Uniform Circular Motion. Key Terms and Equations. Kinematics of UCM. Topics of Uniform Circular Motion (UCM) Kinematics of Uniform Circular Motion

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1 opics of Unifom icu Motion (UM) Kinemtics of UM ick on the topic to go to tht section Unifom icu Motion 2009 b Goodmn & Zvootni Peiod, Fequenc, nd Rottion Veocit nmics of UM Vetic UM uckets of Wte Roecostes s going ove his nd though ves Hoizont UM Unbnked uves nked uves onic Penduum Ke ems nd qutions entipet Foce qutions: R = v 2 / Kinemtics of UM F = mv 2 / Peiod, Fequenc, & Rottion Veocit qutions: = t/n = 1/f f = n/t = 1/ v = 2π/ = 2πf Retun to be of ontents Kinemtics of Unifom icu Motion Unifom cicu motion: motion in cice of constnt dius t constnt speed Kinemtics of Unifom icu Motion his cceetion is ced the centipet, o di, cceetion, nd it points towds the cente of the cice. Instntneous veocit is ws tngent to cice.

2 Kinemtics of Unifom icu Motion Looking t the chnge in veocit in the imit tht the time intev becomes infinitesim sm, we see tht we get two simi tinges. Kinemtics of Unifom icu Motion If dispcement is equ to veocit mutipied b time, then vt is the dispcement coveed in time t. uing tht sme time, veocit chnged b n mount, Δv. Δ Δ vt v1 v2 v 1 v 2 Δv Δ hese e simi tinges becuse the nges e conguent, so the sides must be in popotion. Δv v Kinemtics of Unifom icu Motion vt = Δv v 2 = t = v 2 v1 v2 Δv vt hese e simi tinges becuse the nges e conguent, so the sides must be in popotion. Δv v Kinemtics of Unifom icu Motion vt = Δv v 2 = t = v 2 ht's the mgnitude of the cceetion. Δv v 1 v 2 vt Kinemtics of Unifom icu Motion v1 v2 nspose v 2 to see the vecto ddition. v1 v2 Δv Kinemtics of Unifom icu Motion his cceetion is ced the centipet, o di, cceetion. It diection is towds the cente of the cice. he chnge in veocit, Δv, shows the diection of the cceetion. In the pictue, ou cn see Δv points towds the cente of the cice. It's mgnitude is given b = v 2 /

3 1 Is it possibe fo n object moving with constnt speed to cceete? pin. No, if the speed is constnt then the cceetion is equ to zeo. No, n object cn cceete on if thee is net foce cting on it. Yes, though the speed is constnt, the diection of the veocit cn be chnging. Yes, if n object is moving it is epeiencing cceetion. 2 onside ptice moving with constnt speed such tht its cceetion of constnt mgnitude is ws pependicu to its veocit. It is moving in stight ine. It is moving in cice. It is moving in pbo. None of the bove is definite tue of the time. 3 n object moves in cicu pth t constnt speed. ompe the diection of the object's veocit nd cceetion vectos. oth vectos point in the sme diection. he vectos point in opposite diections. he vectos e pependicu. he question is meningess, since the cceetion is zeo. 4 Wht tpe of cceetion does n object moving with constnt speed in cicu pth epeience? fee f constnt cceetion ine cceetion centipet cceetion 5 n object is tveing with veocit of 6.0 m/s in cicu pth whose dius is 4.0m. Wht is the mgnitude of its centipet cceetion? 4 m/s 2 9 m/s 2 6 m/s m/s 2.67 m/s 2 6 n object is tveing with veocit of 6.0 m/s in cicu pth. Its cceetion is 3.0 m/s 2. Wht is the dius of its pth? 6 m 1 m 2 m 16 m 12 m

4 7 n object is tveing with veocit in cicu pth whose dius is 65 m. Its cceetion is 3.0 m/s 2. Wht is its veocit? m/s m/s m/s m/s 195 m/s Peiod, Fequenc, nd Rottion Veocit Retun to be of ontents Peiod he time it tkes fo n object to compete one tip ound cicu pth is ced its Peiod. he smbo fo Peiod is "" Peiods e mesued in units of time; we wi usu use seconds (s). Often we e given the time (t) it tkes fo n object to mke numbe of tips (n) ound cicu pth. In tht cse, 8 If it tkes 50 seconds fo n object to tve ound cice 5 times, wht is the peiod of its motion? 5 s 10 s 15 s 20 s 25 s = t/n 9 If n object is tveing in cicu motion nd its peiod is 7.0s, how ong wi it tke it to mke 8 compete evoutions? 56 s 54 s 63 s 58 s 48 s Fequenc he numbe of evoutions tht n object competes in given mount of time is ced the fequenc of its motion. he smbo fo fequenc is "f" Peiods e mesued in units of evoutions pe unit time; we wi usu use 1/seconds (s 1 ). nothe nme fo s 1 is Hetz (Hz). Fequenc cn so be mesued in evoutions pe minute (pm), etc. Often we e given the time (t) it tkes fo n object to mke numbe of evoutions (n). In tht cse, f = n/t

5 10 n object tves ound cice 50 times in ten seconds, wht is the fequenc (in Hz) of its motion? 11 If n object is tveing in cicu motion with fequenc of 7.0 Hz, how mn evoutions wi it mke in 20 s? 25 Hz 20 Hz 15 Hz 10 Hz 5 Hz Peiod nd Fequenc Since = t/n nd f = n/t then = 1/f nd f = 1/ 12 n object hs peiod of 4.0s, wht is the fequenc of its motion (in Hetz)? 1/16 Hz 1/8 Hz 1/4 Hz 1/2 Hz 2 Hz 13 n object is evoving with fequenc of 8.0 Hz, wht is its peiod (in seconds)? 1/8 s 1/4 s 1/2 s 2 s 4 s Rottion Veocit ch tip ound cice, n object tves ength equ to the cice's cicumfeence. he cicumfeence of cice is given b: = 2π he time it tkes to go ound once is the peiod,. nd the object's speed is given b s = d/t So the speed must be: s = / = 2π/

6 Rottion Veocit veocit must hve mgnitude nd diection. he mgnitude of n object's instntneous veocit is its speed. So fo n object in unifom cicu motion, the mgnitude of its veocit is: v = / = 2π/ If n object is in unifom cicu motion, the diection of its veocit is tngent to its cicu motion. 14 n object is in cicu motion. he dius of its motion is 2.0m nd its peiod is 5.0s. Wht is its veocit? 1.98 m/s m/s 4.36 m/s 3.25 m/s 2.51 m/s So v = 2π/ tngent to the cice 15 n object is in cicu motion. he dius of its motion is 2.0 m nd its veocit is 20 m/s. Wht is its peiod? 16 n object is in cicu motion. he peiod of its motion is 2.0 s nd its veocit is 20 m/s. Wht is the dius of its motion? 0.38 s 7.87 m 0.63 s 3.56 m 0.78 s 5.61 m 0.89 s 6.36 m 1.43 s 5.67 m Rottion Veocit Since f = 1/, we cn so detemine the veocit of n object in unifom cicu motion b the dius nd fequenc of its motion. v = 2π/ nd f = 1/ so v = 2πf Of couse the diection of its veocit is sti tngent to its cicu motion. 17 n object is in cicu motion. he dius of its motion is 2.0 m nd its fequenc is 8.0 Hz. Wht is its veocit? m/s m/s m/s m/s m/s So v = 2πf tngent to the cice

18 n object is in cicu motion. he dius of its motion is 2.0 m nd its veocit is 30 m/s. Wht is its fequenc? 19 n object is in cicu motion. he fequenc of its motion is 7.0 Hz nd its veocit is 20 m/s.

7 18 n object is in cicu motion. he dius of its motion is 2.0 m nd its veocit is 30 m/s. Wht is its fequenc? 19 n object is in cicu motion. he fequenc of its motion is 7.0 Hz nd its veocit is 20 m/s. Wht is the dius of its motion? 4.12 Hz 0.45 m 2.82 Hz 2.08 m 2.39 Hz 0.33 m 3.67 Hz 1.22 m 1.78 Hz 1.59 m nmics of UM nmics of Unifom icu Motion Fo n object to be in unifom cicu motion, thee must be net foce cting on it. We ed know the cceetion, so we cn wite the foce: Retun to be of ontents nmics of Unifom icu Motion We cn see tht the foce must be inwd b thinking bout b on sting: Foce on hnd eeted b sting Foce on b eeted b sting nmics of Unifom icu Motion hee is no centifug foce pointing outwd; wht hppens is tht the ntu tendenc of the object to move in stight ine must be ovecome. If the centipet foce vnishes, the object fies off tngent to the cice. his hppens. his does NO hppen.

uved Pths entifugtion his concept cn be used fo n object moving ong n cuved pth, s sm segment of the pth wi be ppoimte cicu. centifuge woks b spinning ve fst.

8 uved Pths entifugtion his concept cn be used fo n object moving ong n cuved pth, s sm segment of the pth wi be ppoimte cicu. centifuge woks b spinning ve fst. his mens thee must be ve ge centipet foce. he object t woud go in stight ine but fo this foce; s it is, it winds up t. 20 Wht foce is needed to mke n object move in cice? 21 When n object epeiences unifom cicu motion, the diection of the net foce is kinetic fiction sttic fiction centipet foce weight in the sme diection s the motion of the object in the opposite diection of the motion of the object is diected towd the cente of the cicu pth is diected w fom the cente of the cicu pth 22 c with mss of 1800 kg goes ound n 18 m dius tun t speed of 35 m/s. Wht is the centipet foce on the c? kg mss is ttched to the end of 5.0 m ong met od sting which ottes in hoizont cicu pth. If the mimum foce tht the od cn withstnd is 8500 N. Wht is the mimum speed tht the mss cn ttin without beking the od?

9 Vetic UM on hi od... Retun to be of ontents c is tveing t veocit of 20 m/s. he dive of the c hs mss of 60 kg. he c is octed t the bottom of dip in the od. he dius of the dip is 80m. Wht is the ppent weight of the dive (the nom foce suppied b the set of the c to suppot him) t the bottom of the dip? his is sketch of the pobem. Wht do I do net? 1. w fee bod digm 2. Indicte the diection of cceetion v = 20 m/s = 80 m he dotted ines epesent es with one is pe to cceetion v = 20 m/s = 80 m Net, 3. w es with one is pe to cceetion v = 20 m/s = 80 m foces e pe o pependicu to the es, so I don't hve to esove n vecto into components

10 4. pp Newton's Second Lw ong ech is. diection F = m 0 = 0 diection F = m = m = + m = m (g + ) FN 6. he st step. Substitute numbes. = m (g + v 2 /) = (60 kg) ((9.8 m/s 2 + (20 m/s) 2 / (80m)) v = 20 m/s = 80 m Whie I don't know "", I do know "v" nd "". Wht's m net step? 5. Substitute = v 2 / = m (g + v 2 /) v = 20 m/s = 80 m = (60 kg) (9.8 m/s m/s 2 ) = (60 kg) (14.8 m/s 2 ) = (60 kg) (14.8 m/s 2 ) = 890 N How does this compe to his weight on the ft pt of the od? v = 20 m/s = On the ft od the dive's weight (the nom foce fom the set) is just. = = (60kg)(9.8 m/s 2 ) = 590 N ppent Weight Ft od ottom of dip ( = 80m) 590 N 890 N he gvittion foce on the dive () doesn't chnge, but his ppent weight () does. Is thee sitution whee he wi ppe weightess? t wht veocit must c dive ove hi if the dive (nd the c fo tht mtte) e to ppe weightess? he dive of the c hs mss of 60 kg. he dius of the hi is 80m. his is sketch of the pobem. Wht do I do net? 1. w fee bod digm 2. Indicte the diection of cceetion = 80 m his is the fee bod digm (et's stt out with, knowing we' eventu et it equ zeo) Net, 3. w es with one is pe to cceetion = 80 m

11 = 80 m he dotted ines epesent es with one is pe to cceetion = 80 m 4. pp Newton's Second Lw ong ech is. diection diection FN FN ΣF = m 0 = 0 ΣF = m = m = m = m (g ) ut I'd ike to find the veocit tht the c must be going fo the dive (nd c) to ppe weightess. How? 5. Substitute = v 2 / = m (g v 2 /) = 80 m FN 6. he st step, when does = 0 When the poduct of two vibes is zeo, then one of them must be zeo. Since m is not zeo, the on w F n cn equ zeo is... = m (g v 2 /) 0 = (60 kg) (g v 2 /) 0 = (g v 2 /) Pctice Pobem: c is tveing t veocit of 15 m/s. he dive of the c hs mss of 50 kg. he c is octed t the top of hi in the od. he dius of the hi is 45 m. Wht is the ppent weight of the dive (the nom foce suppied b the set of the c to suppot him) t the top of the hi? 240 N v 2 / = g v = (g) 1/2 v = ((9.8m/s 2 )(80m)) 1/2 v = 28 m/s 24 c is going ove the top of hi whose cuvtue ppoimtes cice of dius 175 m. t wht veocit wi the occupnts of the c ppe to weigh 10% ess thn thei nom weight ( = 0.9)? 25 c is going though dip in the od whose cuvtue ppoimtes cice of dius 175 m. t wht veocit wi the occupnts of the c ppe to weight 10% moe thn thei nom weight? 13.1 m/s 9.8 m/s m/s 12.7 m/s 13.9 m/s 11.9 m/s 14.2 m/s 13.1 m/s 12.7 m/s 14.5 m/s

12 26 he occupnts of c tveing t speed of 25 m/s note tht on pticu pt of od thei ppent weight is 20% highe thn thei weight when diving on ft od. Is tht pt of the od hi, o dip? 27 he occupnts of c tveing t speed of 25 m/s note tht on pticu pt of od thei ppent weight is 20% highe thn thei weight when diving on ft od. Wht is the vetic cuvtue of the od? Hi ip m m m m m bucket of wte is being spun is vetic cice of dius 0.80m. Wht is the smest veocit which wi esut in the wte not eving the bucket? Wht do ou know nd wht e ou ooking fo? uckets nd Roecostes... = 0.80 m g = 9.8 m/s 2 downwd v =? = 0.80 m g = 9.8 m/s 2 downwd v minimum =? = 0.80 m g = 9.8 m/s 2 downwd v minimum =?

13 ΣF = m ΣF = mv 2 / = 0.80 m g = 9.8 m/s 2 downwd v minimum =? = 0.80 m g = 9.8 m/s 2 downwd v minimum =? ΣF = m + = m(v 2 /) ssuming constnt speed, nd tht the mss of the bucket is 2.5kg, wht wi the tension in the sting be t the bottom of the cice? = m(v 2 /) ΣF = mv 2 / = m(v 2 / g) = 0 fo minimum v v 2 / g = 0 v 2 / = g v = (g) 1/2 = ((9.8)(0.8)) 1/2 = 2.8 m/s = 0.80 m g = 9.8 m/s 2 downwd vminimum =? = 0.80 m g = 9.8 m/s 2 downwd ssuming constnt speed, nd tht the mss of the bucket is 2.5kg, wht wi the tension in the sting be t the bottom of the cice? ΣF = m = m Pctice Pobem: bucket of wte is being spun is vetic cice of dius 1.2 m. Wht is the smest veocit which wi esut in the wte not eving the bucket? 3.43 m/s ΣF = mv 2 / = m(v 2 /) = m(v 2 /) + = m(v 2 / + g) ssuming constnt speed, nd tht the mss of the bucket is 1.3 kg, wht wi the tension in the sting be t the bottom of the cice? = 2.5kg (9.8 m/s m/s 2 ) = 2.5kg (19.6 m/s 2 ) = 490 N N = 0.80 m g = 9.8 m/s 2 downwd

14 28 b is ttched to the end of sting. It is swung in vetic cice of dius 10 m. Wht is the minimum veocit tht the b must hve to mke it ound the cice? 29 b is ttched to the end of sting. It is swung in vetic cice of dius 2.25 m. Wht is the minimum veocit tht the b must hve to mke it ound the cice? 2.8 m/s 3.87 m/s 9.9 m/s 4.34 m/s 7.5 m/s 4.69 m/s 2.1 m/s 5.12 m/s 3.9 m/s 5.39 m/s 30 oe coste c is on tck tht foms cicu oop in the vetic pne. If the c is to just mintin contct with tck t the top of the oop, wht is the minimum vue fo its centipet cceetion t this point? 31 oe coste c (mss = M) is on tck tht foms cicu oop (dius = ) in the vetic pne. If the c is to just mintin contct with the tck t the top of the oop, wht is the minimum vue fo its speed t tht point? g downwd 0.5g downwd g upwd 2g upwd g (g) 1/2 (2g) 1/2 (0.5g) 1/2 32 piot eecutes vetic dive then foows semi cicu c unti it is going stight up. Just s the pne is t its owest point, the foce on him is Hoizont UM ess thn nd pointing up ess thn nd pointing down moe thn nd pointing up moe thn nd pointing down Retun to be of ontents

nked nd Unbnked uves When c goes ound cuve, thee must be net foce towds the cente of the cice of which the cuve is n c. If the od is ft, tht foce is suppied b fiction.

If the ties do stt to sip, the fiction is kinetic, which is bd in two ws: he kinetic fiction foce is sme thn the sttic.

15 nked nd Unbnked uves When c goes ound cuve, thee must be net foce towds the cente of the cice of which the cuve is n c. If the od is ft, tht foce is suppied b fiction. nked nd Unbnked uves If the fiction foce is insufficient, the c wi tend to move moe ne in stight ine, s the skid mks show. nked nd Unbnked uves s ong s the ties do not sip, the fiction is sttic. If the ties do stt to sip, the fiction is kinetic, which is bd in two ws: he kinetic fiction foce is sme thn the sttic. he sttic fiction foce cn point towds the cente of the cice, but the kinetic fiction foce opposes the diection of motion, mking it ve difficut to egin conto of the c nd continue ound the cuve. Unbnked uves c is going ound tck with veocit of 20 m/s. he dius of the tck is 150 m. Wht is the minimum coefficient of sttic fiction tht woud mke tht possibe? Wht do ou know nd wht e ou ooking fo? v = 20 m/s = 150 m μ =? Unbnked uves Unbnked uves op View Font View (the c heding towds ou) op View Font View (the c heding towds ou) f s v v = 20 m/s = 150 m μ =? v = 20 m/s = 150 m μ =?

16 Unbnked uves vetic vetic Unbnked uves vetic diection di diection ΣF = m ΣF = m FN = 0 f s = m v f s di f s FN = di μ s = m(v 2 /) μ s = mv 2 / μ s = v 2 /g μ s = (20m/s) 2 /((9.8 m/s2)(150m)) μ s = 0.27 v = 20 m/s = 150 m μ =? v = 20 m/s = 150 m μ =? 33 c goes ound cuve of dius t constnt speed v. hen it goes ound the sme cuve t hf of the oigin speed. Wht is the centipet foce on the c s it goes ound the cuve fo the second time, comped to the fist time? twice s big fou times s big hf s big one fouth s big nked uves nking cuves cn hep keep cs fom skidding. In fct, fo eve bnked cuve, thee is one speed whee the entie centipet foce is suppied b the hoizont component of the nom foce, nd no fiction is equied. Let's figue out wht tht speed is fo given nge nd dius of cuvtue. v op View nked uves Font View (the c heding towds ou) We know the diection of ou cceetion, so now we hve to cete es. vetic nked uves di Net, do the fee bod digm. Let's ssume tht no fiction is necess t the speed we e soving fo. Note tht these es wi be diffeent thn fo n Incined Pne, becuse the c is not epected to side down the pne. Insted, it must hve hoizont cceetion to go in hoizont cice.

17 nked uves nked uves vetic di Net, decompose the foces tht don't ign with n is,. cos vetic sin di Now et's sove fo the veocit such tht no fiction is necess to keep c on the od whie going ound cuve of dius nd bnking nge. nked uves nked uves vetic vetic diection ΣF = m di diection ΣF = m vetic vetic diection ΣF = m di diection ΣF = m cos sin di FNcos = 0 FN = /cos FNsin = m (/cos)(sin) = m(v 2 /) (g/cos)(sin) = (v 2 /) (gtn) = (v 2 /) cos sin di FNcos = 0 FN = /cos FNsin = m (/cos)(sin) = m(v 2 /) (g/cos)(sin) = (v 2 /) (gtn) = (v 2 /) v = (gtn) 1/2 v = (gtn) 1/2 Pctice Pobem: etemine the veocit tht c shoud hve whie tveing ound fictioness cuve with dius 250 m is bnked t n nge of m/s onic Penduum onic Penduum is penduum tht sweeps out cice, the thn just bck nd foth pth. Since the penduum bob is moving in hoizont cice, we cn stud it s nothe empe of unifom cicu motion. Howeve, it wi pove necess to decompose the foces.

18 onic Penduum onic Penduum w sketch of the pobem, uness one is povided. hen dw fee bod digm nd indicte the diection of the cceetion. hen dw es with one is pe to cceetion onic Penduum onic Penduum cos sin hen decompose foces so tht components ie on n is...in this cse, decompose. hen sove fo the cceetion of the bob, bsed on the nge, b pping Newton's Second Lw ong ech is. cos sin onic Penduum diection F = m sin = m diection ivide these two esuts sin = m cos = tn = /g F = m cos = 0 cos = onic Penduum n tentive ppoch is to sove this s vecto eqution using ΣF=m. (his wi wok wheneve on two foces e pesent.) Just tnste the oigin vectos (befoe decomposing them) to fom ight tinge so tht the sum of the two foces equs the new vecto "m". = tn 1 (/g) O = g tn

19 onic Penduum hen, since tn = opposite/djcent tn = m/ = /g tn = (v 2 /)/g = v 2 /g m v 2 = gtn 0.5 kg b on sting is whied in cice with dius of 0.75 m with speed of 3 m/s. Find the tension in the sting. the sme esut we found befoe 0.5 kg b on sting is whied in cice with dius of 0.75 m with speed of 3 m/s. Find the tension in the sting. diection diection ΣF = m ΣF = m = m = 0 = mv 2 / = 1.5 kg b on sting is whied in cice with dius of 2.25 m with speed of 6 m/s. Find the tension in the sting. 2 = = (mv 2 /) 2 + () N 2 = m 2 (v 4 / 2 + g 2 ) 2 = (0.5kg) 2 ((3m/s) 4 /(0.75m) 2 + (9.8m/s 2 ) 2 ) 2 = 60 N 2 = 7.7 N Nonunifom icu Motion If n object is moving in cicu pth but t ving speeds, it must hve tngenti component to its cceetion s we s the di one.

Algebra Based Physics. Gravitational Force. PSI Honors universal gravitation presentation Update Fall 2016.notebookNovember 10, 2016

Algebra Based Physics. Gravitational Force. PSI Honors universal gravitation presentation Update Fall 2016.notebookNovember 10, 2016 Newton's Lw of Univesl Gvittion Gvittionl Foce lick on the topic to go to tht section Gvittionl Field lgeb sed Physics Newton's Lw of Univesl Gvittion Sufce Gvity Gvittionl Field in Spce Keple's Thid Lw