Version 001 HW#6 - Circulr & ottionl Motion rts (00223) 1 This print-out should hve 14 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. Circling Hwk 12 001 (prt 1 of 2) 10.0 points Ahwkfliesinhorizontlrcofrdius6.9m t constnt speed of 6.2 m/s. Find its centripetl ccelertion. Correct nswer: 5.57101 m/s 2. For constnt speed long circulr pth, r v2 r (6.2 m/s)2 6.9 m 5.57101 m/s 2. 002 (prt 2 of 2) 10.0 points It continues to fly long the sme horizontl rc but increses its speed t the rte of 1.07 m/s 2. Find the mgnitude of ccelertion under these new conditions. Correct nswer: 5.67284 m/s 2. The ddition of tngentil ccelertion increses the mgnitude of ccelertion by the vector sum of both ccelertions, so 2 r +2 t (5.57101 m/s 2 ) 2 +(1.07 m/s 2 ) 2 5.67284 m/s 2. Holt SF 07B 04 003 (prt 1 of 4) 10.0 points Consider the following vlues ω vg θ t +2.8 rd 9.34 s +0.68 rev/s b 0.69 s c 1.2 turns 1.4 s +1.9π rd/s +1.9π rd d Wht is the vlue of? Correct nswer: 0.299786 rd/s. ω vg θ t Let : θ +2.8 nd t 9.34 s. +2.8 rd 9.34 s 0.299786 rd/s. 004 (prt 2 of 4) 10.0 points Wht is the vlue of b? Correct nswer: 2.94807 rd. Let : ω vg +0.68 rev/s nd t 0.69 s. θ ω vg t ( ) 2π rd (+0.68 rev/s) (0.69 s) 1 rev 2.94807 rd. 005 (prt 3 of 4) 10.0 points Wht is the vlue of c? Correct nswer: 5.38559 rd/s. Let : θ 1.2 turns nd t 1.4 s. ω vg θ t 1.2 turns 1.4 s 2π rd 1 turn 5.38559 rd/s.
Version 001 HW#6 - Circulr & ottionl Motion rts (00223) 2 006 (prt 4 of 4) 10.0 points Wht is the vlue of d? Correct nswer: 1 s. Let : ω vg +1.9π rd/s, nd θ +1.9π rd. t θ ω vg +1.9π rd +1.9π rd/s 1 s. Serwy CP 07 05 007 (prt 1 of 2) 10.0 points A dentist s drill strts from rest. After 4.96 s of constnt ngulr ccelertion, it turns t rte of 27200 rev/min. Find the drill s ngulr ccelertion. Correct nswer: 574.27 rd/s 2. Let : ω f 27200 rev/min nd t 4.96 s. Since ω i 0, α ω f ω i ω f t t 27200 rev/min rd 2π 4.96 s 1 rev 1 min 60 s 574.27 rd/s 2. 008 (prt 2 of 2) 10.0 points Find the ngle through which the drill rottes during this period. Correct nswer: 7063.98 rd. Since ω i 0, Disk nd Mss 009 10.0 points A circulr disk with moment of inerti 1 2 m2, mss m nd rdius is mounted t its center, bout which it cn rotte freely. A light cord wrpped round it supports weight mg. g Find the totl kinetic energy of the system when the weight is moving t speed v. 1. K 1 3 mv2 2. K 5 4 mv2 3. K 3 2 mv2 4. K mv 2 5. K 4 5 mv2 6. K 2 3 mv2 7. K 5 2 mv2 8. K 1 2 mv2 9. K 3 4 mv2 correct I m T ω θ ω i t+ 1 2 αt2 1 2 αt2 1 2 ( 574.27 rd/s 2 ) (4.96 s) 2 7063.98 rd.
Version 001 HW#6 - Circulr & ottionl Motion rts (00223) 3 v rω, so the totl kinetic energy is K tot K m +K rot T2 T1 1 2 mv2 + 1 2 I ω2 1 2 mv2 + 1 ( ) 1 (v 2 2 m2 ) 2 M2g M 2 m 1 m1g 3 4 mv2. Atwood with Mssive Pulley 03 010 10.0 points A pulley (in the form of uniform disk) with mss 64 kg nd rdius 15 cm is ttched to the ceiling in uniform grvittionl field nd rottes with no friction bout its pivot. The msses re connected by mssless inextensible cord nd T 1, T 2, nd T 3 re mgnitudes of the tensions. The equtionsofmotionfor M 1 nd m 2 re T 1 m 1 g m 1 T 1 m 1 (g +) (1) M 2 g T 2 M 2 T 2 M 2 (g ) (2) nd since α r nd I 1 2 M pr 2 for disk, the eqution of motion for the torque is T 2 47 kg T 3 64 kg 2.8 m T 1 30 kg 15 cm Wht is the mgnitude of the tension T 1? The ccelertion due to grvity is 9.8 m/s 2. Assume up is positive. Correct nswer: 339.853 N. ω Let : M 2 47 kg, m 1 30 kg, M p 64 kg, nd l 2.8 m. Consider the free body digrms Iα (T 2 T 1 )r [M 2 (g ) m 1 (g +)]r ( ) 1 2 M pr 2 r [(M 2 m 1 )g (M 2 +m 1 )]r [M p +2(M 2 +m 1 )] 2(M 2 m 1 )g Thus 2(M 2 m 1 )g M p +2(M 2 +m 1 ) 2(47 kg 30 kg)(9.8 m/s2 ) 64 kg+2(47 kg+30 kg) 1.52844 m/s 2. T 1 m 1 (g +) (1) (30 kg)(9.8 m/s 2 +1.52844 m/s 2 ) 339.853 N. Blocks nd Wedge 011 (prt 1 of 2) 10.0 points A block of mss 2 kg nd one of mss 6 kg re connected by mssless string over pulley tht is in the shpe of disk hving rdius
Version 001 HW#6 - Circulr & ottionl Motion rts (00223) 4 of 0.2 m, nd mss of 5 kg. In ddition, the blocks re llowed to move on fixed blockwedge of ngle 37, s shown. The coefficient of kinetic friction is 0.18 for both blocks. 0.2 m 2 kg 5 kg For the mss m 2, pplying Newton s lw perpendiculr to the slnted surfce N 2 m 2 g cosθ m 2 perp 0 N 2 m 2 g cosθ, so the force of friction is 37 6 kg f 2 µn 2 µm 2 g cosθ (0.18)(6 kg)(9.8 m/s 2 ) cos37 8.45276 N. Applying Newton s lw prllel to the surfce, Wht is the ccelertion of the two blocks? The ccelertion of grvity is 9.8 m/s 2. Assume the positive direction is to the right. Correct nswer: 2.22914 m/s 2. Let : m 1 2 kg, m 2 6 kg, M 5 kg, nd 0.2 m. m 1 T 1 M T 2 m 2 g sinθ f 2 T 2 m 2 T 2 m 2 +m 2 g sinθ f 2 (2) Subtrction eq. (1) from, eq. (2), T 2 T 1 m 2 g sinθ (m 1 +m 2 ) f 1 f 2. Applying Newton s lw to the pulley, τ I α +T 1 T 2 M 2 2 T 1 +T 2 1 2 M m 2 m 2 g sinθ (m 1 +m 2 ) f 1 f 2 1 2 M Applying Newton s lw to m 1, N 1 m 1 g m 1 y 0 N 1 m 1 g, where the force of friction on m 1 is f 1 µn 1 µm 1 g (0.18)(2 kg)(9.8 m/s 2 ) 3.528 N nd T 1 f 1 m 1 T 1 m 1 +f 1. (1) θ 2m 2g sinθ 2f 1 2f 2 2m 1 +2m 2 +M. Since 2m 2 g sinθ 2(f 1 +f 2 ) 2(6 kg)(9.8 m/s 2 )(sin37 ) 2(3.528 N+8.45276 N) 46.8119 kg m/s 2 nd 2m 1 +2m 2 +M 2(2 kg)+2(6 kg)+5 kg 21 kg, then 2m 2g sinθ 2f 1 2f 2 2m 1 +2m 2 +M 46.8119 kg m/s2 21 kg 2.22914 m/s 2.
Version 001 HW#6 - Circulr & ottionl Motion rts (00223) 5 012 (prt 2 of 2) 10.0 points Find the tension in the horizontl prt of the string. Correct nswer: 7.98628 N. From eq. (1), T 1 f 1 +m 1 3.528 N+(2 kg)(2.22914 m/s 2 ) 7.98628 N. Holt SF 08B 01 013 (prt 1 of 2) 10.0 points A uniform 7.18 m long horizontl bem tht weighs 492 N is ttched to wll by pin connection tht llows the bem to rotte. Its fr end is supported by cble tht mkes n ngle of 58 with the horizontl, nd 645 N person is stnding 1.34 m from the pin. 1.34 m 645 N 492 N 58 F T Applying the second (rottionl) condition of equilibrium (with xis of rottion through the center of the bem), τ net (F T sinθ) L 2 +W p ( ) L 2 d ( ) L y 2 0 0 F T L sinθ+w p (L 2d) y L F T L sinθ+w p (L 2d) (W p +W b F T sinθ)l 2F T L sinθ 2W p d W b L F T 2W pd+w b L 2Lsinθ 2(645 N)(1.34 m)+(492 N)(7.18 m) 2(7.18 m)sin58 432.023 N. 014 (prt 2 of 2) 10.0 points WITHDAWN 7.18 m Find the force F T in the cble by ssuming tht the origin of our coordinte system is t the rod s center of mss. Correct nswer: 432.023 N. Let : L 7.18 m, W b 492 N, W p 645 N, θ 58, nd d 1.34 m. Applying the first (trnsltionl) condition of equilibrium verticlly, F y y +F T sinθ W p W b 0 so y W p +W b F T sinθ.