PHYS 1441 Secton 001 Lecture #15 Wednesday, July 8, 2015 Concept of the Center of Mass Center of Mass & Center of Gravty Fundamentals of the Rotatonal Moton Rotatonal Knematcs Equatons of Rotatonal Knematcs Relatonshp Between Angular and Lnear Quanttes 1
Announcements Readng Assgnments: CH7.9 and 7.10 Fnal exam 10:30am 12:30pm, Monday, July 13, n ths room Comprehensve exam, covers from CH1.1 what we fnsh ths Thursday, July 8, plus appendces A1 A8 Brng your calculator but DO NOT nput formula nto t! Your phones or portable computers are NOT allowed as a replacement! You can prepare a one 8.5x11.5 sheet (front and back) of handwrtten formulae and values of constants for the exam è no solutons, dervatons or defntons! No addtonal formulae or values of constants wll be provded! 2
Remnder: Specal Project #6 Make a lst of the rated power of all electrc and electronc devces at your home and compled them n a table. (2 ponts each for the frst 10 tems and 1 pont for each addtonal tem.) What s an tem? Smlar electrc devces count as one tem. All lght bulbs make up one tem, computers another, refrgerators, TVs, dryers (har and clothes), electrc cooktops, heaters, mcrowave ovens, electrc ovens, dshwashers, etc. All you have to do s to count add all wattages of the lght bulbs together as the power of the tem Estmate the cost of electrcty for each of the tems (takng nto account the number of hours you use the devce) on the table usng the electrcty cost per kwh of the power company that serves you and put them n a separate column n the above table for each of the tems. (2 ponts each for the frst 10 tems and 1 pont each addtonal tems). Clearly wrte down what the unt cost of the power s per kwh above the table. Estmate the the total amount of energy n Joules and the total electrcty cost per month and per year for your home. (5 ponts) Due: Begnnng of the class Monday, July 13 3
Specal Project Spread Sheet PHYS1441-001, Summer 15, Specal Project #6 Download ths spread sheet from URL: http://www-hep.uta.edu/~yu/teachng/summer15-1441-001/ Just clck the fle wth the name: sp6-spreadsheet.xlsx Wrte down at the top your name and the charge per kwh by your electrcty company Item Names Lght Bulbs Heaters Fans Ar Condto ner Rated power (W) 30, 40, 60, 100, etc Number of devces 40 Average usage: Number of Hours per day Power Consumpton (kwh) Daly Energy Usage (J) Energy cost ($) Power Consumpton (kwh) Monthly Energy Usage (J) Energy cost ($) Power Consumpton (kwh) Yearly Energy Usage (J) Energy cost ($) Frdgers, Freezers Comput ers Game consoles Total 0 0 0 0 0 0 0 0 0 4
Center of Mass We ve been solvng physcal problems treatng objects as szeless ponts wth masses, but n realstc stuatons objects have shapes wth masses dstrbuted throughout the body. Center of mass of a system s the average poston of the system s mass and represents the moton of the system as f all the mass s on that pont. What does above statement tell you concernng the forces beng exerted on the system? The total external force exerted on the system of total mass M causes the center of mass to move at an acceleraton gven by a F / M as f the entre mass of the system s on the center of mass. Consder a massless rod wth two balls attached at ether end. m 1 m 2 The poston of the center of mass of ths system s x 1 x 2 the mass averaged poston of the system x CM mx 1 1+ mx 2 2 xcm CM s closer to the m1+ m2 heaver object 5
Moton of a Dver and the Center of Mass Dver performs a smple dve. The moton of the center of mass follows a parabola snce t s a projectle moton. Dver performs a complcated dve. The moton of the center of mass stll follows the same parabola snce t stll s a projectle moton. The moton of the center of mass of the dver s always the same. 6
Ex. 7 12 Center of Mass Thee people of roughly equvalent mass M on a lghtweght (ar-flled) banana boat st along the x axs at postons x 1 1.0m, x 2 5.0m, and x 3 6.0m. Fnd the poston of CM. CM M 1.0 + M 5.0 + M 6.0 12.0M M + M + M 3M x Usng the formula for CM m x m 7 4.0( m)
y2 m 1 (0,2) Example for Center of Mass n 2-D A system conssts of three partcles as shown n the fgure. Fnd the poston of the center of mass of ths system. r CM x CM y CM (x CM,y CM ) (1,0) m 2 x1 m x m m y m (2,0) m 3 x2 m1x 1 + m2x2 + m3x m + m + m 1 m1 y1 + m2 y2 + m3 y m + m + m 1 2 2 3 Usng the formula for CM for each poston vector component x One obtans 3 3 m m + m 3 1 m 2 + 2 1 2 CM m3 + m r CM 3 2m1 + m + m 2 3 m x m x CM If y CM + y CM j m1 2kg; m2 m3 1kg r CM 3 + 4 j 0.75 + j 4 m y m ( m + 2m ) + 2m 2 3 1 j m 1 + m 2 + m 3 8
Velocty of the Center of Mass Δ x cm v cm Δ mδ x + m Δx m + m 1 1 2 2 x cm Δt 1 2 m1δx1 Δ t+ m2δx2 Δ t m + m 1 2 mv m + mv + m 1 1 2 2 1 2 In an solated system, the total lnear momentum does not change, therefore the velocty of the center of mass does not change. 9
Another Look at the Ice Skater Problem Startng from rest, two skaters push off aganst each other on ce where frcton s neglgble. One s a 54-kg woman and one s a 88-kg man. The woman moves away wth a velocty of +2.5 m/s. What are the man s velocty and that of the CM? v 10 0 v cm0 m s mv m + mv + m 1 1 2 2 1 2 v 20 0 0 m s v1 f + 2.5m s v2 f 1.5m s mv 1 1f + mv 2 2 f v cmf m1+ m2 54 + ( 2.5) + 88 ( 1.5) 3 0.02 0ms 54 + 88 142 10
Center of Mass and Center of Gravty The center of mass of any symmetrc object les on the axs of symmetry and on any plane of symmetry, f the object s mass s evenly dstrbuted throughout the body. How do you thnk you can determne the CM of the objects that are not symmetrc? Center of Gravty Δm g Δm F g CM Axs of One can use gravty to locate CM. symmetry 1. Hang the object by one pont and draw a vertcal lne followng a plum-bob. 2. Hang the object by another pont and do the same. 3. The pont where the two lnes meet s the CM. Snce a rgd object can be consdered as a collecton of small masses, one can see the total gravtatonal force exerted on the object as What does ths equaton tell you? F Δm g M g The net effect of these small gravtatonal forces s equvalent to a sngle force actng on a pont (Center of Gravty) wth mass M. The CoG s the pont n an object as f all the gravtatonal force s actng on! 11
Rotatonal Moton and Angular Dsplacement In the smplest knd of rotaton, a pont on a rgd object move on crcular paths around an axs of rotaton. The angle swept out by the lne passng through any pont on the body and ntersectng the axs of rotaton perpendcularly s called the angular dsplacement. Δ θ θ θ o It s a vector!! So there must be a drecton How do we defne drectons? +:f counter-clockwse -:f clockwse The drecton vector ponts gets determned based on the rght-hand rule. These are just conventons!! 12
SI Unt of the Angular Dsplacement θ (n radans) Arc length Radus s r θ 2π r r For one full revoluton: Snce the crcumference of a crcle s 2πr 2 π rad Dmenson? None 2 π rad 360 o One radan s an angle subtended by an arc of the same length as the radus! 13
Unt of the Angular Dsplacement How many degrees are n one radan? 1 radan s 1 rad 360 2πrad How radans s one degree? And one degrees s 2π 1 o 1 o 360 1rad 180 π π o 180 1 o o 180 3.14 3.14 o 1 180 How many radans are n 10.5 revolutons? 10.5rev 10.5rev 2π rad rev o o 57.3 0.0175rad 21π rad ( ) Very mportant: In solvng angular problems, all unts, degrees or revolutons, must be converted to radans. 14
Example 8-2 A partcular brd s eyes can just dstngush objects that subtend an angle no smaller than about 3x10-4 rad. (a) How many degrees s ths? (b) How small an object can the brd just dstngush when flyng at a heght of 100m? (a) One radan s 360 o /2π. Thus ( 3 10 4 rad ) ( 360 o 2π rad ) 4 3 10 rad l 0.017 (b) Snce lrθ and for small angle arc length s approxmately the same as the chord length. rθ 100m 3 10 4 rad 2 3 10 m 3 cm 15 o
Ex. Adjacent Synchronous Satelltes Synchronous satelltes are put nto an orbt whose radus s 4.23 10 7 m. If the angular separaton of the two satelltes s 2.00 degrees, fnd the arc length that separates them. What do we need to fnd out? The arc length!!! Convert degrees to radans s θ (n radans) Arc length Radus s r 2 π rad 2.00deg 0.0349 rad 360 deg rθ ( 7 4.23 10 m)( 0.0349 rad) 6 1.48 10 m (920 mles) 16
Ex. A Total Eclpse of the Sun The dameter of the sun s about 400 tmes greater than that of the moon. By concdence, the sun s also about 400 tmes farther from the earth than s the moon. For an observer on the earth, compare the angle subtended by the moon to the angle subtended by the sun and explan why ths result leads to a total solar eclpse. θ (n radans) Arc length Radus s r I can even cover the entre sun wth my thumb!! Why? Because the dstance (r) from my eyes to my thumb s far shorter than that to the sun. 17
Angular Dsplacement, Velocty, and Acceleraton Angular dsplacement s defned as Δθ How about the average angular velocty, the rate of change of angular dsplacement? Unt? rad/s θ θ f By the same token, the average angular acceleraton, rate of change of the angular velocty, s defned as Unt? rad/s 2 Dmenson? [T -1 ] Dmenson? [T -2 ] ω α θ f θ θ t ω t f f f f θ t ω t Δθ Δt Δω Δt When rotatng about a fxed axs, every partcle on a rgd object rotates through the same angle and has the same angular speed and angular acceleraton. 18
Ex. Gymnast on a Hgh Bar A gymnast on a hgh bar swngs through two revolutons n a tme of 1.90 s. Fnd the average angular velocty of the gymnast. What s the angular dsplacement? 2 π rad Δ θ 2.00 rev 1 rev 12.6 rad Why negatve? Because he s rotatng clockwse!! ω 12.6 rad 1.90 s 6.63rad s 19
Ex. A Jet Revvng Its Engnes As seen from the front of the engne, the fan blades are rotatng wth an angular velocty of -110 rad/s. As the plane takes off, the angular velocty of the blades reaches -330 rad/s n a tme of 14 s. Fnd the angular acceleraton, assumng t to be constant. α ω t f f ω t Δω Δt ( 330rad s) ( 110rad s) 2 16rad s 14 s 20
Rotatonal Knematcs The frst type of moton we have learned n lnear knematcs was under a constant acceleraton. We wll learn about the rotatonal moton under constant angular acceleraton (α), because these are the smplest motons n both cases. Just lke the case n lnear moton, one can obtan Angular velocty under constant angular acceleraton: Translatonal knematcs v vo + at Angular dsplacement under ω ω + αt f 0 θ θ + ω t + 1 constant angular acceleraton: f 0 0 2 αt2 Translatonal knematcs x f x + v t + at 1 0 o 2 One can also obtan ω 2 Translatonal 2 2 v f f v + 2a( x x ) knematcs o f 2 ω 2 0 + ( ) 2α θ θ f 0 21
Rotatonal Knematcs Problem Solvng Strategy Vsualze the problem by drawng a pcture. Wrte down the values that are gven for any of the fve knematc varables and convert them to SI unts. Remember that the unt of the angle must be n radans!! Verfy that the nformaton contans values for at least three of the fve knematc varables. Select the approprate equaton. When the moton s dvded nto segments, remember that the fnal angular velocty of one segment s the ntal velocty for the next. Keep n mnd that there may be two possble answers to a knematcs problem. 22
Example for Rotatonal Knematcs A wheel rotates wth a constant angular acceleraton of +3.50 rad/s 2. If the angular velocty of the wheel s +2.00 rad/s at t 0, a) through what angle does the wheel rotate n 2.00s? Usng the angular dsplacement formula n the prevous slde, one gets θ f θ ωt+ 1 2 αt2 2.00 2.00 + 1 2 3.50 ( 2.00) 2 11.0rad 11.0 2π rev. 1.75rev. 23
Example for Rotatonal Knematcs cnt d What s the angular velocty at t2.00s? Usng the angular speed and acceleraton relatonshp ω f ω + αt Fnd the angle through whch the wheel rotates between t2.00 s and t3.00 s. Usng the angular knematc formula At t2.00s At t3.00s Angular dsplacement θ t 2 2.00 + 3.50 2.00 +9.00rad / s θ t 3 2.00 3.00 + 1 3.50 3.00 2 Δθ θ 3 θ 2 10.8rad rev. 1.72rev. θ ωt + 1 2 αt 2 f θ 2.00 2.00 + 1 2 3.50 2.00 11.0rad ( ) 2 21.8rad 10.8 2 π 24
Ex. Blendng wth a Blender The blade s whrlng wth an angular velocty of +375 rad/s when the puree button s pushed n. When the blend button s pushed, the blade accelerates and reaches a greater angular velocty after the blade has rotated through an angular dsplacement of +44.0 rad. The angular acceleraton has a constant value of +1740 rad/s 2. Fnd the fnal angular velocty of the blade. θ α ω ω o t +44.0rad +1740rad/s 2? +375rad/s Whch knematc eq? ω ± ω o 2 + 2αθ ω ω + 2αθ 2 2 o ± ( 375rad s) 2 + 2( 1740rad s 2 )( 44.0rad) ±542rad s Whch sgn? ω 542rad s + Why? Because the blade s acceleratng n counter-clockwse! 25