Section 3.2: Centripetal Acceleration Tutorial 1 Practice .

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Section 3.2: Centripetal AccelerationTutorial 1 Practice, page 1181. Given: r 25 km 2.5 104 m; v 50.0 m/sRequired: acv2Analysis: ac rv2Solution: ac r (50.0 m/s)2( 2.5 ! 10 m )4ac 0.10 m/s 2Statement: The magnitude of the centripetal acceleration is 0.10 m/s2.2. Given: r 1.2 m; v 4.24 m/s!Required: acv2Analysis: ac ; Centripetal acceleration is always directed toward the centre of rotation.rSince the hammer’s velocity is directed south and it is spinning clockwise, the centre of rotationis west of the hammer.v2Solution: ac r( 4.24 m/s) (1.2 m )2ac 15 m/s 2Statement: The centripetal acceleration is 15 m/s2 [W].3. Given: r 1.4 m; ac 12 m/s2Required: vv2Analysis: ac rv ac rSolution: v ac r (12 m/s )(1.4 m )2v 4.1 m/sStatement: The speed of the ball is 4.1 m/s.4. (a) Given: r 1.08 1011 m; ac 1.12 10–2 m/s2Required: vCopyright 2012 Nelson Education Ltd.Chapter 3: Uniform Circular Motion3.2-1

Analysis: ac 4! 2 rT2T 4! 2 racSolution: T 4! 2 rac 4! 2 (1.08 " 1011 m ) #2 m '&% 1.12 " 10 s 2 )(T 1.95 " 107 sStatement: The period of Venus is 1.95 107 s.(b) Convert the period in seconds to days:1 min1h1dT 1.95 ! 107 s !!!60 s60 min 24 hT 226 daysThe period of Venus is 226 days.5. Given: v 7.27 103 m.s; r 7.54 106 mRequired: acv2Analysis: ac rv2Solution: ac r(7.27 ! 103 m/s )2 (7.54 ! 106 m )ac 7.01 m/s 2Statement: The magnitude of the centripetal acceleration is 7.01 m/s2.6. (a) Given: ac 3.3 106 m/s2; r 8.4 cm 8.4 10–2 mRequired: fAnalysis: ac 4! 2 rf 2f ac4! 2 rCopyright 2012 Nelson Education Ltd.Chapter 3: Uniform Circular Motion3.2-2

Solution: f ac4! 2 r#6 m &% 3.3" 10 2 ('s 2)24! (8.4 " 10 m )f 1.0 " 104 HzStatement: The frequency of the centrifuge is 1.0 104 Hz.(b) Convert the frequency in hertz to revolutions per minute:160 sf 1.0 ! 104!1 s 1 minf 6.0 ! 105 rpmThe frequency of the centrifuge is 6.0 105 rpm.Section 3.2 Questions, page 1191. (a) The tension in the string provides the force to keep the puck in its circular path at constantspeed, and so provides the acceleration of the puck.(b) The centripetal acceleration is half as large because centripetal acceleration depends on thev21inverse of the radius: ac .22r(c) The centripetal acceleration is four times as great because centripetal acceleration depends on(2v)2the square of the speed: 4ac .r2. The centripetal acceleration for the first athlete’s hammer is four times greater than that of thev2second athlete. Centripetal acceleration depends on the square of the speed: ac . So if therhammer spins two times as fast, the centripetal acceleration is 22, or 4, times larger: 4a.3. Given: r 0.42 m; T 1.5 sRequired: ac4! 2 rAnalysis: ac 2T4! 2 rSolution: ac 2T4! 2 ( 0.42 m ) (1.5 s )2ac 7.4 m/s 2Statement: The magnitude of the centripetal acceleration of the lasso is 7.4 m/s2.4. Given: v 28 m/s; r 135 mRequired: acv2Analysis: ac rCopyright 2012 Nelson Education Ltd.Chapter 3: Uniform Circular Motion3.2-3

Solution: ac v2r( 28 m/s )2 (135 m )ac 5.8 m/s 2Statement: The magnitude of the centripetal acceleration is 5.8 m/s2.5. Given: r 6.38 106 m; T 1 day or 86 400 sRequired: ac4! 2 rAnalysis: ac 2T4! 2 rSolution: ac 2T4! 2 ( 6.38 " 106 m ) (86 400 s )2ac 3.37 " 10#2 m/s 2Statement: The centripetal acceleration at Earth’s equator is 3.37 10–2 m/s2.6. Given: ac 25 m/s2; r 2.0 mRequired: fAnalysis: ac 4! 2 rf 2f ac4! 2 rSolution: f ac4! 2 r"m% # 25 2 '&s 24! ( 2.0 m )f 0.56 HzStatement: The minimum frequency of the cylinder is 0.56 Hz.7. Given: v 22 m/s; ac 7.8 m/s2Required: rv2Analysis: ac rv2r acCopyright 2012 Nelson Education Ltd.Chapter 3: Uniform Circular Motion3.2-4

Solution: r v2ac( 22 m/s )2(7.8 m/s2 )r 62 mStatement: The radius of the curve is 62 m.8. Given: C 478 m; ac 0.146 m/s2Required: vC;Analysis: C 2πr or r 2!v2ac rv ac r" C%v ac '# 2! &" C%Solution: v ac '# 2! & (0.146 m/s2 )( 478 m )2!m60 s60 min1 km(((s 1 min1000 m1hv 12.0 km/hStatement: The speed of the jogger is 12.0 km/h.9. (a) Given: r 0.300 m; f 60.0 rpmRequired: T1Analysis: T fSolution: Convert the frequency to hertz:f 60.0 rpm 3.333 60.011 minf 1.00 Hz!1 min60 sCopyright 2012 Nelson Education Ltd.Chapter 3: Uniform Circular Motion3.2-5

Determine the period:1T f11.00 HzT 1.00 sStatement: The period of the bicycle wheel is 1.00 s.(b) Given: r 0.300 m; f 1.00 Hz!Required: ac Analysis: ac 4! 2 rf 2 ; Centripetal acceleration is always directed toward the centre of rotation.Since the wheel’s velocity is directed west and it is spinning clockwise, the centre of rotation isnorth of the point.Solution: ac 4! 2 rf 2 4! 2 ( 0.300 m )(1.00 Hz )ac 11.8 m/s 2Statement: The centripetal acceleration of a point on the edge of that wheel is 11.8 m/s2 [N] if itis moving westward at that instant.10. (a) Given: T 27.3 days; ac 2.7 10–3 m/s2Required: r4! 2 rAnalysis: ac 2TSolution: Convert the period to seconds:60 s24 h 60 min!!T 27.3 days !1h1 day1 min2 2.3587 ! 106 s (two extra digits carried)T 2.36 ! 106 sDetermine the radius:4! 2 rac 2TacT 2r 4! 22 #3 m '6()s2.7"102.3587"10&%s 2 )( 4! 2r 3.8 " 108 mStatement: The radius of the curve is 3.8 108 m.(b) The values are the same to two significant digits. Any difference beyond that may be becausethe orbit is not perfectly circular or the speed is not constant.Copyright 2012 Nelson Education Ltd.Chapter 3: Uniform Circular Motion3.2-6

11. (a) Given: ac 711 m/s2; r 1.21 mRequired: vv2Analysis: ac ; v ac rrSolution: v ac r (711 m/s2 )(1.21 m )v 29.3 m/sStatement: The speed of the hammer is 29.3 m/s.(b) Given: d –2.0 m; vi 29.3 m/s; θ 42 Required: dxAnalysis: Use vf2 vi2 2a d to calculate the y-component of the final speed, then calculate thetime of flight vf vi a t. Finally, calculate the range using d v t.Solution: Determine the y-component of the final speed:vf2 vi2 2a ! dvfy2 viy2 2g ! dvfy viy2 2g ! d (( 29.3 m/s )(sin 42 ))2 2 (9.8 m/s2 )( –2.0 m ) 18.58 m/s (two extra digits carried)vfy 19 m/sDetermine the time of flight:vfy viy a !t!t vfy ! viyam "m%! !29.3'& ( sin 42 )#ss m9.8 2s!t 3.896 s (two extra digits carried)Determine the range of the ball:! d x vx !t18.58 vi !t cos!"m% 29.3 ' ( 3.896 s ) ( cos 42 )s&#! d x 85 mStatement: The range of the ball is 85 m.Copyright 2012 Nelson Education Ltd.Chapter 3: Uniform Circular Motion3.2-7

(b) The centripetal acceleration is half as large because centripetal acceleration depends on the inverse of the radius: 1 2 a c v2 2r. (c) The centripetal acceleration is four times as great because centripetal acceleration depends on the square of the speed: 4a c (2v)2 r. 2.

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