) so, K+ U= 0 Kf Ki + Uf Ui = 0 Uf = Ki 3 gh = 4 3 h = h = g θ
18 Toque Toque (geek sybol τ) is the otational analog to foce. To gie an object of ass a linea acceleation a equies a foce F and obeys, F = a To gie an object, of otational inetia I (about that axis) an angula acceleation α about that axis equies a toque τ and obeys τ = Iα The (linea) foce needed to geneate a toque (otation about an axis) is, τ=f
19 τ=±f Whee is the distance fo the axis that the foce F is applied and F is the coponent of the foce pependicula to the line to the axis. Conside a disk fee to otate about its cente axis and a foce F acting at the adius, diectly away fo the axis as shown. F This foce clealy cannot cause otation about the axis. If instead the foce F acts in the plane of the otation, but with a coponent of the foce pependicula to then the disk will angulaly acceleate. F F F
20 Since the coponent of the applied foce F that eains paallel to can t contibute to this angula acceleation (obeying τ= Iα) only plays a ole in geneating the toque, F τ=±f The sign depends on the diection of the angula acceleation, α, (not the diection of the otation ω). By conention if the foce would cause a CCW α the toque is positie while othewise it is negatie. F Toques that act about the sae axis add but as signed alues such that equal toques in opposite diections yield a net toque of zeo (so α too is zeo). F F F
21 Exaple C On a teete totte. piot point L F C = C g τ =+ F =+ g C C C C C L F L = g τ = F = g L L L L L Linus alone would cause a CW angula acceleation about the piot so that toque is negatie ( τ L = g L L ) while Chalie alone would cause an CCW angula acceleation so his toque is positie ( τ C =+ CCg) if they hae the sae ass, C = L =, and they sit the sae distance fo the piot point, C = L = thei toques ae equal with opposite signs so the net toque about the piot is, τ net =τ C +τ L =+ g + ( g) = 0
22 All of us ae esponsible fo causing toque seeal ties a day. Eey tie we go though a closed doo, swinging the doo open geneates a toque about an axis at the doo pins. Most of us hae leaned that to open a doo with iniu foce we should push as fa fo the hinge as possible. This is because the toque τ=±f on the lee a,, o distance at which F is applied F fo the axis. also depends F Fo the sae foce F applied at diffeent lee as we get diffeent angula acceleations: τ = F = Iα τ = F = Iα
23 Fo GRR Table 8. fo a ectangula plate otating about one edge as the axis (a good odel fo a doo) I = ML 3 Whee L is length fo the axis to the opposite edge of the plate. If the doo has a ass of 5 kg and τ= Iα= F is wide its otational inetia about F α= its hinges is I I = (5kg)() = 5kg 3 If we push with a foce of 5 N at = 0 c α = = = = I 5kg s ad π s o F (0.0)(5N) ad 360 deg F F
24 While the sae foce applied at = 90 c gies α = = = = I 5kg s ad π s o F (0.90)(5N) ad 360 deg That s a high angula acceleation fo a sall foce. F F Usually doos in public buildings hae a etun echaniss that applies a counte toque to keep it fo slaing open and close the doo once you ae though it.
25 Fo a foce that is not pependicula to the lee a thee ae two equialent ways to intepet the toque geneated by the foce. Fo the situation below if the angle between the diection of F and is θ as shown, then, τ= F = (Fsin θ) Since = Fsin θ Altenatiely, we can daw a line back along the diection of the foce until we can daw a pependicula to the axis. The inteio angle between the lee a and this line is also θ, while the line dawn to the axis has length = sin θ so in witing τ= F we = (Fsin θ) can just as well wite that τ= F = ( sin θ )F = F, i.e. the full foce acting on the pependicula coponent of the lee a.
26 Which intepetation we use depends on the physical situation and infoation gien. If the foce F slopes the othe way elatie to as shown below we still extend the line of the foce until we can daw the pependicula to the axis and we again hae that, τ= F = (Fsin θ ) = ( sin θ )F = F
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