Chapter 8. Accelerated Circular Motion

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1 Chapte 8 Acceleated Cicula Motion

2 8.1 Rotational Motion and Angula Displacement A new unit, adians, is eally useful fo angles. Radian measue θ(adians) = s = θ s (ac length) (adius) (s in same units as ) θ s Full cicle θ = s = 2π = 2π (adians) Convesion of degee to adian measue θ(ad) = θ(deg.) 2π 360 2π 360 ad deg. = 1 ad deg.

3 8.1 Rotational Motion and Angula Displacement Example: Adjacent Synchonous Satellites Synchonous satellites ae put into an obit whose adius is m. If the angula sepaation of the two satellites is 2.00 degees, find the ac length that sepaates them. Convet degee to adian measue 2.00deg Detemine ac length 2π ad 360deg s = θ = m = ad ( ) ad = m (920 miles) ( )

4 8.1 Rotational Motion and Angula Displacement Fo an obseve on the eath, an eclipse can occu because angles subtended by the sun and the moon ae the same. θ = s Sun Sun s Moon Moon 9.3 mad

8.1 Rotational Motion and Angula Displacement The angle though which the object otates is called the angula displacement vecto Δθ = θ θ o SI unit of angula

5 8.1 Rotational Motion and Angula Displacement The angle though which the object otates is called the angula displacement vecto Δθ = θ θ o SI unit of angula displacement, adian (ad) Simplified using θ o = 0, and Δθ = θ, angula displacement vecto. Δθ Vecto Counte-clockwise is + displacement Clockwise is displacement θ 0 = 0

6 8.2 Angula Velocity and Angula Acceleation DEFINITION OF AVERAGE ANGULAR VELOCITY ω = Δθ Δt whee Δt = t t o SI Unit of Angula Velocity: adian pe second (ad/s) Δθ is the same at all adii. Δt is the same at all adii. ω = Δθ Δt is the same at all adii. 2 1 Δθ in time Δt Δθ Angle change θ 0 = 0

7 8.2 Angula Velocity and Angula Acceleation ω Case 1: Constant angula velocity, ω. ω = Δθ Δt Δθ = ω Δt Δθ θ 0 = 0 Example: A disk otates with a constant angula velocity of +1 ad/s. What is the angula displacement of the disk in 13 seconds? How many otations has the disk made in that time? Δθ = ω Δt = (+1 ad/s)(13 s) = +13 ad 2π adians = 1 otation 2π ad/ot. n ot = Δθ 2π ad/ot. = 13 ad 6.3 ad/ot = 2.1 ot.

8 8.2 Angula Velocity and Angula Acceleation Case 2: Angula velocity, ω, changes in time. ω Instantaneous angula velocity at time t. Δθ ω = lim Δt =0 Δt ω 0 DEFINITION OF AVERAGE ANGULAR ACCELERATION SI Unit of Angula acceleation: adian pe second squaed (ad/s 2 )

8.2 Angula Velocity and Angula Acceleation Example: A Jet Revving Its Engines As seen fom the font of the engine, the fan blades ae otating with an angula speed of 110 ad/s.

9 8.2 Angula Velocity and Angula Acceleation Example: A Jet Revving Its Engines As seen fom the font of the engine, the fan blades ae otating with an angula speed of 110 ad/s. As the plane takes off, the angula velocity of the blades eaches 330 ad/s in a time of 14 s. Rotation is clockwise (negative) Find the angula acceleation, assuming it to be constant. ( ) ( 110ad s) α = 330ad s 14 s = 16ad s 2

10 8.2 The Equations of Rotational Kinematics The equations of otational kinematics fo constant angula acceleation: ANGULAR ACCELERATION ANGULAR VELOCITY ω = ω 0 + αt θ = 1 2 ( ω 0 + ω ) t TIME ANGULAR DISPLACEMENT ω 2 = ω αθ θ = ω 0 t αt 2

11 8.2 The Equations of Rotational Kinematics

12 8.2 The Equations of Rotational Kinematics Reasoning Stategy 1. Make a dawing. 2. Decide which diections ae to be called positive (+) and negative ( ). 3. Wite down the values that ae given fo any of the five kinematic vaiables. 4. Veify that the infomation contains values fo at least thee of the five kinematic vaiables. Select the appopiate equation. 5. When the motion is divided into segments, emembe that the final angula velocity of one segment is the initial angula velocity fo the next. 6. Keep in mind that thee may be two possible answes to a kinematics poblem.

13 8.2 The Equations of Rotational Kinematics Example: A disk has an initial angula velocity of +375 ad/s. The disk acceleates and eaches a geate angula velocity afte otating ω though an angula displacement of ad. ω 0 If the angula acceleation has a constant value of ad/s 2, find the final angula velocity of the disk. Given: ω 0 = +375 ad/s,θ = +44 ad, α = 1740 ad/s 2 Want: final angula velocity, ω. No time! ω 2 = ω αθ = (375 ad/s) 2 + 2(1740 ad/s 2 )(+44 ad) ω = 542 ad/s

14 8.3 Angula Vaiables and Tangential Vaiables ω = angula velocity - same at all adii (adians/s) α = angula acceleation - same at all adii (adians/s 2 ) v T = tangential velocity - diffeent at each adius a T = tangential acceleation - diffeent at each adius Diection is tangent to cicle at that θ v T = ω a T = α v T (m/s) ω (ad/s) (m) a T (m/s 2 ) α (ad/s 2 ) (m)

15 8.3 Angula Vaiables and Tangential Vaiables Example: A Helicopte Blade A helicopte blade has an angula speed of 6.50 ev/s and an angula acceleation of 1.30 ev/s 2. Fo point 1 on the blade, find the magnitude of (a) the tangential speed and (b) the tangential acceleation. Convet evolutions to adians ( )( 2" ad ev) = 40.8 ad s! = 6.50 ev s ( ) 2" ad ev! = 1.30 ev s 2 ( ) = 8.17 ad s 2 v T =! = ( 40.8ad s) ( 3.00 m) = 122m s a T = " = ( 8.17ad s 2 )( 3.00 m) = 24.5m s 2

16 8.3 Rolling Motion The tangential speed of a point on the oute edge of the tie is equal to the speed of the ca ove the gound. v = v T =! a = a T =!

17 8.3 Centipetal Acceleation with Tangential Acceleation! a C v T =!! a!! a C! a T Unifom cicula motion! in ad/s constant Non-unifom cicula motion Changing ω = ω 0 + αt ( ) 2 a C = v 2 T =! =! 2 a T =! a total = a C 2 +! 2 2

18 Chapte 4.5 Foce Geneating Unifom Cicula Motion

19 4.5 Centipetal Foce Newton s Second Law When a net extenal foce acts on an object of mass m, the acceleation that esults is diectly popotional to the net foce and has a magnitude that is invesely popotional to the mass. The diection of the acceleation is the same as the diection of the net foce. Vecto Equations

20 4.5 Centipetal Foce Thus, in unifom cicula motion thee must be a net foce to poduce the centipetal acceleation. The centipetal foce is the name given to the net foce equied to keep an object moving on a cicula path. The diection of the centipetal foce always points towad the cente of the cicle and continually changes diection as the object moves. F C = ma C = m v2 Magnitudes

21 4.5 Centipetal Foce Example: The Effect of Speed on Centipetal Foce The model aiplane has a mass of 0.90 kg and moves at constant speed on a cicle that is paallel to the gound. The path of the aiplane and the guideline lie in the same hoizontal plane because the weight of the plane is balanced by the lift geneated by its wings. Find the tension in the 17 m guideline fo a speed of 19 m/s. Tension vecto points inwad! T = F C = m v2 T

4.5 Centipetal Foce Example: The Effect of Speed on Centipetal Foce The model aiplane has a mass of 0.90 kg and moves at constant speed on a cicle that is paallel to the gound.

22 4.5 Centipetal Foce Example: The Effect of Speed on Centipetal Foce The model aiplane has a mass of 0.90 kg and moves at constant speed on a cicle that is paallel to the gound. The path of the aiplane and the guideline lie in the same hoizontal plane because the weight of the plane is balanced by the lift geneated by its wings. Find the tension in the 17 m guideline fo a speed of 19 m/s. Tension is the centipetal foce necessay to maintain aiplane in the cicle T = F C = m v2 engine keeps speed up wings keep it fom falling Tension vecto points inwad! T

23 4.5 Centipetal Foce Conceptual Example: A Tapeze Act In a cicus, a man hangs upside down fom a tapeze, legs bent ove and ams downwad, holding his patne. Is it hade fo the man to hold his patne when the patne hangs staight down and is stationay o when the patne is swinging though the staight-down position? Tension in ams maintains cicula motion but also must counte the gavitational foce (weight)

4.5 Centipetal Foce Conceptual Example: A Tapeze Act In a cicus, a man hangs upside down fom a tapeze, legs bent ove and ams downwad, holding his patne.

24 4.5 Centipetal Foce Conceptual Example: A Tapeze Act In a cicus, a man hangs upside down fom a tapeze, legs bent ove and ams downwad, holding his patne. Is it hade fo the man to hold his patne when the patne hangs staight down and is stationay o when the patne is swinging though the staight-down position? Tension in ams maintains cicula motion but also must counte the gavitational foce (weight)! F! = +T "W = F C T = W + F C v T v W = mg

25 4.5 Banked Cuves On an unbanked cuve, the static fictional foce povides the centipetal foce.

26 4.5 Banked Cuves On a fictionless banked cuve, the centipetal foce is the hoizontal component of the nomal foce. The vetical component of the nomal foce balances the ca s weight. Compession of the banked oad povides the nomal foce. The nomal foce pushes against the ca to 1) suppot the weight and 2) povides the centipetal foce equied fo the ca to move in a cicle.

27 4.5 Banked Cuves F C = F N sin! = m v2 Combining the two elationships can detemine the speed necessay to keep the ca on the tack with the given angle

28 4.5 Atificial Gavity Example: Atificial Gavity At what speed must the suface of a space station move so that an astonaut expeiences a nomal foce on the feet equal to the weight on eath? The adius is 1700 m. v = g = ( 1700 m) ( 9.80m s 2 ) = 130 m/s

29 4.5 Vetical Cicula Motion Nomal foces ae ceated by stetching of the hoop. v 3 2 must be > g to stay on the tack F N 4 = m v 2 4

Chapter 5: Uniform Circular Motion

Chapter 5: Uniform Circular Motion Chapte 5: Unifom Cicula Motion Motion at constant speed in a cicle Centipetal acceleation Banked cuves Obital motion Weightlessness, atificial gavity Vetical cicula motion Centipetal Foce Acceleation towad