Understanding the Concepts

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Kimberly Turner
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1 Chistian Bache Phsics Depatment Bn Maw College Undestanding the Concepts PHYSICS Homewok Assignment #5 - Solutions 5.7. A cclist making a tun must make use of a centipetal foce, one that is pependicula to his diection of motion. The onl etenal foce able to do this on a flat oad is the foce of fiction between tie and oad, and in leaning into the cuve the cclist is attempting to incease this foce b ting to put the bike into a position whee it would slide out if thee wee no fiction. If the cuve is banked, then the cclist can make use of the nomal foce that keeps the biccle fom moving into the oad suface. B tilting the biccle so that it is pependicula to the oad, the cclist avoids intoducing a sidewas fiction foce which in this case would be undesiable All of the statements ae tue. An net foce, fiction o not, will cause an object to acceleate. Fictions can also be used to incease the speed of an object. Static fiction eeted b the oad on the ties, fo eample, is esponsible fo the incease in speed of a ca when its acceleato pedal is pushed. As an eample of kinetic fiction causing the speed of an object to incease, conside a bo that ests on a stationa cat. If the cat is suddenl pulled fowad with sufficient acceleation, the bo will stat to slide fowad as well; although its speed is not as fast as that of the cat, it does incease its speed fom zeo, due to the kinetic fiction it eceives fom the cat A centipetal foce is needed to keep ou moving in a cicle along with the ca. You seat belt o a contact foce fom the ca doo and seat povides this foce. When it is not pesent, ou would move in a staight line; that is, ou would be thown fom the ca The net foce on Tazan is such that he moves in a cicle. The tension in the vine is the onl foce that can act to move Tazan in this wa. The tension at the bottom of the swing must also compensate fo the entie foce of gavit on Tazan, and moeove, the speed is geatest at the bottom of the swing, because the movement has a tangential component due to gavit that inceases Tazan s speed to the bottom. Thus the tension must be geatest at the bottom, and this is the point of geatest dange, meaning the point whee the tension is most likel to eceed the vine s beaking stength Wheneve ou make contact with the cushion thee is alwas an upwad nomal foce acting on ou. The magnitude of this nomal foce depends on ou acceleation. As ou sink into the seat cushion with a cetain initial downwad velocit ou must soon be bought to est, so thee is a net upwad foce on ou (causing an upwad acceleation), so the nomal foce eeted b the seat cushion eceeds ou weight. As ou settle into the seat and stops sinking the net foce on ou becomes zeo, and the nomal foce fom the cushion is now equal to ou weight (assuming, fo simplicit, that ou feet do not touch the gound, as that would intoduce anothe nomal foce fom the floo).

2 Solve ouself poblems The angula speed can be found fom ω = v/ = (18 m/s)/0.35 m = 51.4 ad/s. (a) The angle of otation is θ = ωt = (51.4 ad/s)(0.1) s = 5.14 ad = 95. Then (, ) = ( cos θ, sin θ = [(0.35 m) cos 95, (0.35 m) sin 95 ] = (0.15 m, 0.3 m). (b) The magnitude of the acceleation is v / = (18 m/s) /(0.35 m) = m/s and its diection is centipetal. At t = 0 s, the acceleation is a 0 = ( m/s ) i. (c) At θ = 90, the acceleation is a = ( m/s ) j Because F ρ AC v, at teminal speed we have mg ρ AC v, so D = 1 D A = mg/ρc D v = (0.50 kg)(9.8 m/s )/(1.5 kg/m 3 )(0.4)(18 m/s) = m = 600 cm. = 1 D Because the ock has a centipetal acceleation povided b the tension, we can wite: F = T = ma = mv / ; thus the speed will be maimum when the tension is maimum: 6 N = (0.0 kg)v ma /(0.35 m), which gives v ma = 6.4 m/s We wite F = ma fom the foce diagam fo the hanging mass, with down

3 positive: m 1 g T = m 1 a 1 = 0; which gives T = m 1 g = (1.00 kg)(9.8 m/s ) = 9.80 N. Fo the otating puck, the tension povides the centipetal acceleation, F = ma : T = m v /; 9.80 N = (0.400 kg)v /(0.80 m), which gives T T F N v = 4.43 m/s. m 1 g m g puck (a)no. The most likel point whee the sting will beak is at the bottom of the cicula tack, not at the top. The least likel point fo the sting to beak is at the top of the tack, whee the tension in the sting is the lowest in magnitude. (b) At the bottom of the tack T mg = ma = mv /R. Set T = 40 N to obtain v = (TR/m gr) 1/ = [(40 N)(1.0 m)/(0.150 kg) (9.8 m/s )(1.0 m)] 1/ = 16 m/s At the top of the loop, both the nomal foce and the weight ae downwad. We wite ΣF = ma fom the foce diagam fo the motoccle: -component: F + mg = N mv /R. The speed v will be minimum when the nomal foce is minimum. The nomal foce can onl push awa fom the amp, that is, with ou coodinate sstem it must be positive, so F = 0. N min

4 Thus we have v min = gr, o v min = (gr) 1/ m g F N = [(9.8 m/s )(1 m)] 1/ = 11 m/s ( 4 mi/h). R Hand-in Poblems The adius of the cicula path is R = R E + h = km + 0 km = km. (a) The speed is v = πr/t = π( m)/[(89 min)(60 s/min)] = m/s = km/h. (b) The acceleation is a = v /R = ( m/s) /( m) = 9.1 m/s towad Eath s cente We assume that the static fiction foce is the maimum, without skidding. The acceleation can be found fom the ca s one-dimensional motion: v = v 0 + at; 0 = 5 m/s + a(4. s), which gives a = 5.6 m/s. We wite ΣF = ma fom the foce diagam fo the ca: -component: f s = ma -component: FN mg = 0. F N f m g s Thus FN = Mg and µ s = a/g = ( 5.6 m/s )/(9.8 m/s ) = 0.61.

5 5.51. If the automobile does not skid, the fiction is static, with f s µ s FN. At high speed f s will be down the incline. Note that we take a coodinate sstem with the -ais in the diection of the f s θ F N m g θ a centipetal acceleation. We wite F = ma fom the foce diagam fo the auto: -component: FN sin θ + f s cos θ = ma = mv /R; -component: FN cos θ f s sin θ mg = 0. The speed is maimum when f s = f s,ma = µ s N. Fom the -equation we get FN cos θ µ s FN sin θ = mg, o FN = mg/(cos θ µ s sin θ). Fom the -equation we get v ma /R = g(sin θ + µs cos θ)/(cos θ µ s sin θ). v ma = (150 m)(9.8 m/s )(sin cos 18 )/(cos sin 18 ), which gives v ma = 3 m/s. At low speed, the automobile will tend to slide down the incline, so f s will be up the incline. The speed is minimum when f s = f s,ma = µ s FN. If we change the sign of f s in the equations, we get FN = mg/(cos θ + µ s sin θ) and v min /R = mg(sin θ µs cos θ)/(cos θ + µ s sin θ).

6 Thus v min = (150 m)(9.8 m/s )(sin cos 18 )/(cos sin 18 ), which gives v min = 5.8 m/s (a) θ T m g (b) The mass moves in a cicle of adius L sin θ and has a centipetal acceleation. We wite F = ma fom the foce diagam fo the mass: -component: T sin θ = mv /; -component: T cos θ mg = 0. Combining these, we get v g tan θ = ( sin θ) g tan θ = 1 4 = (0.5 m)(si n 10 )(9.8 m /s )(tan 10 ), which gives v = 0.4 m/s The static fiction foce povides the centipetal acceleation. We wite ΣF = ma fom the foce diagam fo the biccle: -component: f s = mv /R; -component: FN mg = 0. m g F N a The shotest tun (smallest R) equies f s,ma = µ s FN.

7 Thus R min = mv /f s,ma = mv /µ s mg = v /µ s g = (10 m/s) /0.4(9.8 m/s ) = 6 m.

Physics 101 Lecture 6 Circular Motion

Physics 101 Lecture 6 Circular Motion Physics 101 Lectue 6 Cicula Motion Assist. Pof. D. Ali ÖVGÜN EMU Physics Depatment www.aovgun.com Equilibium, Example 1 q What is the smallest value of the foce F such that the.0-kg block will not slide