Chapte 7 Rotational Motion Angles, Angula Velocity and Angula Acceleation Univesal Law of Gavitation Keple s Laws
Angula Displacement Cicula motion about AXIS Thee diffeent measues of angles: 1. Degees. Revolutions (1 ev. 360 deg.) 3. Radians (π ad.s 360 deg.)
Angula Displacement, cont. Change in distance of a point: s π θ N (N counts evolutions) ( θ is in adians)
Example An automobile wheel has a adius of 4 cm. If a ca dives 10 km, though what angle has the wheel otated? a) In evolutions b) In adians c) In degees
Solution Note distance ca moves distance outside of wheel moves a) Find N: Basic fomula s πn θ b) Find θ in adians Known: N c) Find θ in degees Known: N Known: s 10 000 m, 0.4 m N s π 3 789 θ π (adians/evolution) N θ.38 x 10 4 ad. θ 360(degees/evolution) N θ 1.36 x 10 6 deg
Angula Speed Can be given in Revolutions/s Radians/s --> Called ω Degees/s ω θ f θ i in adians t Linea Speed at v θ π π 360 π π f θ θ f f θ i θ θ i i (in evolutions) t (in degees) t (in adians) ω t
Example A ace ca engine can tun at a maximum ate of 1 000 pm. (evolutions pe minute). a) What is the angula velocity in adians pe second. b) If helipcopte blades wee attached to the cankshaft while it tuns with this angula velocity, what is the maximum adius of a blade such that the speed of the blade tips stays below the speed of sound. DATA: The speed of sound is 343 m/s
Solution a) Convet pm to adians pe second ev. min 1000 sec 60 min π ad ev b) Known: v 343 m/s, ω 156 ad./s Find Basic fomula v ω v ω 156 adians/s.7 m
Angula Acceleation Denoted by α α ω ω f i t ω must be in adians pe sec. Units of angula acceleation ae ad/s² Evey potion of the object has same angula speed and same angula acceleation
Analogies Between Linea and Rotational Motion Rotational Motion θ ( ω ) i + ω f t 1 θ ω it + α t ω ω + αt i ω ω + α θ x Linea Motion ( v v ) i + f t x v t + v v a x i i + i 1 v v + i at at
Linea movement of a otating point Distance Speed s θ v ω Acceleation a α Diffeent points on the same object have diffeent linea motions! Only woks when θ, ω and α ae in adians!
Example A pottey wheel is acceleated unifomly fom est to a ate of 10 pm in 30 seconds. a.) What was the angula acceleation? (in ad/s ) b.)how many evolutions did the wheel undego duing that time?
Solution Fist, find the final angula velocity in adians/s. ω f ev. 1 ad 10 π min 60( sec/ min) ev a) Find angula acceleation Basic fomula ω ω + αt f i α ω f ω i t 1.047 ad sec 0.0349 ad./s b) Find numbe of evolutions: Known ω i 0, ω f 1.047, and t 30 Fist find θ in adians Basic fomula θ ω i + ω f t ω f θ t 15.7 ad. θ (ad.) N π (ad./ev.).5 ev.
Solution b) Find numbe of evolutions: Known ω i 0, ω f 1.047, and t 30, Fist find θ in adians Basic fomula θ ω i + ω f t ω f θ t 15.7 ad. θ (ad.) N π (ad./ev.).5 ev.
Example A coin of adius 1.5 cm is initially olling with a otational speed of 3.0 adians pe second, and comes to a est afte expeiencing a slowing down of α 0.05 ad/s. a.) Ove what angle (in adians) did the coin otate? b.) What linea distance did the coin move?
Solution ω a) Find θ, Given ω i 3.0 ad/s, ω f 0, α -0.05 ad/s Basic fomula f i ω + α θ θ ωi α 90 adians 90/π evolutions b) Find s, the distance the coin olled Given: 1.5 cm and θ 90 ad Basic fomula s θ s θ, ( θ is in ad.s) 135 cm
Centipetal Acceleation Moving in cicle at constant SPEED does not mean constant VELOCITY Centipetal acceleation esults fom CHANGING DIRECTION of the velocity
Centipetal Acceleation, cont. Acceleation is diected towad the cente of the cicle of motion Basic fomula a v t
Deivation: a ω v / Fom the geomety of the Figue v v sin( θ / ) v θ fo small θ Fom the definition of angula velocity θ v a ω ω t v t v t v ω ω v
Foces Causing Centipetal Acceleation Newton s Second Law F ma Radial acceleation equies adial foce Examples of foces Spinning ball on a sting Gavity Electic foces, e.g. atoms
Example A space-station is constucted like a babell with two 1000-kg compatments sepaated by 50 metes that spin in a cicle (5 m). The compatments spins once evey 10 seconds. a) What is the acceleation at the exteme end of the compatment? Give answe in tems of g s. b) If the two compatments ae held togethe by a cable, what is the tension in the cable?
Solution a) Find acceleation a Given: T 10 s, 5 m Basic fomula ω π ad ev N ev s Fist, find ω in ad/s ω π 0.1 Basic fomula a v ω Then, find acceleation a ω 9.87 m/s 1.006 g
Solution b) Find the tension Given m 1000 kg, a 1.006 g Basic fomula F ma T ma 9870 N
Example A ace ca speeds aound a cicula tack. a) If the coefficient of fiction with the ties is 1.1, what is the maximum centipetal acceleation (in g s) that the ace ca can expeience? b) What is the minimum cicumfeence of the tack that would pemit the ace ca to tavel at 300 km/h?
Solution a) Find the maximum centipetal acceleation Known: µ 1.1 Remembe, only conside foces towads cente Basic fomula f µn F ma f µ mg ma µ mg a µ g Maximum a 1.1 g
Solution b) Find the minumum cicumfeence Known: v 300 km/h 83.33 m/s, a 1.1 g Basic fomula a v ω Fist, find adius v a Then, find cicumfeence L π 4 043 m In the eal wold: tacks ae banked
Example A yo-yo is spun in a cicle as shown. If the length of the sting is L 35 cm and the cicula path is epeated 1.5 times pe second, at what angle θ (with espect to the vetical) does the sting bend?
Solution Basic fomula F ma Apply Fma fo both the hoizontal and vetical components. may Fy T cos θ 0 T cosθ mg mg Basic fomula a ω max Fx mω L mω T sinθ T mω Lsinθ Lsinθ
Solution We want to find θ, given ωπ 1.5 & L0.35 max Fx mω L mω T sinθ T mω Lsinθ may Fy T cos θ 0 T cosθ mg mg eq.s & unknowns (T and θ) mg g cosθ T ω L θ 71 degees
Acceleating Refeence Fames Conside a fame that is acceleating with a f F F ma f ma m ( a a f ) Fictitious foce Looks like gavitational foce If fame acceleation g, fictitious foce cancels eal gavity. Examples: Falling elevato, planetay obit otating space stations
DEMO: FLYING POKER CHIPS
Example Which of these astonauts expeiences zeo gavity? a) An astonaut billions of light yeas fom any planet. b) An astonaut falling feely in a boken elevato. c) An astonaut obiting the Eath in a low obit. d) An astonaut fa fom any significant stella object in a apidly otating space station
Newton s Law of Univesal Gavitation Foce is always attactive Foce is popotional to both masses Foce is invesely popotional to sepaation squaed G m1m F G 3 11 m 6.67 10 kg s
Gavitation Constant Detemined expeimentally Heny Cavendish, 1798 Light beam / mio amplify motion
Example Given: In SI units, G 6.67x10-11, g9.81 and the adius of Eath is 6.38 x10 6. Find Eath s mass: F Basic fomula Mm G mg mg G M gr G Mm R 5.99x10 4 kg
Example Given: The mass of Jupite is 1.73x10 7 kg and Peiod of Io s obit is 17 days Find: Radius of Io s obit
Solution Given: T 17 4 36001.47x10 6, M1.73x10 7, G6.67x10-11 Find: Basic fomula ω π F N Fist, find ω fom the peiod 1 6 ω π 4.8 10 s t T Basic fomula ma m Basic fomula F G Mm ω Next, solve fo M ω G 3 GM ω 1.84x10 9 m
Tycho Bahe (1546-1601) 1601) Lost pat of nose in a duel EXTREMELY ACCURATE astonomical obsevations, nealy 10X impovement, coected fo atmosphee Believed in Retogade Motion Hied Keple to wok as mathematician
Johannes Keple (1571-1630) 1630) Fist to: Explain planetay motion Investigate the fomation of pictues with a pin hole camea; Explain the pocess of vision by efaction within the eye Fomulate eyeglass designed fo neasightedness and fasightedness; Explain the use of both eyes fo depth peception. Fist to descibe: eal, vitual, upight and inveted images and magnification
Johannes Keple (1571-1630) 1630) Fist to: explain the pinciples of how a telescope woks discove and descibe total intenal eflection. explain that tides ae caused by the Moon. He tied to use stella paallax caused by the Eath's obit to measue the distance to the stas; the same pinciple as depth peception. Today this banch of eseach is called astomety. suggest that the Sun otates about its axis deive the bith yea of Chist, that is now univesally accepted. deive logaithms puely based on mathematics,
Isaac Newton (164-177) 177) Invented Calculus Fomulated the univesal law of gavitation Showed how Keple s laws could be deived fom an invese-squae-law foce Invented Wave Mechanics Numeous advances to mathematics and geomety
Keple s Laws 1. All planets move in elliptical obits with the Sun at one of the focal points.. A line dawn fom the Sun to any planet sweeps out equal aeas in equal time intevals. 3. The squae of the obital peiod of any planet is popotional to cube of the aveage distance fom the Sun to the planet.
Keple s Fist Law All planets move in elliptical obits with the Sun at one focus. Any object bound to anothe by an invese squae law will move in an elliptical path Second focus is empty
Keple s Second Law A line dawn fom the Sun to any planet will sweep out equal aeas in equal times Aea fom A to B and C to D ae the same This is tue fo any cental foce due to angula momentum consevation (next chapte)
Keple s Thid Law The squae of the obital peiod of any planet is popotional to cube of the aveage distance fom the Sun to the planet. T 3 K sun Fo obit aound the Sun, K S.97x10-19 s /m 3 K is independent of the mass of the planet
Deivation of Keple s Thid Law Basic fomula F a ω ma Basic fomula ω π T G Mm ma mω 3 T 3 G G GM T 4π 4π GM Mm Mm K sun
Example Data: Radius of Eath s obit 1.0 A.U. Peiod of Jupite s obit 11.9 yeas Peiod of Eath s obit 1.0 yeas Find: Radius of Jupite s obit Basic fomula T K 3 sun T Eath 3 Eath 3 Jupite T Jupite Jupite 3 Jupite 3 Eath Eath T T Jupite Eath T T Jupite Eath /3 5. A.U.
Gavitational Potential Enegy PE mgy is valid only nea the Eath s suface Fo abitay altitude PE G M Em Zeo efeence level is infinitely fa fom the eath
Gaphing PE vs. position PE G M Em
Example You wish to hul a pojectile is huled fom the suface of the Eath (R e 6.3x10 6 m) to an altitude of 0x10 6 m above the suface of the Eath. Ignoe the otation of the Eath and ai esistance. a) What initial velocity is equied? b) What velocity would be equied in ode fo the pojectile to each infinitely high? I.e., what is the escape velocity?
Solution Given: R 0 6.3x10 6, R 6.3x10 6, G, M 6.0x10 4 Find: v 0 Fist, get expession fo change in PE Basic fomula PE G Mm PE 1 GMm 0 Then, apply enegy consevation Basic fomula KE 1 mv PE KE 1 1 GMm 0 1 1 mv Finally, solve fo v 0 6 600 m/s 0
Solution Fo Escape Velocity Given: R 0 6.3x10 6, R, G, M 6.0x10 4 Find: v 0 Fist, get expession fo change in PE Basic fomula PE G Mm 1 PE GMm Then, apply enegy consevation Basic fomula KE 1 mv PE KE 1 1 GMm 0 0 mv Solve fo v 0 11 70 m/s 0