AP Physics C Sping, 2017 Cicula-Rotational Motion Mock Exam Name: Answe Key M. Leonad Instuctions: (92 points) Answe the following questions. SHOW ALL OF YOUR WORK. ( ) 1. A stuntman dives a motocycle though a vetical loop which has a adius of 2.5 m. The motocycle s engine keeps the motocycle moving at a constant speed as it dives though the loop. (a) (6 pts) Daw a fee-body diagam of the motocycle when bike is at the top of the loop. Solution: Remembe, the centipetal foce is not a foce in and of itself. Theefoe, you should not include it in a fee-body diagam. F g F N (b) (8 pts) What is the minimum speed that the motocycle must maintain to make it though the loop? Solution: When the motocycle eaches the top of the loop, both the gavitational foce and the nomal foce point towads the cente of the cicula path. Theefoe F g + F N = m v 2 /. Fom this fomula we can see that thee is a elationship between F N and v. In geneal, the faste the motocycle tavels though the loop, the lage the nomal foce. Since thee is a minimum value fo the nomal foce (F N = 0 N), thee is likewise a minimum value fo v. Setting F N equal to zeo and solving fo v gives: F g = m v2 m g = m v2 g = v2 v 2 = g v = g = 5 m s (c) (8 pts) Suppose the stuntman dives though the loop at 10 m. If the motocycle and the stuntman s have a 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? Solution: Since F g + F N = m v 2 /,
AP Physics C/Cicula-Rotational Motion Mock Exam Page 2 of 6 Name: Answe Key F g + F N = m v2 F N = m v2 m g ( ) v 2 = m g = 130 N ( ) 2. A coin is placed 20 cm fom the cente of a tuntable. The tun table stats fom est and begins to otate with an angula acceleation of α = 12 ad.. The coefficient of fiction between the coin and the tuntable is 0.4. (a) (4 pts) What is the tangential acceleation of the coin? s 2 Solution: Recall that tangential acceleation is given by a t = α. Theefoe, a t = (0.2 m)(12ad/s 2 ) = 2.4 m/s 2. (b) (8 pts) What is the maximum centipetal acceleation the coin can withstand without sliding? Solution: The foce esponsible fo holding the coin on the tuntable is the static fictional foce. Since thee is a maximum static fictional foce, thee is a maximum acceleation. We can find the magnitude of the maximum acceleation by setting the fictional foce equal to m a. f = m a µ m g = m a a = µ g Because the tuntable is speeding up as it spins, the coin s acceleation can be boken into two non-zeo components: tangential and centipetal. Because these two components ae othogonal, the magnitude of the coin s total acceleation is a 2 t + a 2 c. Setting this equal to the maximum acceleation we found above, and solving fo a c gives: a 2 t + a 2 c = µ g a 2 t + a 2 c = (µ g) 2 a 2 c = (µ g) 2 a 2 t a c = (µ g) 2 a 2 t ) = 3.2 m s 2 (c) (4 pts) What is the maximum angula velocity the coin can withstand without sliding? Solution: The centipetal acceleation is elated to the angula velocity by a c = ω 2. So, ω = ac / = 4 ad./s. (d) (6 pts) How long afte the tuntable stated spinning was it befoe the coin stated sliding? Solution: Using α = ω/ t gives:
AP Physics C/Cicula-Rotational Motion Mock Exam Page 3 of 6 Name: Answe Key α = ω t t = ω α = 0.333 s ( ) 3. A ginding wheel has a adius of 0.2 m. The wheel stats fom est when a moto begins otating the wheel with an angula acceleation of 1.2 ad s 2. (a) (6 pts) What is the angula velocity of the tuntable the instant it completes one full evolution? Solution: Using ω 2 f = ω 2 0 + 2 α θ gives: 0 ωf 2 = ω 2 f = 2 ω0 2 + 2 α θ ( 1.2 ad s 2 ω f = 3.88 ad s ) (2π) (b) (4 pts) How fast is a point on the oute edge of the wheel moving the instant it completes one full evolution? Solution: Using v = ω gives v = 0.777 m/s. (c) (8 pts) What is the magnitude of the acceleation of a point on the oute edge of the wheel? Solution: Recall that when looking at a point on a otationally acceleating object, the acceleation has two components: centipetal, and tangential. The centipetal acceleation is given by: a c = ω 2, while the tangential acceleation is given by a t = α. Since these two components ae othogonal, the magnitude of the acceleation is given by: a = a 2 c + a 2 t = (ω 2 ) 2 + ( α) 2 = ω 4 + α 2 = 15.1 m s 2 (d) (4 pts) Afte the wheel completes one full otation, the moto dies. It takes 10 seconds fo the wheel to come to est. What was the angula acceleation of the wheel afte the moto died? Solution: Using α = ω/ t, we see that α = 0.388 ad s 2.
AP Physics C/Cicula-Rotational Motion Mock Exam Page 4 of 6 Name: Answe Key (12 pts ) 4. A conical pendulum is fomed by attaching a 500 gam ball to the end of a 1.0 m sting. If the mass moves in a cicula path with a adius of 20 cm, how fast is the mass moving? 12 pts 1.0 m 0.2 m Solution: Thee ae two foces acting on the pendulum bob: gavity and tension. T θ F g Because the bob does not move vetically up o down, we can conclude that F y = 0 N. Theefoe, the vetical component of the tension foce must cancel the gavitational foce: T y = F g = 4.9 N. When the bob swings in a cicula path as shown above, a component of the tension points towads the cente of the cicula path. So, T x = m v 2 /. If we could find the x-component of the tension, we could use this elationship to solve fo v. Since we know the vetical component of the tension, we can use the geomety of the poblem to solve fo T x. Fom the fee-body diagam, we can see that tan θ = T x /T y. We can also see fom the fee-body diagam that sin θ = 0.2; theefoe, θ = 11.5 and tan θ = 0.204. Setting T x /T y = tan θ = 0.204 and solving fo T x gives: Setting this equal to m v 2 / gives: T x T y = 0.204 T x = 0.204T y = 1.00 12 pts
AP Physics C/Cicula-Rotational Motion Mock Exam Page 5 of 6 Name: Answe Key m v2 = T x Tx v = m (1 N)(0.2 m) v = 0.5 kg v = 0.632 m s (8 pts ) 5. Two satellites, A and B, ae in diffeent cicula obits about the Eath. The obital speed of the satellite A is thee times that of satellite B. Find the atio T A /T B. Solution: Since we ae given the atio of the obital speeds and we ae asked to find the atio of the obital peiods, the easiest way to solve this poblem is using popotions. To do this, we must find a fomula which elates the peiod, T, to the velocity, v. When a satellite is in a cicula obit aound the Eath, the centipetal foce equied to keep the satellite in a cicula path is geneated by gavitational foce. Theefoe: 8 pts F c = F g m v2 = GM E m 2 v 2 = GM E We can elate this to the peiod of the motion, by ecalling that v = 2π. Since we want an expession T elating v to T, we need to solve this equation fo and plug it inso the expession above. Solving fo gives: = vt. Plugging this into the expession above gives: 2π v 2 = GM E ( ) 2π v 2 = GM E vt v 3 = 2πGM E T Theefoe T is popotional to v 3. Theefoe, if v inceases by a facto of 3, T will decease by a facto of 3 3 = 27. Theefoe: T A T B = 1 27 8 pts
AP Physics C/Cicula-Rotational Motion Mock Exam Page 6 of 6 Name: Answe Key (6 pts ) 6. Planet X obits the sta Omega with an obital peiod of 200 Eath days. Planet Y obits Omega at a distance that is fou times lage than Planet X. How long is a yea on Planet Y? Solution: Since we ae given R Y /R X, and we ae asked about the peiods, we should use Keple s Thid Law. Recall that Keple s Thid Law says: 6 pts O eaanging to solve fo T Y : R 3 X T 2 X = R3 Y T 2 Y RX 3 TX 2 = R3 Y TY 2 ( ( ) 3 TY RY ) 2 = T Y = R X ( ) 3/2 RY R X T Y = ( ) 3/2 RX R Y T Y = ( RX R Y ) 3/2 = 8 (200 days) = 1600 days 6 pts