Chpter E - Problems Blinn Collee - Physic425 - Terry Honn Problem E.1 () Wht is the centripetl (rdil) ccelertion of point on the erth's equtor? (b) Give n expression for the centripetl ccelertion s function of the ltitude nle, q L. Wht is this t the ltitude of Bryn, Texs, t q L = 30.7? (c) Wht re the speeds of point t the equtor nd t Bryn, Texs due to the erth's rottion? (d) The erth's rottion is slowin t rte of 2.2 s every 100,000 yers.. T t = D T D t = 2.3 s 100,000 yers Wht is the (very smll) tnentil component of the ccelertion of point on the erth's equtor? Hint: use the chin rule. 2 p r t T = - 2 p r T T 2 t Solution to E.1 = v2 r nd v = 2 p r T ï = J 2 p T N2 r For the erth: T = 1 dy = 24ÿ3600 s () At the equtor: 2 p r = R E ï = K 24ÿ3600 O2 6.37µ10 6 = 0.0337 m Comment: Compred to the rvittionl ccelertion of, this is smll but not neliible. (b) r is the rdius of the circle but somethin on the rottin erth moves lon pth tht trces ltitude line. At some point other tht the equtor this is: r = R E cos q L ï = 0.0337 m µcos q L q L = 30.7 ï = 0.0290 m (c) The speed is found form the period: v = 2 p r T. At equtor: v = At Bryn: v = (d) To find the tnentil ccelertion use t = vê t nd the hints. 2 p 24ÿ3600 6.37µ106 = 463 m s 2 p 24ÿ3600 6.37µ106 cos 30.7 = 398 m s
2 Chpter E - Problems t = v t = 2 p r t T = - 2 p r T T 2 = - 2 p r t T 2 = - 2 p R E H1 dyl 2 D T D t = - 2 p R E H1 dyl 2 2.3 s 100,000 yers 2.3 s 100,000 yer = - 2 p 6.37µ106 H24µ3600L 2 2.3 100,000µH365.24µ24µ3600L = 3.91µ10-15 m Problem E.2 A cr drives round 200 m rdius circle with speed tht decreses uniformly from 30 mês to 20 mês in 8 s. At the instnt the speed i5 mês then: () wht is the centripetl ccelertion, (b) wht is the tnentil ccelertion nd (c) wht is the mnitude of the ccelertion? Solution to E.2 () = v2 r = 252 200 = 3.125 m (b) t = v t = D v D t = 20-30 = -1.25 m 8 (c) = 2 + t 2 = 3.37 m Problem E.3 While movin in ircle with 12 m rdius the speed of prticle vries with time by vhtl = 2 + 10 t - 4 t 2 in SI units. At t = 2 s wht re the centripetl nd tnentil components of th ccelertion? Also ive the mnitude of the totl ccelertion nd the nle of this ccelertion mesured reltive to the centripetl direction. Solution to E.3 vhtl = 2 + 10 t - 4 t 2 ï vh2 sl = 6 ï = v2 r = 62 12 = 3 m v v HtL = 10-8 t ï t t H2 sl = -6 ï t = v t = -6 m = 2 + t 2 = 6.71 m q = tn -1 t = -63.4 Problem E.4 In the Bohr model of the hydroen tom n electron moves in ircle of rdius 5.29µ10-11 m with speed of 2.20µ10 6 mês. Wht net force is needed to produce this motion.
Chpter E - Problems 3 Solution to E.4 = v2 r = I2.20µ106 M 2 5.29µ10-11 = 9.14µ1022 m ï F net = m = 9.11µ10-31 = 8.34µ10-8 N Problem E.5 A smll block sits on turntble tht rottes with period of 3 s. If the coefficient of sttic friction between the turntble nd the block is 0.4 then wht is the lrest distnce the block cn be from the center without slippin? Solution to E.5 Newton's second lw in the centripetl direction ives: nd in the verticl direction F net,c = m ï f s = m J 2 p T N2 r F net,ver = m ver = 0 ï N = m. The constrint force of sttic friction stisfies the inequlity f s m s N. Insertin our informtion from bove into this inequlity ives The lrest rdius r mx sturtes this inequlity. m J 2 p T N2 r m s m. m J 2 p T N2 r mx = m s m ï r mx = m s K T 2 p O2 = 0.4µ9.80 K 3 2 p O2 = 0.894 m Problem E.6 A Ferris Wheel hs rdius of 30 m nd rottes once every 40 s. Wht re the minimum nd mximum norml force of the set on 160 lb mn. A Ferris wheel rottes in verticl circle nd rider lwys sits upriht. Solution to E.6 The extreme cses should be t the bottom nd t the top so tht is ll we need to consider. The mnitude of the ccelertion is the sme everywhere in the circle. Since we re iven the weiht the mss is W ê nd we cn write = J 2 p T N2 r = K 2 p 40 O2 30 = 0.74022
4 Chpter E - Problems m = W 0.74022 = 160 µ = 12.085 lb 9.80 N N W W t top t bottom At the top the ccelertion is downwrd. F net = m ï W - N = m ï N = W - m = 160-12.085 = 148 lb At the bottom the ccelertion is upwrd. F net = m ï N - W = m ï N = W + m = 160 + 12.085 = 172 lb It follows tht the minimum nd mximum norml forces re 148 lb nd 162 lb. Problem E.7 A mss m is moves in verticl circle t the end of strin of lenth L. () If t the bottom the mss hs speed v then wht is the tension T in the strin? (b) If t the top the mss hs speed v then wht is the tension T in the strin? Wht is the minimum speed the mss cn hve without the strin losin its tension? Solution to E.7 T m T m t bottom t top () The ccelertion is directed towrd the center of the circle, which is upwrd in this cse. F net = m ï T - m = m ï T = m + m = m v2 L + (b) At the top the ccelertion is downwrd. F net = m ï T + m = m ï T = m - m = m v2 L - The minimum speed t the top is when the tension is zero.
Chpter E - Problems 5 T = 0 ï v min = L Problem E.8 A mss m is moves in verticl circle t the end of strin of lenth L. If the mss hs speed v, t nle q mesured from verticl s shown, then wht is the tension T in the strin t tht position. Also, find the tnentil componet of the ccelertion t tht position. L q m T =? v Solution to E.8 t m cos q T q m sin q c m Here there re both tnentil nd centripetl components of the ccelertion. The tension force is in the centripetl direction, so tht is wht we must consider. F net,c = m ï T +m cos q = m v2 r ï T = m v2 r - m cos q The tnentil component of the ccelertion comes from the other component of the weiht. The tnentil direction is the direction of the velocity; this ives netive component. F net,t = m t ï -m sin q = m t ï t = - sin q
6 Chpter E - Problems Problem E.9 We sw in the Chpter D notes tht pendulum in n ccelertin crt will hn t n nle iven by tn q =. Tretin the ccelertin crt s n ccelertin frme renlyze the problem to derive the sme result. Wht is the tension in the strin? Solution to E.9 rt y x q eff A forwrd ccelertion corresponds to bckwrd rtificil rvity. A suspended object will hn in the direction of the effective rvity eff = + rt. The nle is iven by: q = tn -1 The tension is just the effective weiht. T = W eff = m eff = m 2 + 2 Problem E.10 This is n extension of problem E.1. Define true verticl to be the direction pointin towrd the center of the erth nd ssume the erth is perfect sphere. A plumb bob will not hn in the direction of true verticl but will hn in the dirction of the effective rvity eff. The ltitude of Bryn Texs is q L = 30.7.
Chpter E - Problems 7 () Wht is the strenth of the effective rvity eff t Bryn Texs, tkin = 9.80 më s exct? (b) Wht is the nle between true verticl nd the direction of eff t Bryn Texs? Solution to E.10 rt = q L f eff q L rt = () Use the lw of cosines to find eff. eff = 2 + 2 rt - 2 rt cos q L = 9.80 2 + 0.0290 2-2µ9.80µ0.0289 cos 30.7 = 9.78 m (b) To et the nle, f in the dirm, use the lw of sines. sin f rt = sin q L eff ï sin f 0.0290 = sin 30.7 9.78 ï f = 0.0867 Problem E.11 HAL HBL HCL HDL HEL The bove rphics represent lss of wter slidin down n incline. View the lss s n ccelerted frme nd nswer the followin: () Which describes lss slidin down frictionless incline? (b) Which describes lss slidin down n incline in the cse where the friction is such tht the speed of the lss is constnt? (c) Which describes lss slidin down n incline in the cse where there is sliht friction, so tht the speed of the lss is incresin? (d) Which describes lss slidin down n incline in the cse where there is hih friction, so tht the speed of the lss is decresin? Solution to E.11
8 Chpter E - Problems The surfce of the wter will be perpendiculr to the effective rvity eff t tht position. () If the incline is frictionless then the ccelertion is = sin q nd wter level is prllel to the surfce. This ives cse (D). sin q rt eff (b) If the friction cuses onstnt velocity, then the ccelertion is zero nd eff =. This ives horizontl wter level nd cse (B). (c) If the friction cuses n incresin velocity, then eff is between norml to the surfce nd verticl, nd thus ives cse (C). rt sin q eff (c) If the friction cuses decresin velocity, then the ccelertion is up the incline nd rt is down the incline. This skews the wter level to cse (A). sin q rt eff Note tht cse (E) would correspond to some externl force, for exmple tension, pullin the lss down the incline with n ccelertion reter thn sin q.
Chpter E - Problems 9 Problem E.12 The outside rim of rottin spce sttion hs 150 m rdius. It is desired to crete n rtificil rvity where 150 lb mn hs n rtificil weiht of 50lb. Wht period of rottion is needed to do this? Solution to E.12 Usin W = m, we cn relte the rtificil weiht to the erth weiht. W rt W = rt ï 50 150 = rt ï rt = 1 3 Since rt = - we hve rt = nd = J 2 p T N2 r. rt = = J 2 p T N2 r ï T = 2 p r rt = 2 p 150 9.80ê3 = 42.6 s