( ) ( ) 1.4 m ( ) Section 3.2: Centripetal Acceleration Tutorial 1 Practice, page Given: r = 25 km = m; v = 50.0 m/s. Required: a!

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1 Section 3.2: Centipetal Acceleation Tutoial 1 Pactice, page Given: 25 km m; v 50.0 m/s Requied: Analysis: Solution: ( 50.0 m/s) 2 2.5!10 4 m 0.10 m/s 2 Statement: The magnitude of the centipetal acceleation is 0.10 m/s Given: 1.2 m; v 4.24 m/s Requied: a! c Analysis: ; 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: ( 4.24 m/s) 2 ( 1.2 m) 15 m/s 2 Statement: The centipetal acceleation is 15 m/s 2 [W]. 3. Given: 1.4 m; 12 m/s 2 Requied: v Analysis: v Solution: v 1.4 m 12 m/s 2 v 4.1 m/s Statement: The speed of the ball is 4.1 m/s. 4. (a) Given: m; m/s 2 Requied: v Copyight 2012 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.2-1

2 Analysis: 4! 2 T Solution: T 4! 2 4! 2 4! "10 11 m m 1.12 "10 2 s 2 ( ) 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 24 h T 226 days The peiod of Venus is 226 days. 5. Given: v m.s; m Requied: Analysis: Solution: !103 m/s ( 7.54!10 6 m) 7.01 m/s 2 Statement: The magnitude of the centipetal acceleation is 7.01 m/s (a) Given: m/s 2 ; 8.4 cm m Requied: f Analysis: 4! 2 f 2 f 4! 2 Copyight 2012 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.2-2

3 Solution: f 4! 2 m 3.3"10 6 ( s 2 4! "10 )2 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 3.2 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: (c) The centipetal acceleation is fou times as geat because centipetal acceleation depends on the squae of the speed: 4 (2v)2. 2. 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:. So if the hamme spins two times as fast, the centipetal acceleation is 2 2, o 4, times lage: 4a. 3. Given: 0.42 m; T 1.5 s Requied: Analysis: 4! 2 Solution: 4! 2 4! 2 ( 0.42 m) ( 1.5 s) m/s 2 Statement: The magnitude of the centipetal acceleation of the lasso is 7.4 m/s Given: v 28 m/s; 135 m Requied: Analysis: Copyight 2012 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.2-3

4 Solution: 2 28 m/s 135 m 5.8 m/s 2 Statement: The magnitude of the centipetal acceleation is 5.8 m/s Given: m; T 1 day o s Requied: Analysis: 4! 2 Solution: 4! 2 4! "10 6 m ( s) "10 2 m/s 2 Statement: The centipetal acceleation at Eath s equato is m/s Given: 25 m/s 2 ; 2.0 m Requied: f Analysis: 4! 2 f 2 f 4! 2 Solution: f 4! 2 " 25 m s 2 4! 2 ( 2.0 m ) f 0.56 Hz Statement: The minimum fequency of the cylinde is 0.56 Hz. 7. Given: v 22 m/s; 7.8 m/s 2 Requied: Analysis: Copyight 2012 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.2-4

5 Solution: ( 22 m/s) m/s 2 62 m Statement: The adius of the cuve is 62 m. 8. Given: C 478 m; m/s 2 Requied: v Analysis: C 2π o C 2! ; v " C v 2! " C Solution: v 2! ( 478 m ) m/s 2 2! m s ( 60 s 1 min ( 60 min 1 h ( 1 km 1000 m v 12.0 km/h Statement: The speed of the jogge is 12.0 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 2012 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.2-5

6 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! 2 f 2 ; 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! 2 f 2 ( 1.00 Hz ) 2 4! m 11.8 m/s 2 Statement: The centipetal acceleation of a point on the edge of that wheel is 11.8 m/s 2 [N] if it is moving westwad at that instant. 10. (a) Given: T 27.3 days; m/s 2 Requied: Analysis: 4! 2 Solution: Convet the peiod to seconds: T 27.3 days! 24 h 1 day! 60 min 1 h! 60 s 1 min !10 6 s (two exta digits caied) T 2.36!10 6 s Detemine the adius: 4! 2 a c 4! 2 m 2.7 "10 3 ( s 2 ( ) "106 s ) 2 4! "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 2012 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.2-6

7 11. (a) Given: 711 m/s 2 ; 1.21 m Requied: v Analysis: ; v Solution: v ( 711 m/s 2 )( 1.21 m ) v 29.3 m/s Statement: The speed of the hamme is 29.3 m/s. (b) Given: d 2.0 m; v i 29.3 m/s; θ 42 Requied: d x Analysis: Use v f 2 v i 2 + 2a 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 2 v i 2 + 2a! d v fy 2 v iy 2 + 2g! d v fy v iy 2 + 2g! d (( sin42 )) 2 + 2( 9.8 m/s 2 )( 2.0 m ) 29.3 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! "!29.3 m s sin m s 2!t s (two exta digits caied) Detemine the ange of the ball:! d x v x!t v i!t cos! " 29.3 m s s! d x 85 m ( cos42 ) Statement: The ange of the ball is 85 m. Copyight 2012 Nelson Education Ltd. Chapte 3: Unifom Cicula Motion 3.2-7

3.2 Centripetal Acceleration

3.2 Centripetal Acceleration unifom cicula motion the motion of an object with onstant speed along a cicula path of constant adius 3.2 Centipetal Acceleation The hamme thow is a tack-and-field event in which an athlete thows a hamme