Honos Physics Fall, 2016 Cicula Motion & Toque Test Review Name: M. Leonad Instuctions: Complete the following woksheet. SHOW ALL OF YOUR WORK ON A SEPARATE SHEET OF PAPER. 1. Detemine whethe each statement is Tue o False. If the statement is false, coect the statement so that it is tue. a) When diving aound a tun, the fictional foce fom the ca s ties cancels the centipetal foce as shown in the fee-body diagam below. f F c b) The peiod is the amount of time it takes fo an object to tavel aound a cicula path once. c) Newton s Law of Univesal Gavity says that evey object is attacted to evey othe object by a foce given by: F g = G m 1 m 2 2 d) An object is in equilibium when F = 0 N.
Honos Physics/Cicula Motion & Toque Test Review Page 2 of 7 Name: 2. The figue below shows fou foces acting on a ba which otates aound a pivot on the left end of the ba. Rank the fou foces in tems of the toque each foce geneates about the pivot, list the foces fom smallest to lagest toque. F A F C F B F D 3. A ca dives ove a lage bump in a oad. The bump has a adius of cuvatue of 20.0 m. a) Daw a fee-body diagam of the ca when the ca eaches the top of the hill. b) How does the magnitude of the Nomal Foce compae the the magnitude of the ca s Weight when the ca dives ove the top of the bump? c) How fast can the ca dive ove the bump without losing contact with the oad? 4. A stuntman dives a motocycle though a vetical loop which has a adius of 2.5 m. a) Daw a fee-body diagam of the motocycle when bike is at the top of the loop. b) What is the minimum speed the bike can have and still make it though the vetical loop? c) Suppose the stuntman dives though the loop at 10 m. If the motocycle and the stuntman have a s combined mass of 150 kg, what was the Nomal Foce exeted by the loop on the motocycle when the bike eached the top of the loop? 5. A ecod playe playing a single spins at 45 pm. a) What is the peiod of the ecod? b) What is the angula fequency of the ecod? c) A typical single has a adius of 9.0 cm. What is the acceleation of a point on the oute edge of the ecod? d) Suppose a coin is placed on the oute edge of the ecod. The coefficient of static fiction between the coin and the ecod is µ s = 0.3. Will the coin slide off the edge of the ecod?
Honos Physics/Cicula Motion & Toque Test Review Page 3 of 7 Name: 6. Satellite A obits the Eath in a cicula obit with a peiod of 8.0 hous. The adius of this obit is 2.0 10 7) m. Satellite B has a peiod of 24.0 hous. What is the distance fom the cente of the Eath to satellite B? 7. A gymnast stands 0.3 m fom the left edge of a balance beam. The balance beam has a suppot on each end of the beam. The beam has a mass of 500 kg while the gymnast has a mass of 50 kg. 0.3 m a) What is the total foce exeted by both suppots? b) What is the foce exeted by the suppot on the ight side of the balance beam? c) What is the foce exeted on the left end of the beam?
Honos Physics/Cicula Motion & Toque Test Review Page 4 of 7 SOLUTIONS Solutions to Cicula Motion & Toque Test Review 1. a) When an object moves with unifom cicula motion, the net foce exeted on the object points towads the cente of the cicula path. Theefoe the fictional foce exeted on the ca points towads the cente of the cicula path. Thee ae no foces pointing away fom the cente of the cicula path. 1. b) Tue. 1. c) Tue. 1. d) False. An object is in equilibium when F = 0 N and τ = 0 N m. 2. C > A > B > D. Recall that the fomula fo the toque is τ = F l, whee F is the component of the foce that is pependicula to the leve am and l is the length of the leve am. The leve am is a vecto that points fom the pivot axis of otation) to the point whee the foce is applied. Since F D points paallel to its leve am, F D, = 0 N. Theefoe F D geneates the smallest toque. FA and F B ae closest to the pivot, and theefoe have the shotest leve am. FA points pependicula to the leve am, while F B points at an angle, theefoe F B, < F A, and, as a esult, τ B < τ A. FC has the lagest leve am aside fom F D ) and points pependicula to its leve am, as a esult F C geneates the lagest toque. 3. a) F N F g 3. b) When the ca dives ove the top of the bump, the ca is taveling along a cicula path. Wheneve an object tavels along a cicula path, the net foce acting on the object must point towads the cente of the cicula path neve away). Since the gavitational foce points towads the cente of the cicula path and the nomal foce points away fom the cente, the gavitational foce must be geate than the nomal foce so that the net foce acting on the ca points towads the cente of the cicula path. 3. c) The ca will just baely lose contact with the oad when the Nomal Foce exeted by the oad on the ca goes to zeo. When this happens the only foce acting on the ca is the gavitational foce. Since the gavitational foce points towads the cente of the cicula path, we can identify it as the centipetal foce and set it equal to m v 2 /. Using this to solve fo v gives: F g = m v2 m g = m v2 g = v2 v = g = 14 m s 4. a)
Honos Physics/Cicula Motion & Toque Test Review Page 5 of 7 SOLUTIONS F N F g 4. b) When the motocycle goes though the loop at the slowest possible speed, the bike momentaily loses contact with the gound at the vey top of the loop. At this point, the only foce exeted on the motocycle is the gavitational foce. Since the foce is pointing towads the cente of the cicula path, we may set it equal to m v 2 /. Using this to solve fo v gives: F g = m v2 m g = m v2 g = v2 v = g = 4.95 m s 4. c) When the motocycle exceeds the minimum speed, the motocycle pushes against the top of the loop. As a esult, the loop exets a downwad nomal foce on the motocycle. Since both of these foces point towads the cente of the cicula path, the nomal foce and gavitational foce combine to cause the cicula motion. Theefoe, F g + F N = m v 2 /. Reaanging this fomula to solve fo F N gives: F g + F N = m v2 F N = m v2 m g = 4530 N 5. a) Revolutions pe minute is a fequency. Since we will be doing calculations with this fequency, we should convet this numbe into SI units of evolutions pe second. Since thee ae 60 seconds in a evolution, we see that the fequency is: 45 ev. min 1 min ) 60 s = 0.75 ev. s The fequency is elated to the peiod accoding to: f = 1/T. Reaanging this to solve fo T gives: T = 1 f = 1.33 s
Honos Physics/Cicula Motion & Toque Test Review Page 6 of 7 SOLUTIONS 5. b) The angula fequency is simply ω = 2π f. So, ω = 4.71 adians second. 5. c) a = ω 2 = 0.09 m) = 2.0 m s 2 4.71 adians ) 2 second 5. d) f When the coin sits on a spinning ecod, the fictional foce causes the coin to undego unifom cicula motion. Theefoe, the fictional foce equals m a c. The maximum acceleation is given by the maximum static fictional foce. Using f = m a c along with f = µ m g allows us to solve fo the maximum acceleation. f max s = m a c,max µ s m g = m a c,max a c,max = µ s g = 2.94 m s 2 Since the static fictional foce can geneate a maximum acceleation of 2.94 m s 2 and the centipetal acceleation of the coin esting on the oute edge of the ecod is only 2.0 m, the coin does not slide off s2 the edge of the ecod. 6. The poblem asks us to elate the peiod and adius of two obits. Keple s Thid Law of Planetay Motion says: R 3 /T 2 is the same fo any two satellites obiting the same body. Theefoe: R 3 A T 2 A R 3 B = R B = = R3 B TB 2 TB = T A TB T A ) 2 R 3 A 24.0 h 8.0 h ) 2 3 RA = 4.2 10 7) m ) 2 3 2.0 10 7 ) m)
Honos Physics/Cicula Motion & Toque Test Review Page 7 of 7 SOLUTIONS 7. a) The balance beam does not move, theefoe the beam must satisfy the conditions fo equilibium: F = 0 N and τ = 0 N. To find the total foce exeted by both of the suppots we use the fist condition. F F l F g,g F g,b F = 0 N F l + F F g,g F g,b = 0 N F l + F = F g,g + F g,b = m g g + m b g = 50 kg) 9.80 m ) s 2 + 500 kg) 9.80 m ) s 2 = 5390 N 7. b) To find the foce exeted by the suppot on the ight end of the balance beam, we use τ = 0 N m. Recall that τ = F l, whee l is the length of the leve am. Also emembe that the gavitational foce exeted on the beam is applied to the cente of the beam 0.25 m fom eithe end). Since we ae inteested in the foce exeted on the ight end of the balance beam, I will choose to daw the axis of otation stating fom the the left end of the beam this way the toque caused by the foce on the ight side of the beam factos into the equation). As a esult, the toque geneated by the foce exeted on the left end of the beam is zeo since the leve am fo this foce will be zeo). Since the foce exeted by the suppot on the ight end of the beam tends to otate the beam counte-clockwise, while the gavitational foces of the beam and the gymnast end to otate the beam clockwise, these toques will have opposite signs. Futhemoe, since the beam is in equilibium, these toques must cancel so that τ = 0 N m. τ = τ g,g + τ g,b F ) = F g,g 0.3 m) + F g,b 0.25 m) ) ) 0.3 m 0.25 m F = F g,g + F g,b ) ) 0.3 m 0.25 m = m g g) + m b g) = 50 kg) 9.8 m ) ) 0.3 m s 2 + 500 kg) 9.8 m ) ) 0.25 m s 2 = 2744 N 7. c) In pat a) we found that F l + F = 5390 N. Combining this with the esult we found fom pat b), we see that F l = 2646 N.