AP Physics 1- Circular Motion and Rotation Practice Problems

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AP Physics 1- Cicula Motion and Rotation Pactice Poblems FACT: The motion of an object in a cicula path at a constant speed is known as unifom cicula motion (UCM).

1 AP Physics 1- Cicula Motion and Rotation Pactice Poblems FACT: The motion of an object in a cicula path at a constant speed is known as unifom cicula motion (UCM). Because an object in UCM is constantly changing diection, it is also constantly changing its velocity. This means that an object in UCM is constantly acceleating. This is called centipetal acceleation (a c). FACT: Recall, that a = Δv = v f v 0, divided by the change in time. We can show this vectoially in the diagams to the ight. Recall, that when we subtact vectos, we ae eally adding a negative vecto. You can epeat this pocess aound a cicle and the acceleation vecto will always point towad the cente of the cicle. Centipetal means cente-seeking. FACT: Now that we know the diection of the object s acceleation is towads the cente, we need to find the magnitude. The fomula fo the object s centipetal acceleation is: a c = v2. The magnitude of centipetal acceleation clealy depends on the objects velocity (v) and the adius () of the cicula path. The visual intuition fo this idea is descibed in podcast #2 and it is impotant to acknowledge that diffeential calculus is needed fo the deivation. Howeve, hee is anothe explanation that does not involve calculus. One otation takes 2π/v 2 units of time (t). And the velocity changes fom v to -v and back to v again (total change = 4 v). So acceleation = Δv/Δt = 2v 2 /π. Again, podcast #2 povides a visual epesentation of this explanation. Q1. The diagam shows the top view of a 65 kg student at a point A on an amusement pak ide. The ide spins the student in a hoizontal cicle of adius 2.5 m, at a constant speed of 8.6 m/s. The floo is loweed and the student emains against the wall without falling to the floo. (a) daw the vecto fo the centipetal accleation on the diagam (b) Which diection is the foce nomal? (c) Which diection is the foce of fiction on the student s body? Q2. Which gaph best epesents the elationship between the magnitude of the centipetal acceleation and the speed of an object moving in a cicle of constant adius? Q3. A 0.50 kg object moves in a hoizontal cicula path with a adius of 0.25 m at a constant speed of 4.0 m/s. What is the magnitude of the object s acceleation? FACT: Since an object taveling in a cicula path has an inwad acceleation, Newton s 2 nd Law states thee must also be an inwad net foce. This centipetal foce can be caused by a gavitational foce, fictional foce, tension foce, o foce applied. Now go back to Q1 and identify the centipetal foce (F N). Let s looks at the expession fo the centipetal foce using F c = ma c and substituting in the fomula fo a c we now get F c = m v2. Please note that it is incoect to label the centipetal foce F c on a FBD. You must identify the foce actually causing the a c and use that label in you FBD (e.g., F g, mg, F T, F a, F f). Always call the inwad foces positive (+) when summing the foces.

2 Q4. An object of mass 5 kg moves at a constant speed of 6 m/s in a cicula path of adius 2 m. Find the object s acceleation and the net foce esponsible fo the motion. Q5. A 10.0 kg mass is attached to a sting that has a beaking stength of 200N. If the mass is whiled in a hoizontal cicle of adius 80 cm, what maximum speed can it have? Q6. An athlete who weighs 800 N is unning aound a cuve at a speed of 5.0 m/s in an ac whose adius of cuvatue,, is 5.0 m. Find the centipetal foce acting on him. What povides the centipetal foce? What could happen to him if wee smalle? Q7. A 1200-kg ca taveling at a constant speed of 9 m/s tuns at an intesection. The ca follows a hoizontal cicula path with a adius of 25 m to a point, P. At point P, the ca hits an aea of ice and loses all fictional foce on its ties. (a) Which path does the ca follow on the ice? (b) What was the magnitude of the centipetal foce the instant befoe the ca hit point P? (c) What is the coefficient of static fiction fo the ca s ties on the pavement befoe hitting point P? Q8. The maximum speed with which a 945-kg ca makes a 180-degee tun is 10.0 m/s. The adius of the cicle though which the ca is tuning is 25.0m. Calculate the foce of fiction and the coefficient of fiction acting upon the ca. FACT: Fo objects moving in cicula paths, you can chaacteize thei motion aound the cicle in tems of fequency (f) and peiod (T). The fequency is the numbe of evolutions the object makes in a second (1/s, Hz). The peiod is the time it takes to complete one evolution. The peiod and fequency ae expessed as: f = 1 T and T = 1 f. Q9. Sophia makes 38 complete evolutions on the playgound mey-go-ound in 30 seconds. If the adius of the meygo-ound is 1 mete and he mass is 16 kg, detemine the a). peiod of motion; b). fequency of motion; c). speed at which she evolves; d). centipetal foce FACT: Thus fa, we have only exploed centipetal motion in the context of hoizontal cicula paths. When objects ae manipulated into vetical cicula paths, we need to account fo the foce of gavity, o weight (mg) of the object duing cetain points along the cicle. Always, daw a FBD and sum the foces to avoid caeless mistakes. Q10. A ollecoaste ca entes the cicula-loop potion of the ide. At the vey top of the cicle, the speed of the ca is 15 m/s, and the acceleation points downwad. If the diamete of the loop is 40 m and the total mass of the ca is 1200 kg, find the magnitude of the nomal foce exeted by the tack on the ca. Would the peson in the ca feel lighte o heavie? Q11. Refeing to Q10, how would the nomal foce change if the ca was at the bottom of the loop? Please explain using qualitative and quantitative easoning. FACT: Many students ae cuious about the concept of the g-foce felt by a peson. Although, this concept is pobably not on the AP exam, it is still inteesting to conside. In the peceding question, you simplified the expession to F N = m( v 2 +g). In paenthesis you have the acceleation due to gavity, g, and an additional tem. It is this tem (ac) that is the additional g-foce felt by a peson. If that tem equals 9.8 m/s 2, then the peson expeiences two g s. Now go back to Q10 and detemine how many g s ae the people in the ollecoaste ca expeience at the bottom of the loop?

3 Q12. Sophia expeiences a downwad acceleation of 15.6 m/s 2 at the top of a olle coaste loop and an upwad acceleation of 26.3 m/s 2 at the bottom of the loop. Use Newton's second law to detemine the nomal foce acting upon Sophia's 864 kg olle coaste ca. Q13. Sophia is iding on a olle coaste. Sophia is moving at 18.9 m/s ove the top of a hill that has a adius of cuvatue of 12.7 m. Use Newton's second law to detemine the magnitude of the applied foce of the tack pulling down upon Sophia's 621 kg olle coaste ca. Q14. In an expeiment, a ubbe stoppe is attached to one end of a sting that is passed though a plastic tube befoe weights ae attached to the othe end. The stoppe is whiled in a hoizontal cicula path at a constant speed. (a) Descibe what would happen to the adius of the cicle if the student whils the stoppe at a geate speed without changing the balancing weights. (b) The ubbe stoppe is now whiled in a vetical cicle at the same speed. Daw a FBD fo this scenaio. Q15. Amedeo dives his ca clockwise aound a cicula tack with a adius of 30 metes. He completes 10 laps aound the tack in 2 minutes. a). Find Amedeo s total distance taveled, aveage speed, and centipetal acceleation. b). What is his displacement and aveage velocity? Q16. Banking allows fo cas to tavel aound a cuve at o below a posted speed limit. In the diagam, the adius of cuvatue is 60 metes and the cuve is banked at Ѳ = 11.8 degees. What ecommended speed limit should be posted? FACT: So fa in this couse we have only consideed tanslational motion. Tanslational motion of a body can be eithe ectilinea o cuvilinea, as we have studied in pevious units, including an object manipulated into a cicula path. Howeve, many objects also otate aound an axis, known as otational, o angula, motion. Thus, some objects have both tanslational and otational motion. FACT: Befoe poceeding, we need to eview some basics about the unit cicle. In otational motion, and depending on the situation, we will use adians (ad), degees, and evolutions (ev). Recall that a adian measues a distance aound an ac (s = ac length in metes) equal to the length of the ac s adius (). Also ecall that π = C and eaanging we get d C= 2π. FACT: In discussing angula displacements (Δθ), we must tansition to descibing the tanslational displacement aound an ac (Δs) in tems of Δθ. In othe wods, we need to be able to convet between tanslational and otational motion. We will use the following convesions to accomplish this: Δs= C= 2π=2π ad =1 ev= 360. Lastly, ecognize that adians ae dimensionless and a ad is meely a placeholde to emind you that you'e talking about an angle (θ). Fo example, C = 2π ad = 360 and when solving fo a tanslational quantity we will cancel out the ad and use the units we ae left with, in this case metes (m). Let s do some examples to pactice these convesions and tun off adians mode. Q17. Convet 90 to adians; convet 6 adians to degees; convet 1.5 evolutions to both adians and degees. Q18. An insect sits at the top of a bike wheel with diamete of 0.75 m. Assuming the wheel tuns counteclockwise, what is the angula displacement of the insect befoe it is squashed unde the wheel? What ac length does the beetle tavel though befoe it is squashed?

FACT: Rotational kinematics is simila to tanslational kinematics. You simply need to lean the otational vesions of the kinematic vaiables and equations (see tables below).

The angula acceleation (α; alpha) is given in units of adians pe second 2 (ad/s 2 ) and is equal to the change in omega ove the change in time (α = ω t ). t Q19. An 18.5 kg bag of sand sits 1.

4 FACT: Rotational kinematics is simila to tanslational kinematics. You simply need to lean the otational vesions of the kinematic vaiables and equations (see tables below). The angula velocity (ω; omega) is given in units of adians pe second (ad/s) and is equal to the change in theta ove the change in time (ω = θ ). The angula acceleation (α; alpha) is given in units of adians pe second 2 (ad/s 2 ) and is equal to the change in omega ove the change in time (α = ω t ). t Q19. An 18.5 kg bag of sand sits 1.5 m fom cente of mey-go-ound. If mey-go-ound otates at a constant angula speed of 6.58 adians/s. Find the coefficient of static fiction between the bag of sand and the mey go ound. FACT: The diection of angula vectos is detemined by the ight-hand ule. Wap the finges of you ight hand in the diection of the otational displacement, velocity, o acceleation, and you thumb will point in the vecto s diection with up (counteclockwise) being positive. See the diagam to the ight. Q20. A ca s tie otates counteclockwise 3.5 times in 0.75 seconds. What is the angula speed of the tie? Now convet into pm to ensue you ae using the calculato coectly. Q21. A cylindical space station is otating counteclockwise at ad/s with a 115 mete ladde extending fom the middle to the outside. Find the centipetal acceleation at thee points extending fom the middle, half-way, and the outside (= 115 m on outside and 0 m in middle). Q22. Answe the following questions using both qualitative and quantitative easoning. Can an object move in a cicle without having centipetal acceleation? Can an object move in a cicle without having tangential acceleation? Q23. Fou childen climb on a caousel that initially at est. The caousel acceleates to 0.4 ad/s within 10 seconds, what is the angula acceleation? What is the linea acceleation (tangential acceleation) 3 metes fom the axis of otation? Q24. Sophia ides a unicycle. If the wheel begins at est, and acceleates unifomly in a counte-clockwise diection to an angula velocity of 15 pms in a time 6 sec. find the angula acceleation of the wheel. Q25. Ca is taveling at 23 m/s aound a flat cuve with a adius of cuvatue of 100 m. The ca is going as fast as it can without skidding. What minimum coefficient of fiction is necessay fo the ca to ound the cuve? Q26. A capente cuts a piece of wood with a cicula saw. The saw blade acceleates fom est with an angula acceleation of 14 ad/s/s to a max speed of 15,000 pms. 1). What is the maximum speed of the saw in adians/second? 2). How long does it take to each this maximum speed? 3). How many complete otations does the saw make while eaching the maximum speed.

5 Q27. An amusement pak ide of adius x allows childen to sit in a spinning swing held by a cable of length L. At maximum angula speed, the cable makes an angle Ѳ with the vetical as shown in the diagam. Deive an expession fo the maximum angula speed (ω) of the ide in tems of g, Ѳ, x, and L. FACT: All objects that have mass attact each othe with a gavitational foce. The magnitude of that foce, F g, can be calculated using Newton s Law of Univesal Gavitation: F g = G m 1m 2. The value fo the constant, G, is 6.67 x N m 2 /kg 2. Please note that the distance between the masses,, in the denominato is the distance between the cente of mass of the objects. It is not necessaily the adius of the planet o object. Some textbooks and tutoials eplace the with d. Fo example, if compaing two elatively massive objects like the Eath and Moon, then the is the distance between the cente of mass of the two objects. But, when calculating the foce between a human and Eath, the is effectively the adius of Eath, o close enough. As usual, let s fist stat with some simple poblems: Q28. What is the magnitude of the gavitational foce of attaction between two asteoids in space, each with a mass of 50,000 kg, sepaated by a distance of 3800 metes? Q29. Given that the adius of the Eath is about 6.37 x 10 6 metes, detemine the mass of the Eath. Q30. The Moon obits Eath in a cicula path at a constant speed. If M is the mass of Eath, m is the mass of the Moon, and is the Moon s adius, deive an expession fo the Moon s obit speed. The Moon obits Eath in a cicula path at a constant speed. If M is the mass of Eath, m is the mass of the Moon, and is the Moon s adius, deive an expession fo the Moon s obit speed. R is distance to moon FACT: As I have mentioned epeatedly, on the AP Exam you will likely be asked a mathematical easoning question. When using the Law of Univesal Gavitation, note that if eithe of the masses doubles, then the gavitational foce doubles. If the distance between the objects () is doubled, then the foce is quateed. If distance () inceases by thee, then the weight deceases by 3 2 = 9 fold. Mass is an intinsic popety of an object and does not change. Q31. A dwaf planet has 1/200 the mass and 1/5 the adius of Eath. What is the value fo g (m/s/s) on the suface of the planet? Q32. At the suface of Eath, an object of mass m has a weight w. If this object is tanspoted to an altitude that is twice the adius of Eath, what can be said about m and w? Q33. (a) Find the magnitude of the acceleation due to gavity of an object in the toposphee given the mass of the Eath is 5.97 x kg and the adius of the Eath is 6371 km. Remembe that G = 6.67 x (b) What about a space station 400 km above Eath? How fast does the space station need to move to stay in obit? Q34. A 2.00 kg object weighs 19.6 N on Eath. If the acceleation due to gavity on Mas is 3.71 m/s/s, what is the object s mass on Mas? FACT: As two objects exet an equal and opposite gavitational foce, they cause a displacement. So the gavitational foce is doing wok (W = F x d) on the objects, which coesponds with the gavitational potential enegy. You ae familia with the fact that U g = mgh close to the suface of the Eath, but what about objects in deep space? You must use the equation fo the law of univesal gavitation as the foce and the displacement as. Theefoe, this can be expessed as U g = G m 1m 2 x, which simplifies to 2 Ug = Gm 1m 2 and again, you may see d used in place of.

6 Q35. A satellite is cuently obiting Eath in a cicula path of adius R and its kinetic enegy K. If the satellite is moved and entes a new cicula obit of adius 2R, deive an expession fo its new kinetic enegy. Q36. A binay sta is a celestial phenomenon in which two stas obit aound thei mutual cente of mass. If the mass of the pimay sta is 4m, and the mass of the seconday sta is m, what is the gavitational potential enegy of the stat system if the stas ae sepaated by a distance of? FACT: Gavitational mass indicates how an object esponds to a gavitational field. Inetial mass indicates how an object acceleates in esponse to a net foce. These values ae equal. You will be asked a question on the AP Exam that will equie you to distinguish between these two types of mass. Please view the two Bozeman podcasts fo a wondeful explanation with examples. Q37. Design an expeiment that would allow you to detemine which object, object A o object B, has the lagest gavitational mass given they ae in the same gavitational field. Next, descibe an expeiment that would allow you to detemine which object, object A o object B, has the lagest inetial mass egadless of the gavitational field. Q38. Visit the Unit 5 webpage on and scoll down to the poblem set fo this unit. You will find an accompanying video of a levitating neodymium magnet. The mass of the magnet is 11 gams and the diamete is 4.45 cm. (a). What is the aveage acceleation duing the 15-minute inteval shown on the clock? (b). Based on this value, how long would it take fo the disk to stop spinning? (c). If you obseved that the disk continues to spin longe than you pedicted time, suggest possible easons.

PS113 Chapter 5 Dynamics of Uniform Circular Motion

PS113 Chapter 5 Dynamics of Uniform Circular Motion PS113 Chapte 5 Dynamics of Unifom Cicula Motion 1 Unifom cicula motion Unifom cicula motion is the motion of an object taveling at a constant (unifom) speed on a cicula path. The peiod T is the time equied