As observed from the frame of reference of the sidewalk:

Transcription

1 Section 3.1: Inetial and Non-inetial Fames of Refeence Tutoial 1 Pactice, page (a) When the ca is moving with constant velocity, I see the ball lie still on the floo. I would see the same situation when the ca is at est. (b) To an obseve on the sidewalk, the ball appeas to be moving with onstant velocity of 14 m/s [E]. (c) If the ca acceleates fowad, I see the ball oll backwad on the floo. (d) As obseved fom the fame of efeence of the ca: As obseved fom the fame of efeence of the sidewalk: The ca s fame of efeence is non-inetial. I obseve a fictitious foce (the foce pushing the ball backwad, west) in the ca s fame of efeence.. (a) Given: m.0 g 0.00 kg; θ 3.5 Requied: a Analysis: Look at the situation fom an Eath (inetial) fame of efeence. The hoizontal component of the tension F T balances the acceleation and the vetical component of the tension F T balances the gavitational foce. Expess the components of the tension in tems of the hoizontal and vetical applied foces. Then calculate the magnitude of the acceleation. Solution: Vetical component of foce:!f y 0 F T cos" mg 0 F T mg cos" Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.1-1

2 Hoizontal component of foce:!f x ma F T sin" ma mg & cos" ' ( sin" ma g sin" & cos" ' ( a g tan" a a 9.8 m/s tan3.5 a 6. m/s Statement: The magnitude of the boat s acceleation is 6. m/s. I did not need to know the mass of the ball to make the calculation because that value was cancelled out to obtain the foces. (b) Given: m.0 g 0.00 kg; θ 3.5 Requied: F T Analysis: Fom pat (a), F T mg cos!. Solution: F T mg cos! 0.00 kg ( 9.8 m/s ) cos3.5 F T 0.6 N Statement: The magnitude of the tension in the sting is 0.6 N. I need to know the mass to make this calculation because the foce is the mass multiplied by the acceleation. 3. (a) When the subway is moving at onstant velocity, thee is no tension in the stap. (b) Given: m 14 kg; θ 35 ; a 1.4 m/s Requied: F T Analysis: Look at the situation fom an Eath (inetial) fame of efeence. The hoizontal component of the tension F T balances the acceleation, so expess the x-components of the tension in tems of the hoizontal applied foces. The vetical foces of gavity and the nomal foce balance each othe.!f x ma F T sin" ma F T ma sin" Solution: F T ma sin! ( 14 kg) ( 1.4 m/s ) tan35 F T 34 N Statement: The tension on the stap duing acceleation is 34 N. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.1-

3 4. Given: µ s 0.4 Requied: maximum a Analysis: The foce due to the acceleation of the tain, F ma, must be less than the foce of static fiction, F s µ s, whee mg mg. So the maximum acceleation occus when F F s. F F s ma! s ma! s ( mg) a! s g Solution: a! s g 9.8 m/s 0.4 a 4.1 m/s Statement: The maximum acceleation of the tain befoe the passenge begins to slip along the floo is 4.1 m/s. Tutoial Pactice, page (a) Given: m 55 kg; a!.9 m/s [up] Requied: Analysis: Use up as positive and solve fo the nomal foce when + + ( mg) ma. + + (!mg) ma ma + mg Solution: ma + mg.9 m/s 55 kg + 55 kg 9.8 m/s 7.0!10 N Statement: The student s appaent weight is N. (b) Given: m 55 kg; a!.9 m/s [down] Requied: Analysis: Use down as positive and solve fo the nomal foce when + (mg) ma. + (mg) ma ma mg Solution: mg! ma 9.8 m/s 55 kg! 55 kg.9 m/s 3.8 "10 N Statement: The student s appaent weight is N. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.1-3

4 . (a) Given: m kg; m.5 kg; 70.0 N Requied: a! Analysis: Use up as positive and solve fo the acceleation when + + ( mg) ma. + + (!mg) ma a F! mg N m F! m N ( + m 1 )g m 1 + m Solution: a F! m N ( + m 1 )g m 1 + m 9.8 m/s 70.0 kg " m! 9.5 kg +.5 kg s 9.5 kg +.5 kg!3.967 m/s (two exta digits caied) a!4.0 m/s Statement: The acceleation of the elevato is 4.0 m/s [down]. (b) Given: m.5 kg;! a m/s [down] Requied:! Analysis: Use down as positive and solve fo the nomal foce when + (mg) ma. Solution:! + (mg) ma mg! ma 9.8 m/s.5 kg 15 N!.5 kg m/s Statement: The foce on the smalle box is 15 N [up]. 3. (a) Given: m 1 4. kg; m.6 kg; a 0 Requied: F TA ; F TB Analysis: Since this is an inetial fame of efeence, the tension on ope B balances the foce of gavity on block. The tension on ope A balances the foce of gavity on block 1 and the tension on ope B. F g mg Solution: Detemine the foce on ope B: F TB m g ( 9.8 m/s ).6 kg 5.48 N (two exta digits caied) F TB 5 N Detemine the foce on ope A: F TA m 1 g + F TB 4. kg ( 9.8 m/s ) N F TA 67 N Statement: The tension on the ope A is 67 N and the tension on the ope B is 5 N. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.1-4

5 (b) Given: m 1 4. kg; m.6 kg a! 1. m/s [up] Requied: F TA ; F TB Analysis: This is now a non-inetial fame of efeence, so instead of just gavity, the acceleation is g ( a), o g + a, whee down is positive. Solution: Detemine the foce on ope B: F TB m g + a ( 9.8 m/s + 1. m/s ).6 kg 8.6 N (one exta digit caied) F TB 9 N Detemine the foce on ope A: F TA m 1 g + a + F TB ( 9.8 m/s + 1. m/s ) N 4. kg F TA 75 N Statement: The tension on the ope A is 75 N and the tension on the ope B is 9 N. 4. Given:! a 0.98 m/s [down]; m 61 kg Requied: Analysis: Use down as positive and solve fo the nomal foce when + (mg) ma. Solution:! + (mg) ma mg! ma 9.8 m/s 61 kg! 61 kg 0.98 m/s 5.4 "10 N Statement: The passenge s appaent weight is m/s. Section 3.1 Questions, page (a) The ball would appea to move staight up and down because I am moving with the same velocity as the othe tain. Fom my viewpoint, the othe tain and passenge ae standing still, and the ball is not affected by the tain s motion. (b) If the tains moved in opposite diections, the ball would appea to have hoizontal motion, so I would see the path as a paabola.. Given: a 1.5 m/s Requied: θ Analysis: The hoizontal component of the tension F T balances the acceleation, and the vetical component of the tension F T balances the gavitational foce. Expess the tangent atio of the angle in tems of the applied foce and the gavitational foce, then solve fo the angle. tan! F a F g ma &! tan "1 mg ' ( a &! tan "1 g ' ( Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.1-5

6 a & Solution:! tan "1 g ' ( m & 1.5 tan "1 s ( 9.8 m ( ( s ' (! 8.7 Statement: The sting makes an 8.7 angle with the vetical. 3. Given: v f 55 km/h; t 10.0 s Requied: θ Analysis: Detemine the acceleation using v f a t o a v f. The hoizontal component of the!t tension F T balances the acceleation and the vetical component of the tension F T balances the gavitational foce. Expess the tangent atio of the angle in tems of the applied foce and the gavitational foce, then solve fo the angle. Solution: Detemine the acceleation of the plane: a v f!t 55 km h! 1000 m 10.0 s 1 km! 1 h 60 min! 1 min 60 s m/s (two exta digits caied) a 7.08 m/s Detemine the angle the sting makes with the vetical: tan! F a F g ma &! tan "1 mg ' ( a & tan "1 g ' ( m & tan "1 s ( 9.8 m ( ( s ' (! 35.9 Statement: The sting makes an 35.9 angle with the vetical. 4. Given: θ 16 Requied: a Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.1-6

7 Analysis: Look at the situation fom an Eath (inetial) fame of efeence. The hoizontal component of the tension F T balances the acceleation and the vetical component of the tension F T must balance the gavitational foce since the cok ball does not move. Expess the components of the tension in tems of the hoizontal and vetical applied foces. Then calculate the magnitude of the acceleation. Vetical component of foce:!f y 0 F T cos" mg 0 F T mg cos" Hoizontal component of foce:!f x ma F T sin" ma mg & cos" ' ( sin" ma g sin" & cos" ' ( a g tan" a Solution: a g tan! ( 9.8 m/s )tan16 a.8 m/s Statement: The magnitude of the ca s acceleation is.8 m/s. 5. Given: v f 6.0 m/s; t 10.0 s; m 64 kg Requied: Analysis: Detemine the upwad acceleation using v f a t o a v f. Use up as positive and!t solve fo the nomal foce when + + ( mg) ma. + +!mg ma ma + mg Solution: Detemine the upwad acceleation: a v f!t 6.0 m/s 10.0 s a 0.60 m/s Detemine the appaent weight: ma + mg 0.6 m/s 64 kg + 64 kg 9.8 m/s 6.7!10 N Statement: The passenge s appaent weight is N. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.1-7

8 6. Given: 55 N; m 5 kg Requied: a Analysis: Use up as positive and solve fo the acceleation when + + ( mg) ma. + +!mg ma a! mg m Solution: a! mg m 9.8 m/s 55 kg " m! 5 kg s 5 kg a!4.9 m/s Statement: The acceleation of the ide is 4.9 m/s [down]. 7. (a) At est: Acceleating downhill: When the ca is at est, the dice ae aligned with the vetical (with espect to level gound), and make an angle of 17 with espect to the nomal (pependicula to the oof). The only foces acting on the dice ae gavity and tension. When the ca is acceleating, a hoizontal fictitious foce pointing to the ea of the ca deflects the dice so that they ae aligned with the nomal, and make an angle of 17 with espect to the vetical, as viewed fom the level gound. (b) Given: θ 17 Requied: a Analysis: Look at the situation fom an inetial fame of efeence on the same angle as the hill. The vetical component of the gavitational foce F g balances the tension and the hoizontal component of the gavitational foce F g balances the foce applied by the acceleation. Expess the hoizontal component of F g in tems of the applied foces. Then calculate the magnitude of the acceleation.!f x 0 F g sin" ma 0 mg sin" ma 0 g sin" a 0 a g sin" Solution: a g sin! 9.8 m/s sin17 a.9 m/s Statement: The magnitude of the ca s acceleation is.9 m/s. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.1-8

9 8. (a) Given: m kg; m 1. kg; m 1. kg; 70.0 N Requied: F a Analysis: Since mass 1 does not slide, the acceleation due to the hoizontal foce must balance the acceleation due to the tension, which equals m g. Use F ma to detemine the acceleation. F a F T m 1 + m + m 3 m 1 F a m g m 1 + m + m 3 m 1 F a m g ( m m 1 + m + m 3 ) 1 Solution: F a m g ( m m 1 + m + m 3 ) 1 ( 1. kg) 9.8 m/s 1.8 kg ( 1.8 kg +1. kg kg) F a 39 N Statement: The applied foce is 39 N. (b) Given: m 1 1. kg; m.8 kg; θ 5 Requied: F a Analysis: Since mass 1 does not slide, the applied foce on mass 1, the gavitational foce, and the nomal foce must balance each othe: ΣF 0. Use F ma to detemine the acceleation. Since the applied foce is entiely hoizontal and the gavitational foce is entiely vetical, use the tangent atio to elate them. Detemine the applied acceleation: F ma F a ( m 1 + m )a F a a m 1 + m Detemine the hoizontal acceleation of mass 1: tan! m a 1 m 1 g a g a g tan! Detemine the applied foce: F a g tan! m 1 + m F a g tan! ( m 1 + m ) Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.1-9

10 Solution: F a g tan! m 1 + m ( 9.8 m/s ) tan5 F a 17 N ( 1. kg +.8 kg) Statement: The applied foce is 17 N. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion

11 Section 3.: Centipetal Acceleation Tutoial 1 Pactice, page Given: 5 km m; v 50.0 m/s Requied: Analysis: v Solution: v ( 50.0 m/s).5!10 4 m 0.10 m/s Statement: The magnitude of the centipetal acceleation is 0.10 m/s.. Given: 1. m; v 4.4 m/s Requied: a! c Analysis: v ; Centipetal acceleation is always diected towad the cente of otation. Since the hamme s velocity is diected south and it is spinning clockwise, the cente of otation is west of the hamme. Solution: v ( 4.4 m/s) ( 1. m) 15 m/s Statement: The centipetal acceleation is 15 m/s [W]. 3. Given: 1.4 m; 1 m/s Requied: v Analysis: v v Solution: v 1.4 m 1 m/s v 4.1 m/s Statement: The speed of the ball is 4.1 m/s. 4. (a) Given: m; m/s Requied: v Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.-1

12 Analysis: 4! T T Solution: T 4! 4! 4! 1.08 "10 11 m m ' 1.1 "10 & s ( ) T 1.95"10 7 s Statement: The peiod of Venus is s. (b) Convet the peiod in seconds to days: T 1.95!10 7 s! 1 min 60 s! 1 h 60 min! 1 d 4 h T 6 days The peiod of Venus is 6 days. 5. Given: v m.s; m Requied: Analysis: v Solution: v 7.7!103 m/s ( 7.54!10 6 m) 7.01 m/s Statement: The magnitude of the centipetal acceleation is 7.01 m/s. 6. (a) Given: m/s ; 8.4 cm m Requied: f Analysis: 4! f f 4! Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.-

13 Solution: f 4! m & 3.3"10 6 ' ( s 4! 8.4 "10 ) m f 1.0 "10 4 Hz Statement: The fequency of the centifuge is Hz. (b) Convet the fequency in hetz to evolutions pe minute: 1 f 1.0! s! 60 s 1 min f 6.0!10 5 pm The fequency of the centifuge is pm. Section 3. Questions, page (a) The tension in the sting povides the foce to keep the puck in its cicula path at constant speed, and so povides the acceleation of the puck. (b) The centipetal acceleation is half as lage because centipetal acceleation depends on the invese of the adius: 1 v. (c) The centipetal acceleation is fou times as geat because centipetal acceleation depends on the squae of the speed: 4 (v).. The centipetal acceleation fo the fist athlete s hamme is fou times geate than that of the second athlete. Centipetal acceleation depends on the squae of the speed: v. So if the hamme spins two times as fast, the centipetal acceleation is, o 4, times lage: 4a. 3. Given: 0.4 m; T 1.5 s Requied: Analysis: 4! T Solution: 4! T 4! ( 0.4 m) ( 1.5 s) 7.4 m/s Statement: The magnitude of the centipetal acceleation of the lasso is 7.4 m/s. 4. Given: v 8 m/s; 135 m Requied: Analysis: v Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.-3

14 Solution: v 8 m/s 135 m 5.8 m/s Statement: The magnitude of the centipetal acceleation is 5.8 m/s. 5. Given: m; T 1 day o s Requied: Analysis: 4! T Solution: 4! T 4! 6.38 "10 6 m ( s) 3.37 "10 m/s Statement: The centipetal acceleation at Eath s equato is m/s. 6. Given: 5 m/s ;.0 m Requied: f Analysis: 4! f f 4! Solution: f 4! " & ' 5 m s 4! (.0 m ) f 0.56 Hz Statement: The minimum fequency of the cylinde is 0.56 Hz. 7. Given: v m/s; 7.8 m/s Requied: Analysis: v v Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.-4

15 Solution: v ( m/s) 7.8 m/s 6 m Statement: The adius of the cuve is 6 m. 8. Given: C 478 m; m/s Requied: v Analysis: C π o C! ; v v " C v! & ' " C Solution: v! & ' ( 478 m ) m/s! m s ( 60 s 1 min ( 60 min 1 h ( 1 km 1000 m v 1.0 km/h Statement: The speed of the jogge is 1.0 km/h. 9. (a) Given: m; f 60.0 pm Requied: T Analysis: T 1 f Solution: Convet the fequency to hetz: f 60.0 pm min! 1 min 60 s f 1.00 Hz Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.-5

16 Detemine the peiod: T 1 f Hz T 1.00 s Statement: The peiod of the bicycle wheel is 1.00 s. (b) Given: m; f 1.00 Hz Requied: a! c Analysis: 4! f ; Centipetal acceleation is always diected towad the cente of otation. Since the wheel s velocity is diected west and it is spinning clockwise, the cente of otation is noth of the point. Solution: 4! f ( 1.00 Hz ) 4! m 11.8 m/s Statement: The centipetal acceleation of a point on the edge of that wheel is 11.8 m/s [N] if it is moving westwad at that instant. 10. (a) Given: T 7.3 days; m/s Requied: Analysis: 4! T Solution: Convet the peiod to seconds: T 7.3 days! 4 h 1 day! 60 min 1 h! 60 s 1 min.3587!10 6 s (two exta digits caied) T.36!10 6 s Detemine the adius: 4! T a T c 4! m '.7 "10 3 ( & s ( ).3587 "106 s ) 4! 3.8 "10 8 m Statement: The adius of the cuve is m. (b) The values ae the same to two significant digits. Any diffeence beyond that may be because the obit is not pefectly cicula o the speed is not constant. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.-6

17 11. (a) Given: 711 m/s ; 1.1 m Requied: v Analysis: v ; v Solution: v ( 711 m/s )( 1.1 m ) v 9.3 m/s Statement: The speed of the hamme is 9.3 m/s. (b) Given: d.0 m; v i 9.3 m/s; θ 4 Requied: d x Analysis: Use v f v i + a d to calculate the y-component of the final speed, then calculate the time of flight v f v i + a t. Finally, calculate the ange using d v t. Solution: Detemine the y-component of the final speed: v f v i + a! d v fy v iy + g! d v fy v iy + g! d (( sin4 )) + ( 9.8 m/s )(.0 m ) 9.3 m/s m/s (two exta digits caied) v fy 19 m/s Detemine the time of flight: v fy v iy + a!t!t v! v fy iy a m s! "!9.3 m s & ' sin m s!t s (two exta digits caied) Detemine the ange of the ball:! d x v x!t v i!t cos! " 9.3 m s & ' s! d x 85 m ( cos4 ) Statement: The ange of the ball is 85 m. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.-7

18 Section 3.3: Centipetal Foce Tutoial 1 Pactice, page Given: m 0.11 kg; 5.6 m; v 1.7 m/s Requied: Analysis: Lift is equivalent to the nomal foce, so it is the sum of the centipetal foce and the gavitational foce; mv ; F g mg + F g mv + mg Solution: mv + mg 0.11 kg 5.6 m 5.9 N 1.7 m/s + ( 0.11 kg) ( 9.8 m/s ) Statement: The lift on the plane at the bottom of the ac is 5.9 N.. Given: 450 m; v 97 km/h Requied: θ Analysis: The vetical component of the nomal balances the gavitational foce and the hoizontal component of the nomal epesents the centipetal foce, mv. Use the tangent atio to detemine the angle the nomal makes with the vetical. tan! x y F g " mv mg & ' v g " v! tan (1 g & ' Solution: Convet the speed to metes pe second: v 97 km h! 1000 m 1 km! 1 h 60 min! 1 min 60 s 6.94 m/s (two exta digits caied) v 7 m/s Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.3-1

19 Detemine the angle: v &! tan "1 g ' ( ( 6.94 m/s) & tan "1 ( ( 450 m) 9.8 m/s '! 9.3 Statement: The banking angle is Given: m.0 kg; f.50 Hz; 4.00 m.00 s Requied: F T Analysis: F T m ; 4! f ; 4! mf Solution: 4! mf 4! (.00 kg) ( 4.00 m) (.50 Hz) F T.0 "10 3 N Statement: The magnitude of the tension in the sting is N. 4. Given: 150 m; F g Requied: v Analysis: F g mg; mv F g mv mg v g v g Solution: v g ( 9.8 m/s ) ( 150 m) v 38 m/s Statement: The speed of the ban swallow is 38 m/s. 5. I pedict that the maximum speed will decease because the oad conditions ae slippey. Given: µ s 0.5;.0 10 m; θ 0.0 Requied: maximum v Analysis: Fom Sample Poblem 3: v g sin! +! cos! & s cos! "! s sin! ( ' Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.3-

20 Solution: v g sin! +! cos! & s cos! "! s sin! ( ' sin ( 0.5)cos0.0 & ( 9.8 m/s )(.0 ) 10 m) ( cos 0.0 " ( 0.5)sin 0.0 ' v 36 m/s Statement: The maximum speed in slippey conditions is 36 m/s. Section 3.3 Questions, page Given: d 4 m o 1 m; F net 1 3 F g Requied: v Analysis: mv ; F g mg; F net + F g F net 1 3 F g + F g 1 3 F g! 3 F g mv! 3 mg v! mg 3 m v! 3 g Solution: v! 3 g 1 m!!9.8 m/s 3 v 8.9 m/s Statement: The speed of the olle coaste is 8.9 m/s.. (a) Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.3-3

21 (b) Given: m kg; 40.0 m; v 15 m/s Requied: Analysis: mv ; F g mg; F g + mg + mv Solution: mg + mv ( kg) ( kg) ( 15 m/s) ( 9.8 m/s ) + ( 40.0 m) 1.5!10 4 N Statement: The magnitude of the nomal foce is N. (c) Given: m kg; 40.0 m; v 15 m/s; F net 0 Requied: v Analysis: mv ; F g mg F net F g + 0 mg + mv 0 g + v v!g Solution: v!g!(!9.8 m/s ) ( 40.0 m) v 0 m/s Statement: The speed equied to make the dive feel weightless is 0 m/s. 3. (a) When the banking angle inceases, the maximum speed of a also inceases because the hoizontal component of the nomal foce has inceased. (b) When the coefficient of fiction inceases, the maximum speed of a also inceases because the foce due to fiction, which points into the tun, has inceased. (c) When the mass of the ca inceases, the maximum speed of a also inceases because the nomal foce and the foce due to fiction both incease. 4. Given: m; µ s 0.7; F net 0 Requied: maximum v Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.3-4

22 Analysis: mv ; F g mg; F s µ s F s mv µ s mv µ s mg v µ g s v µ s g Solution: v µ s g 9.8 m/s 0.7 ( 1.!10 m) v 9 m/s Statement: The maximum speed of the ca is 9 m/s. 5. (a) The banking angle ceates a hoizontal component of the nomal foce, which is only vetical on a hoizontal ound. This hoizontal component inceases the net foce pushing into the cuve. Thanks to this foce into the cuve, cas can navigate the tun at highe speeds without losing fiction. (b) Dives must go much moe slowly because the coefficient of static fiction is damatically educed and the net foce pushing into the cuve is much less. (c) Answes may vay. Sample answe: The banking angle must wok fo conditions whee the coefficient of static fiction is high and when it is low. Making the angles significantly lage would allow fo geate speeds but would be much moe dangeous in slippey conditions. 6. Given: m kg; m 0.68 kg; 1. m Requied: v Analysis: mv ; F g mg; The tension in the sting equals the gavitational foce on m and the centipetal foce on m 1. F g m 1 v v m g m g m 1 Solution: v m g m 1 1. m ( 0.68 kg) 9.8 m/s ( 0.6 kg) v 5.5 m/s Statement: The speed of the ai puck is 5.5 m/s. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.3-5

23 Section 3.4: Rotating Fames of Refeence Mini Investigation: Foucault Pendulum, page 18 Answes may vay. Sample answes: A. The otation does not affect the pendulum mass. Fom ou fame of efeence, the mass swings back and foth consistently while the globe otates beneath it. B. Fom ou fame of efeence, the peiod of otation does not affect the mass. To an obseve on the globe, the faste the otation of Eath, the faste the pendulum appeas to move. This implies that the otation of Eath causes the movement of the Foucault pendulum. C. At the equato the pendulum would not shift at all. Tutoial 1 Pactice, page (a) Given: d 34 m o 16 m; F g Requied: v Analysis: mv ; F g mg F g mv mg v g v g Solution: v g ( 9.8 m/s ) ( 16 m) v 39.8 m/s Statement: The elative speed of the astonauts is 39.8 m/s. (b) Given: g; 16 m Requied: T Analysis: 4! T T 4! Solution: T 4! 4! 16 m " 9.8 m s & ' T 6 s Statement: The peiod of the otation of the spacecaft is 6 s. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.4-1

24 . Yes, both would expeience atificial gavity equal to about 30.0 of Eath s gavity, o 0.300g. The mass cancels out in the equation to detemine speed, so the effect is independent of mass. 3. Given: g m/s ; a net 9.70 m/s ; m Requied: T Analysis: 4! ; the centipetal acceleation is the diffeence between the acceleation due T to gavity and the net acceleation expeienced by a falling object. 4! T g " a net 4! T 4! T g " a net Solution: T 4! g " a net 4! ( m ) m s " 9.70 m ' & s ( ) s 1 min 1 h 60 s 60 min T 7.9 h Statement: The length of the day, o peiod of the planet, is 7.9 h. 4. Given: m 1 56 kg; 50 m; m 4 kg Requied: v Analysis: mv ; F g mg; the acceleation on the space station is 4 56 o 3 4 because the scale eads the astonaut s weight as 4 kg instead of 56 kg. mv 3 4 mg v 3 4 g that of Eath v 3 4 g Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.4-

25 Solution: v 3 4 g 16 m m/s 4 v 43 m/s Statement: The space station floo otates at a speed of 43 m/s. 5. Given: m Requied: v Analysis: mv ; F g mg; the speed of the ca would make the centipetal foce geate than the gavitational foce. F g mv mg v g v g Solution: v g ( 9.8 m/s )( 6.38!10 6 m) v 7.9!10 3 m/s Statement: The ca would need a speed of m/s. Section 3.4 Questions, page At just the ight speed, the centifugal acceleation is enough to povide enough foce to keep the wate in the bucket.. The spinning washing machine ceates entifugal acceleation that foces wate in the clothes to the oute wall and though poes in the wall, thus emoving excess wate fom the clothes. 3. (a) (b) Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.4-3

26 (c) (d) Given:.7 m; m 10 g o 0.1 kg; T.9 s Requied: θ Analysis: The hoizontal component of the tension F T balances the centipetal foce and the vetical component of the tension F T balances the gavitational foce. Expess the tangent atio of the angle in tems of the applied foce and the gavitational foce, then solve fo the angle; v! T tan! F g mv &! tan "1 ( ( mg ' v &! tan "1 g ' ( Solution: Detemine the speed.7 m fom the cente: v! T! (.7 m) ( 3.9 s) m/s (two exta digits caied) v 4.3 m/s Detemine the angle the sting makes with the vetical: v &! tan "1 g ' ( m & & tan "1 s ' ( ( ( (.7 m ) 9.8 m & ( s ' ( ( ' (two exta digits caied)! 36 Statement: The sting makes a 36 angle with the vetical. (e) Given:.7 m; m 10 g 0.1 kg; θ Requied: F T Analysis: The vetical component of the tension F T balances the gavitational foce. Expess the cosine atio of the angle in tems of the tension and the gavitational foce. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.4-4

27 cos! F g F T F T F g cos! F T mg cos! Solution: F T mg cos! 9.8 m s 0.1 kg F T 0.96 N " & ' cos35.57 Statement: The tension in the sting is 0.96 N. 4. Given: m; T 4 h Requied: at the equato g Analysis: Eath is a non-inetial fame of efeence. The acceleation of an object at the equato is the diffeence between the acceleation due to gavity and the centifugal acceleation, g. Use 4! to detemine the centifugal acceleation, then calculate its atio with g. T Solution: Detemine the centipetal acceleation at the equato: 4! T 4 h " 4! ( 6.38 "10 6 m ) 60 min 1 h " 60 s 1 min 4! 6.38 "10 6 m ( s) m/s Theefoe, the atio of the centifugal acceleation to g is: & ( ' m/s g ( 9.8 m/s ) g Statement: The acceleation at the equato is 0.34 less that g. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.4-5

28 5. Given: d 10 m o 5 m; T 30 s; d 1.7 m Requied: compae at and d Analysis: 4! T Solution: Detemine the centipetal acceleation at the astonaut s feet: 4! T 4! 5 m ( 30 s) 0.07 m/s Detemine the centipetal acceleation at the astonaut s head: 4! ( "! d) T 4! 5 m "1.7 m ( 30 s) 0.05 m/s Statement: The movie did not get the physics ight. The acceleation expeienced by the astonaut is in the ange of 0.05 m/s to 0.07 m/s instead of 9.8 m/s. 6. (a) Given: 100 m; g Requied: T Analysis: 4! T g 4! T T 4! g Solution: T 4! g 4! 100 m " 9.8 m s & ' 0.07 s (two exta digits caied) T.0 (10 1 s Statement: The peiod of otation is s. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.4-6

29 (b) Given: 100 m; g Requied: v Analysis: v v Solution: v! 9.8 m " s & ( 100 m) m/s (two exta digits caied) v 31 m/s Statement: The speed of otation is 31 m/s. (c) Given: 100 m; v i m/s; v 4. m/s Requied: Analysis: mv ; F N Solution: mv m( m/s! 4. m/s) ( 100 m) m( 7.34 m/s ) 7.3m Statement: The appaent weight is 7.3 times the mass. (d) Given: 100 m; v i m/s; v +4. m/s Requied: Analysis: mv ; F N Solution: mv m( m/s + 4. m/s) ( 100 m) m( 1.6 m/s ) 13m Statement: The appaent weight is 13 times the mass. (e) Running with the diection of the otation is a bette wokout because you expeience a geate centifugal foce and it equies moe effot o exetion. 7. (a) Given: m 65 kg; 150 m; 540 N Requied: Analysis: ; m ; m Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.4-7

30 Solution: m ( 540 N ) ( 65 kg) m/s (two exta digits caied) 8.3 m/s Statement: The acceleation of objects nea the floo of the space station is 8.3 m/s. (b) Given: 150 m; m/s Requied: v Analysis: v v Solution: v! m " s & ( 150 m) m/s (two exta digits caied) v 35 m/s Statement: The speed of otation of the oute im is 35 m/s. (c) Given: 150 m; m/s Requied: T Analysis: 4! T g 4! T T 4! g Solution: T 4! g 4! 150 m " m s & ' T 7 s Statement: The peiod of otation of the space station is 7 s. 8. (a) Given: 3.4 cm o m; f Hz Requied: Analysis: 4! f Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.4-8

31 Solution: 4! f ( 1.1"10 3 Hz) 4! m 1.6 "10 6 m/s Statement: Fom Eath s fame of efeence, the magnitude of the centipetal acceleation is m/s. (b) Answes may vay. Sample answe: Centifuges need high fequencies to get the geatest possible acceleation. A high centifugal foce moves the dense paticles to the bottom of a test tube, pefectly sepaating mixed solutions such as plasma and ed blood cells. (c) Answes may vay. Sample answe: By sepaating paticles, medical eseaches can study the paticles in thei pue fom. 9. Answes may vay. Sample answe: A lage-scale centifuge, like all centifuges, spins to sepaate a mixtue into its components. In a lage-scale centifuge, a wastewate mixtue is spun and wate is sepaated fom the heavie mixtue, often called sludge, which settles on the bottom. The thickened mixtue is moved to anothe facility fo teatment while the wate is sent on fo diffeent teatment befoe etuning to the envionment. By sepaating wate fom the heavie mixtue, these two components of wastewate can eceive the appopiate teatment befoe etuning to the envionment. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.4-9

32 Section 3.5: Physics Jounal: The Physics of Rolle Coastes Section 3.5 Questions, page (a) Given: 18 m;.0f g Requied: v Analysis: mv ; F g mg; The foce felt at the top of the loop is the diffeence between the centifugal foce and the gavitational foce.! F g.0f g 3.0F g mv 3.0mg v 3.0g v 3.0g Solution: v 3.0g 3.0( 9.8 m/s )( 18 m ) v 3 m/s Statement: The speed at the top of the loop is 3 m/s. (b) The acceleomete expeiences the gavitational foce and the centipetal foce. (c) The acceleomete will ead 3g, the value of the centipetal acceleation. (d) Answes may vay. Sample answes: You could expect eos due to vibations of the olle coaste o the difficulty of holding the acceleomete level.. Given: 1 ; F net 1.5g Requied: v 1 v Analysis: mv ; F g mg; The foce felt at the top of the loop is the diffeence between the centifugal foce and the gavitational foce. Solution: Expess the speed at the top of the loop in tems of the adius:! F g 1.5F g mv.5f g v.5mg.5g v.5g Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.5-1

33 Detemine the atio of the speeds:! F g 1.5F g mv.5f g v.5mg.5g v 1.5g 1 v.5g 1 v v Statement: The atio of the speed at the top of the cicula loop to the speed at the top of the clothoid loop is 1.4:1. 3. Answes may vay. Sample answe: The nomal foce must be set to zeo because that epesents the moment when the centifugal foce balances the gavitational foce. At geate speeds, thee will be a nomal foce and the ide stays in he seat. At slowe speeds, the centifugal foce is insufficient to balance gavity and the ides fall out of thei seats. 4. Given: m 6 kg; v m/s; 35 m Requied: Analysis: mv F g + mg + mv ; F g mg Solution: mg + mv 6 kg ( 6 kg) ( 9.8 m/s ) + m/s 35 m 1500 N Statement: The nomal foce is 1500 N [up]. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.5-

34 Chapte 3 Self-Quiz, page (c). (c) 3. (c) 4. (c) 5. (d) 6. (c) 7. (d) 8. (b) 9. False. Duing fee fall, the appaent weight is zeo because objects ae acceleating due to gavity. 10. False. Fo two objects moving in the same cicula path with diffeent speeds, the faste object expeiences geate centipetal acceleation. 11. Tue 1. Tue 13. False. The effect of the Coiolis foce is not noticeable on eveyday objects moving along Eath s suface. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-1

35 Chapte 3 Review, pages Knowledge 1. (a). (a) 3. (b) 4. (b) 5. (c) 6. (d) 7. False. An amusement pak ide moving down with onstant velocity is an example of an inetial fame of efeence. 8. Tue 9. False. The diection of centipetal acceleation fo a on a banked cuve is always hoizontal into the tun. 10. False. The magnitude of an object s centipetal acceleation inceases with the mass and the velocity of the object, but not the adius of the cicula path. 11. Tue 1. False. The Moon is an example of an object in unifom cicula motion. 13. False. Objects moving in a otating fame of efeence expeience a foce pependicula to the velocity of the object in the otating fame. 14. False. A Foucault pendulum demonstates that Eath is a otating fame of efeence. 15. Tue 16. (a) Substitute v new v in the equation fo centipetal acceleation: v new v 4 v The centipetal acceleation is fou times its oiginal value. (b) Substitute new in the equation fo centipetal acceleation: v new v v 1 The centipetal acceleation is half its oiginal value. 17. The faste ca expeiences the geate foce because centipetal foce inceases with speed. 18. Centifuges ae used to sepaate blood into its components pats. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-

36 19. (a) The souce of the centipetal foce on the Moon is Eath s gavitational foce. (b) The souce of the centipetal foce on a tuning one is static fiction. (c) The souce of the centipetal foce on a ock on a sting is tension. Undestanding 0. (a) You must hold the acceleomete exactly hoizontal, so that the angle the bead makes with the vetical is 0 when the ca is at est. If it is tilted slightly, a non-zeo angle will intoduce an eo into all measuements. (b) The bead will be at an angle of 0 with espect to the vetical. (c) The bead will oll to the west side of the acceleomete, o towad you. (d) The bead will be at an angle of 0 with espect to the vetical. (e) The bead will oll to the east side of the acceleomete, o away fom you. (f) Given: θ 13 Requied: a Analysis: Look at the situation fom an Eath (inetial) fame of efeence. The hoizontal component of the tension F T balances the acceleation and the vetical component of the tension F T balances the gavitational foce. Expess the components of the tension in tems of the hoizontal and vetical applied foces. Then calculate the magnitude of the acceleation. Solution: Vetical component of foce:!f y 0 F T cos" mg 0 F T mg cos" Hoizontal component of foce:!f x ma F T sin" ma mg & cos" ' ( sin" ma g sin" & cos" ' ( a g tan" a a ( 9.8 m/s )tan13 a.3 m/s Statement: The magnitude of the ca s acceleation is.3 m/s. 1. (a) Given: 13 cm 0.13 m; f 33.5 pm Requied: Analysis: 4! f Solution: Convet the fequency to hetz: f 33.5 pm min! 1 min 60 s f Hz (two exta digits caied) Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-3

37 Detemine the centipetal acceleation: 4! f ( Hz) 4! 0.13 m 1.6 m/s Statement: The centipetal acceleation is 1.6 m/s. (b) Given: 4.3 m; T 1. s Requied: Analysis: 4! T Solution: 4! T 4! 4.3 m ( 1. s) 1. "10 m/s Statement: The centipetal acceleation of the lasso is m/s. (c) Given: v m/s; d m m Requied: Analysis: v Solution: v.18!106 m/s ( 5.30!10 "11 m) 8.97!10 m/s Statement: The magnitude of the centipetal acceleation is m/s.. Given: 15 m; 33.8 m/s Requied: v Analysis: a v c v Solution: v ( 33.8 m/s )( 15 m ) v 65 m/s Statement: The speed of the ca is 65 m/s. 3. Given: v 50.0 km/h; d 33.5 m o m Requied: Analysis: v Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-4

38 Solution: Convet the speed to metes pe second: v 50.0 km h! 1000 m 1 km! 1 h 60 min! 1 min 60 s m/s (two exta digits caied) v 13.9 m/s Detemine the centipetal acceleation: v m/s ( m) 11.5 m/s Statement: The magnitude of the centipetal acceleation is 11.5 m/s. 4. Given: d 0 m o 10 m; F net 1 3 F g Requied: v Analysis: mv ; F g mg; F net + F g F net 1 3 F g + F g 1 3 F g! 3 F g mv! 3 mg v! mg 3 m v! 3 g Solution: v! 3 g 10 m!!9.8 m/s 3 v 8.1 m/s Statement: The speed of the olle coaste is 8.1 m/s. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-5

39 5. Substitute m locomotive 3m cago ca in the equation fo centipetal foce: F locomotive m locomotive v ( 3m cago ca )v 3 m cago ca v F locomotive 3ago ca The centipetal foce on the locomotive is thee times the foce on the cago ca. 6. (a) Given: m.0 kg; v 0 m/s; 16 m Requied: Analysis: mv Solution: mv (90 kg)(0 m/s) (16 m)!10 3 N Statement: The centipetal foce is 10 3 N. (b) Given: m.0 kg; v 0 m/s; 10 m Requied: Analysis: mv Solution: mv (90 kg)(0 m/s) (10 m) 4!10 3 N Statement: The centipetal foce is N. (c) Given: m.0 kg; v 5 m/s; 10 m Requied: Analysis: mv Solution: mv (90 kg)(5 m/s) (10 m)!10 N Statement: The centipetal foce is 10 N. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-6

40 7. Given: m.0 kg;.8 10 N; 1.00 m Requied: Analysis: mv v F c m Solution: v F c m.8!10 kg " m & s ' ( ( 1.00 m) (.0 kg ) v 1 m/s Statement: The speed of the discus is 1 m/s when eleased. 8. (a) Given: m.0 kg; 0.5 m; f 1.0 pm 1 60 Hz Requied: Analysis: m ; 4! f ; 4! mf Solution: 4! mf " 1 4! (.0 kg)(0.5 m) 60 Hz & ' N Statement: The centipetal foce is N. (b) Given: m.0 kg; 0.5 m; f 5.0 pm 1 1 Hz Requied: Analysis: m ; 4! f ; 4! mf Solution: 4! mf " 1 4! (.0 kg)(0.5 m) 1 Hz & ' 0.7 N Statement: The centipetal foce is 0.7 N. 1 (c) Given: m.0 kg; 0.5 m; f 1.0 pm 10 Hz Requied: Analysis: m ; 4! f ; 4! mf Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-7

41 Solution: 4! mf " 1 4! (.0 kg)(0.5 m) 10 Hz & ' N Statement: The centipetal foce is N. 9. The peson s weight equals mg, so if the nomal foce of N is the esult of g, then half the nomal, o N is the peson s weight. Analysis and Application 30. Given: m; f 60 pm 1.0 Hz Requied: Analysis: 4π f Solution: 4! f ( 1.0 Hz) 4! m 1. m/s Statement: The centipetal acceleation is 1. m/s. 31. Given: 0.50 m; m 1.5 kg; F T 5 N Requied: maximum v Analysis: The maximum speed of otation is when the centipetal foce equals the maximum tension befoe the sting beaks. mv v F c m Solution: v F c m " 5 kg! m s & ' ( 0.50 m ) ( 1.5 kg ) v.9 m/s Statement: The maximum speed of otation befoe beaking the sting is.9 m/s. 3. Given: 3.5F g ; v 6 m/s Requied: Analysis: mv ; F g mg; + F g Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-8

42 Solution: + F g 3.5F g + F.5F g.5mg mv mv.5mg v.5g 6 m/s m/s 8 m Statement: The adius of the tack s cuvatue is 8 m. 33. (a) Given: m 35 kg; d m o 11 m; f 3.5 pm Hz Requied: at the top Analysis: 4π mf ; F g Solution: F g! mg! 4" mf ( 35 kg) ( 9.8 m/s )! 4" ( 35 kg) ( 11 m) ) 10 N Statement: The foce of the seat is.9 10 N. 60 Hz & ' ( (b) Given: m 35 kg; d m o 11 m; f 3.5 pm Hz Requied: at the bottom Analysis: 4π mf ; F g + Solution: F g + mg + 4! mf " ( 35 kg) ( 9.8 m/s ) + 4! ( 35 kg) ( 11 m) Hz & ' 3.9 ( 10 N Statement: The foce of the seat is N. 34. Given: m 1.5 kg;.0 m Requied: F T at the bottom Analysis: Since the tension at the top of the loop equals the gavitational foce, but the speed constant, the centipetal foce must equal the gavitational foce at all points along the cicula path: mg; F T F g + Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-9

43 Solution: F T F g + mg + mg mg 1.5 kg 9.8 m/s F T 9 N Statement: The tension when the ock is at the bottom is 9 N. 35. (a) Ovehead: Side view: (b) (c) The centipetal foce is the sum of the tension and the gavitational foce. Given the angle, the centipetal foce is equal to F T cos θ. (d) Given: 1.5 m; v 10.0 m/s Requied: θ Analysis: mv tan! F g mg mv g v g &! tan "1 ' ( v ; F g mg g & Solution:! tan "1 v ' ( m & 9.8 kg ) s ' ( ( 1.5 m ) & ( tan "1 ( 10.0 m & ( s ' ( ( '! 8.4 Statement: The sting makes an 8.4 angle with the hoizontal. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-10

44 36. Given: m kg; 35 m; v 1 m/s Requied: F s Analysis: mv Solution: mv F s mv ; F s ( 1 m/s) ( 35 m) 1.7! 103 kg F s 7.0! 10 3 N Statement: The fiction foce is N. 37. Given: m 50.0 kg; 0.75 m; F T 50.0 N Requied: maximum v Analysis: The maximum speed of otation is when the centipetal foce equals the maximum tension befoe the sting beaks. mv v F c m Solution: v F c m " 50.0 kg! m s & ' 0.75 m ( 0.30 kg ) v 11 m/s Statement: The maximum speed of otation befoe beaking the sting is 11 m/s. 38. Given: m 70.0 kg; v.0 m/s; 1.0 m Requied: Analysis: mv Solution: mv 70.0 kg (.0 m/s) ( 1.0 m).8! 10 N Statement: The centipetal foce is.8 10 N. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-11

45 39. Given: m 30.0 kg; d 0.0 m o 10.0 m; 3 N Requied: v Analysis: mv v F c m Solution: v F c m " 3 kg! m s & ' 10.0 m 30.0 kg v 3.3 m/s Statement: The speed of the child cycling is 3.3 m/s. 40. Given: v 55 m/s; m 15 kg; 5 m Requied: Analysis: mv Solution: mv 15 kg ( 55 m/s) ( 5 m) 1.5! 10 4 N Statement: The centipetal foce is N. 41. (a) Given: m; v m/s; 16 N Requied: m Analysis: mv m F c v Solution: m F c v " 16 kg! m s " m s & ' m & ' m 8 kg Statement: The mass of the boat is 8 kg. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-1

46 (b) Given: m; m 8 kg; 4 N Requied: v Analysis: mv v F c m Solution: v F c m " 4 kg! m s & ' ( m) ( 8 kg ) v 1 m/s Statement: The speed should be deceased to 1 m/s. 4. (a) Substitute v v 1 in the equation fo centipetal foce: F mv m v 1 4 mv 1 F 4F 1 The centipetal foce on the ca is fou times the foce at half the speed. (b) If thee wee no fiction, the ca would not be able to tun and would continue in the same diection. 43. Given: m 105 kg; v 7.0 m/s; 15 m Requied: Analysis: mv Solution: mv 105 kg ( 7.0 m/s) ( 15 m) 3.4! 10 N Statement: The centipetal foce is N. 44. Given: T.00 s; m kg; m 5.00 kg; m; 6.00 m Requied: Analysis: 4! m T Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-13

47 Solution: Detemine the tension in sting B: F B 4! m T ( 6.00 m) (.00 s) 4! 5.00 kg F B 96 N Detemine the tension in sting A: F A 4! m F B T ( 4.00 m) + 96 N (.00 s) 4! 3.00 kg F A 414 N Statement: The tension in sting A is 414 N and the tension in sting B is 96 N. 45. (a) Given: m 1.0 kg; m 5.0 kg; µ s 0.30; 5.0 m Requied: maximum v Analysis: mv ; F s µ s mg; The maximum speed is the point at which the centipetal acceleation on mass 1 equals the foce of static fiction keeping mass 1 on top of mass. Solution: F s mv! s mg v! mg s m v! s g 9.8 m/s ( 5.0 m ) 0.30 v 3.8 m/s Statement: The maximum speed is 3.8 m/s. (b) Given: m 1.0 kg; m 5.0 kg; µ s 0.30; 5.0 m Requied: F T Analysis: mv Solution: mv ( F T m + m 1 )v.0 kg kg ( m/s) ( 5.0 m) F T 1 N Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-14

48 Statement: The tension in the sting at the maximum speed is 1 N. 46. Given: T 4.0 s; m kg; m 1800 kg; m; 4.0 m m 7.0 m Requied: F T at the bottom fo each suppot Analysis: 4! m ; F T F g + T Solution: Detemine the tension in suppot B: F TB F g + F TB mg + 4! m T ( 1800 kg) 9.8 m/s + 4! ( 1800 kg) ( 7.0 m) ( 4.0 s) 4.873" 10 4 N (two exta digits caied) 4.9 " 10 4 N Detemine the tension in suppot A: F TA F g + + F TB F TA mg + 4! m T ( 100 kg) 9.8 m/s 7. " 10 4 N + 4! ( 100 kg) ( 4.0 m) " 10 4 N ( 4.0 s) Statement: The tension in suppot A is N and the tension in suppot B is N. 47. (a) The ide feels the gavitational foce, centipetal foce, and the foce due to static fiction. (b) Given: 3.0 m; µ s 0.40 Requied: minimum v Analysis: In this scenaio the nomal is hoizontal and opposite the centipetal foce instead of the gavitational foce. Detemine the speed at which the foce due to static fiction equals the gavitational foce; mv ; F s µ s Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-15

49 Solution: F g F s mg! s! mv mg! s " & g! s v v g! s v g! s ( 9.8 m/s )( 3.0 m ) 0.40 v 8.6 m/s Statement: The minimum speed of otation to keep the peson stuck to the wall 8.6 m/s. 48. Given: m 6.0 kg; A 5.0 m; B 5.0 m; h 8.0 m; v 1 m/s Requied: F T in each sting Analysis: In this scenaio the tensions will not be equal because of the gavitational foce. The x- components of the tensions will balance the centipetal foce while the x-components of the tensions balance the gavitational foce; mv ; F s µ s. The adius of the cicula path is one side of a ight tiangle with sting A as the hypotenuse and half the height cylinde as the othe side. Solution: Detemine the adius of the cicula path using the Pythagoean theoem: + 1! " h & A! A ' 1 " h &! A ' 1 " h & 5.0 m! " ' 1 ( 8.0 m ) & 3.0 m Detemine the angle the stings make with the hoizontal: cos! 4.0 m 5.0 m! Also, we can use cos θ 0.8 (and theefoe sin θ 0.6) late in the solution. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-16

50 Detemine the centipetal foce on the mass: mv 6.0 kg ( 1 m/s) ( 3.0 m) 88 N Expess the balancing of the hoizontal foces and isolate one of the tensions: F TAx + F TBx F TA sin! + F TB sin! 88 N F TA 88 N " F TB sin! sin! Expess the balancing of the vetical foces and solve fo the tension on sting B: F TA y! F TBy F g F TA cos"! F TB cos" mg 88 N! F TB sin" & sin" ' ( cos"! F TB cos" 6.0 kg cos" 88 N sin"! F sin" cos" TB! F TB cos" 58.8 N sin" ( 88 N)cos"! F sin" TB cos" 58.8 N ( 88 N) ( 0.8 )! F ( 0.6) TB ( 0.8) 58.8 N 384 N!1.6F TB 58.8 N Solve fo the tension on sting A: F TA 88 N! F sin" TB sin" 88 N! ( 03. N) ( 0.6 ) ( 0.6) F TA.8 10 N 9.8 m/s 58.8 N! 384 N F TB! N F TB.0 )10 N Statement: The tension in sting A is.8 10 N and the tension in sting B is.0 10 N. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-17

51 49. Given: f pm 60 Hz; 30.0 m; m kg Requied: Analysis: m ; 4! f Solution: m m( 4! f ) 4! ( 9.8 " 10 kg) ( 30.0 m) 1.3 " 10 3 N 60 Hz & ' ( Statement: The dive will feel a foce of N. 50. Given: m.0 kg; 0.35 m; f 50.0 pm Hz Requied: Analysis: m ; 4! f Solution: m m 4! f " 4! (.0 kg) ( 0.35 m) N 60 Hz & ' Statement: The foce acting on the clothes is 19 N. 51. Given: f 0.30 ps 0.30 Hz; 15 cm 0.15 m Requied: µ s Analysis: mv ; F s µ s mg; The speed is the point at which the centipetal acceleation on the coin equals the foce of static fiction. Solution: F s 4! mf! s mg 4! f! s g! s 4! f g ( 0.30 Hz) 4! 0.15 m ( 9.8 m/s )! s Statement: The coefficient of static fiction is Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-18

52 5. Given: 0.0 m; 3.00F g Requied: v Analysis: mv ; F g mg; The nomal foce at the bottom of the loop is the sum of the centifugal foce and the gavitational foce. Solution: + F g 3.00F g.00f g mv.00mg v.00g v.00g ( 0.0 m ) m/s v 19.8 m/s Statement: The speed at the bottom of the loop is 19.8 m/s. 53. Given: v 1 m/s; 0.500g Requied: Analysis: mv ; F c m Solution: m mv ma c v 0.500g v 0.500g ( 1 m/s) 0.500( 9.8 m/s ) 9 m Statement: The adius of the station is 9 m. 54. (a) The bucket moves in icle because of the fictitious centifugal foce. (b) The souce of the centifugal foce is the tension in the ope. (c) Given: m 15 kg; v m/s; m Requied: Analysis: mv Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-19

53 Solution: mv 15 kg ( m/s) ( m) 30 N Statement: The magnitude of the foce is 30 N. 55. (a) Given: d 4 m o m; m 65 kg + 95 kg 160 kg; f pm 60 Hz Requied: µ s Analysis: mv ; F s µ s ; The foce of static fiction needs to offset the gavitation foce. Solution: F g F s mg! s mg! s 4! mf g! s 4! f! s g 4! f ( 9.8 m/s ) " 4! m 60 Hz & '! s 0.9 Statement: The coefficient of static fiction is 0.9. (b) The fequency of the ide is Hz o 0.37 loops/s. 60 (c) Given: m; m 65 kg + 95 kg 160 kg; v 6 m/s Requied: at top and bottom Analysis: mv ; The foce at the top of the sphee is the diffeence between the centifugal foce and the gavitational foce: F g. The foce at the bottom of the sphee is the sum of the centifugal foce and the gavitational foce: + F g. Solution: Detemine the foce at the top of the sphee: F s! F g mv! mg 160 kg ( 6.0 m/s) ( m) 1.3" 10 3 N! ( 160 kg) ( 9.8 m/s ) Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-0

54 Detemine the foce at the bottom of the sphee: F s + F g mv + mg 160 kg ( 6.0 m/s) ( m) 4.4! 10 3 N + ( 160 kg) ( 9.8 m/s ) Statement: The nomal foce at the top of the sphee N. The nomal foce at the bottom of the sphee N. 56. Given: 3.0 m; f.0 pm Hz; m 54 kg Requied: Analysis: 4! mf Solution: 4! mf 4! ( 54 kg) ( 3.0 m) 7.1 N 60 " 60 Hz & ' Statement: The magnitude of the centipetal foce is 7.1 N. 57. Given: m 450 kg; f 1 pm 1 60 Hz; 48 N Requied: Analysis: 4! mf Solution: 4! mf F c 4! mf 48 N " 1 4! ( 450 kg) 60 Hz & ' 9.7 m Statement: The length of the lead ope is 9.7 m. 58. Given: 5 m; 1.0g Requied: v Analysis: v Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-1

55 Solution: v v 1.0g ( 5 m ) m/s m s! 60 s 60 min!! 1 km 1 min 1 h 1000 m v 56 km/h Statement: The speed of the hose is 56 km/h. Evaluation 59. (a) The centifugal foce is the fictitious foce that occus in a otating fame of efeence that is diected away fom the cente of otation. (b) The Coiolis foce is the fictitious foce in a otating fame of efeence that acts pependicula to an object s velocity. (c) Fictitious foces ae not actual foces; they only appea when the natual fame of efeence fo a given situation is itself acceleating and they ae always popotional to the mass of the object on which they act. 60. (a) As a peson ides an elevato to the top of the CN Towe, the fame of efeence acceleates upwad. The peson will expeience an incease in the nomal foce and thei appaent weight inceases. (b) As a peson is in fee fall while skydiving, the nomal foce acting on the peson is zeo and the peson also has an appaent weight of zeo. 61. (a) Equations fo Centipetal Acceleation Equation Vaiables in equation Vaiables not in equation v, T, f v 4! T, T v, f 4! f, f v, T (b) The fist equation follows fom the definition of aveage acceleation, simila tiangles, and looking at vey small peiods of time. The second equation follows fom the fist equation by inseting the cicumfeence of the cicula path fo distance and using peiod of evolution fo the time inteval in the basic equation fo speed. The thid equation follows fom the second equation by simply substituting 1 f fo T. 6. (a) Twiling the ope allows he to incease its speed while it tavels in icula path. (b) By spinning the ope faste, she can ensue it has a geate speed when it is eleased. This will help the ope go fathe. (c) The ope will tavel away fom the odeo pefome in a staight-line path in the diection it was moving when she let go. Copyight 01 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3-