Circular Motion. Mr. Velazquez AP/Honors Physics

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1 Cicula Motion M. Velazquez AP/Honos Physics

2 Objects in Cicula Motion Accoding to Newton s Laws, if no foce acts on an object, it will move with constant speed in a constant diection. Theefoe, if an object moves along a cuved o cicula path, thee must be a foce acting on it to keep it on this tajectoy. To keep an object in cicula motion, a foce diected towad the cente of otation is equied. We call this the centipetal foce. It also follows that since thee is a foce diected towad the cente, thee must also be acceleation towad the cente of the cicle. Centipetal Acceleation a c = v2 Centipetal Foce F c = ma c = mv2

3 Objects in Cicula Motion EXAMPLE If a oadway is banked at the pope angle, a ca will be able to ound the cone without any assistance fom fiction between the ties and the oad. Find the appopiate banking angle θ fo a 940-kg ca taveling at 25.0 m/s in a tun with adius 62.5 m. a C

4 Objects in Cicula Motion Given: m = 940 kg v = 25.0 m/s = 62.5 m F x = ma c F N sin θ = ma c mg v2 sin θ = m cos θ g tan θ = v2 tan θ = v2 g F y = 0 F N cos θ mg = 0 F N = mg cos θ a C θ = tan 1 v2 g = 25.0 m s 2 tan m s m = 45. 5

5 Objects in Cicula Motion OTHER EXAMPLES Paticles in a Centifuge Rolle coastes in cicula loops o cuves

6 Angula Kinematics: Position The most of the angula quantities is angula position, defined as the change in angle fom the efeence line. Fo this quantity we will need to look at angles in adians instead of degees. One adian is the angle fo which the ac length on a cicle of adius is equal to the adius of the cicle, o about 57.3 θ θ ev π 180 2π 1 ev = θ ads = θ ads θ = angle measued in adians 360 = 2π ad = 1 ev Ac length: s = θ

7 Angula Kinematics: Velocity Angula velocity descibes how quickly the angle of motion is changing ove time. We can measue this quantity in adians pe second (ad/s) o evolutions pe minute (pm) ω = θ t θ = ωt ω = s t ω = v t Aveage Angula Velocity: ω av = θ t

8 Angula Kinematics: Peiod We efe to the time it takes to complete one evolution as the peiod (T) of motion. Its invese quantity is the numbe of evolutions in one unit of time (usually 1 second); this quantity is known as the fequency (f) Fo one evolution: ω = θ t ω = 2π T = 2πf T = 2π ω = 1 f Units fo T: seconds (s) Units fo f: evolutions pe second o Hetz (Hz = s 1 )

9 = 2.15 m Angula Kinematics A 1.50-kg stone is tied to a 2.15-m cod and swung in a counteclockwise cicle with constant angula speed. In seconds, the stone tavels though an ac of length m. Calculate the angula displacement, angula velocity and the peiod of evolution. s = θ θ = s = m 2.15 m θ = ad ω = θ t = ad s ω = ad s θ T = 2π ω = 2π 1.83 ad s = s

acceleation α of the object as the change in angula velocity ω ove

10 Angula Kinematics: Acceleation If the angula velocity of an object in cicula motion is inceasing ove time, we can descibe the angula acceleation α of the object as the change in angula velocity ω ove time t. α av = ω t If tangential acceleation (a T ) emains constant: α = v t α = a T

11 Rotational Kinematic Equations Each linea kinematic equation has an analogous equation fo angula kinematics. The table below shows these equations: Linea Equations (a = constant) v = v 0 + at x = x 0 + v 0 t at2 Angula Equations (α = constant) ω = ω 0 + αt θ = θ 0 + ω 0 t αt2 v 2 = v a(x x 0 ) ω 2 = ω α(θ θ 0 )

12 Rotational Kinematics To thow a cuve ball, a pitche applies topspin to the ball equivalent to an angula speed of 36.0 ad/s. When the catche gloves the ball seconds late, its angula speed has deceased (due to ai esistance) to 34.2 ad/s. a) What is the ball s angula acceleation, assuming it is constant? b) How many complete evolutions will the ball make befoe being caught by the catche? Angula Equations (α = constant) ω = ω 0 + αt θ = θ 0 + ω 0 t αt2 ω 2 = ω α(θ θ 0 )

13 Tangential Quantities At any instant, an object moving with cicula motion has a diection that is tangential to the cicula path. Fom this, we can define vecto quantities that act tangent to the cicle, which we call tangential quantities. a t Tangential Velocity v t = ω v t a c a t a c Tangential Acceleation a t = α v t θ Centipetal Acceleation a c = v2 = (ω)2 a c = ω 2 = 2 ω 2 a c a t a

14 Tangential Quantities We can conside the centipetal and tangential acceleation to be two components of one single vecto acceleation with a magnitude a and a diection φ (elative to the tangential diection). a t a = Total Acceleation (Magnitude) a c 2 + a t 2 a t a c v t a c Total Acceleation (Diection) v t θ φ = tan 1 a c a t a c a t φ a

15 Exit Ticket fo Pat A (DO NOW) A tuntable, initially spinning at counteclockwise at 45 pm, is switched off and begins to gadually slow down. Two kg mounds of clay sit at diffeent points on the tuntable: Mound A is 4.60 cm fom the cente of the cicle, and Mound B is 11.2 cm fom the cente. If the tuntable takes a total of 8.15 seconds to come to a complete stop, find: a) The angula acceleation of the tuntable b) The angula speed of the tuntable 2.00 seconds afte it is switched off c) The tangential acceleation of each clay mound d) The centipetal foce acting on each clay mound 2.00 seconds afte it is switched off 1 ev = 2π ad B A ω = ω 0 + αt a t = α a c = v2 = ω2 F c = ma c Homewok Poblem Set: Cicula Motion 1 (Due 12/18) Pg #12-44 (mult. of 4)

Rotational Motion. Lecture 6. Chapter 4. Physics I. Course website:

Rotational Motion. Lecture 6. Chapter 4. Physics I. Course website: Lectue 6 Chapte 4 Physics I Rotational Motion Couse website: http://faculty.uml.edu/andiy_danylov/teaching/physicsi Today we ae going to discuss: Chapte 4: Unifom Cicula Motion: Section 4.4 Nonunifom Cicula