CIRCULAR MOTION, HARMONIC MOTION, ROTATIONAL MOTION

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1 CIRCULAR MOTION, HARMONIC MOTION, ROTATIONAL MOTION 1

2 UNIFORM CIRCULAR MOTION path circle distance arc Definition: An object which moves on a circle, travels equal arcs in equal times. Periodic motion 2

3 PARAMETERS OF PERIODICITY Period (T), T = 1s Required time for one complete period Revolution (n), n = 1 1 s Number of rotations per second Connection: The period and the revolution are reciprocal: T = 1 n 3

4 VELOCITY OF CIRCULAR MOTION ORIENTATION Δt big v = r t The orientation of the velocity vector is always the same as the displacement vector s ( r) orientation it s changing constantly secant Δt 0 tangent The direction of velocity is always perpendicular to the radius of the circle so it s tangential. TANGENTIAL VELOCITY (v) 4

5 VELOCITY OF CIRCULAR MOTION MAGNITUDE By the definition: v = s t where i = s v = i t = const If Δt = T Δi = 2Rπ (=perimeter) Then: v = 2Rπ T = 2Rπn 5

6 ANGULAR VELOCITY It s hard to measure arcs (Δi). New definition: An object which moves on a circle, turns equal angles in equal times. i t = const φ Because Δi ~ Δφ t = const ANGULAR VELOCITY: ω = φ t φ = 1rad, ω = 1 1 s If Δt = T Δφ = 2π Then: ω = 2π 6 T = 2πn

7 CONNECTION BETWEEN ANGULAR VELOCITY AND TANGENTIAL VELOCITY v = 2Rπ T v = Rω ω = 2π T 7

8 ACCELERATION OF CIRCULAR MOTION ORIENTATION v changes acceleration The orientation of the acceleration vector is always the same as the v vector s orientation. By right of the figure: Isosceles triangle β = Δφ If Δt 0 then Δφ = β 0 and α = 90 a = v t The direction of acceleration is always perpendicular to the direction of the velocity it s always pointing towards the center of the circle CENTRIPETAL ACCELERATION (a cp ) 8

9 ACCELERATION OF CIRCULAR MOTION MAGNITUDE a cp = v2 R v = Rω a cp = Rω 2 9

10 CENTRIPETAL ACCELERATION IN CENTRIFUGES Generally you can set N (rpm = revolutions per minute) on centrifuges: a cp = Rω 2 where ω = 2πn a cp = R4π 2 n 2 a cp = 4π 2 N 60 2 R = 0,011N 2 R 10

11 RELATIVE ACCELERATION (RCF = relative centrifugal force) a rel = a cp g Physical amount without dimension It s the ratio of the centripetal acceleration and the gravitation acceleration E.g.: N = 3000, R = 10 cm generates 1000 g Ultracentrifuge generates 10 6 g 11

12 DYNAMICS OF CIRCULAR MOTION ΣF = ma Because a cp has constant magnitude radial orientation cause effect Then ΣF has constant magnitude radial orientation Label: ΣF = F cp centripetal force Dynamic condition: The forces, which effect on the object must have constant magnitude and radial orientation. F cp = ΣF = ma cp 12

13 ACCELERATING CIRCULAR MOTION Tangential force accelerated circular motion Characterization: Tangentional acceleration: a t = v t Angular acceleration: β = ω t, β = 1 1 s 2 13

14 ACCELERATING CIRCULAR MOTION Total force and acceleration Editing: parallelogram method Calculation: pythagora s theorem 14

15 CIRCULAR MOTION SUMMARY Period (T) T = 1 Revolution (n) n Tangential velocity (v) orientation: tangential magnitude: v = 2Rπ T = 2Rπn Angular velocity (ω) ω = 2π T = 2πn v = Rω Centripetal acceleration (a cp ) orientation: radial magnitude: a cp = v2 or R = Rω2 Tangential acceleration (a t ) orientation: tangentional magnitude: a t = v t β = ω t Angular acceleration (β) a cp = 0,011N 2 R 15

16 CIRCULAR MOTION PROBLEMS 1) An object moves with a speed of 12 m/s. Calculate the acceleration and the required time for moving 300 meters, if it a) moves straight. b) moves on a circle, which radius is 20 m. 2) A moving object makes a 5 m long half circle in 2 seconds. a) Calculate the tangential velocity and the angular velocity! b) Calculate the centripetal acceleration! 3) In a centrifuge the sample do 3000 rotations per minute. The radius of the centrifuge is 15 cm. Calculate the a) tangential velocity! b) centripetal acceleration! c) relative acceleration! 16

17 SIMPLE HARMONIC MOTION A y(t) terminal position equilibrium position terminal position Periodic motion y(t): displacement y(t) max = A (amplitude) 17

18 PARAMETERS OF PERIODICITY Period (T) T = 1s Required time for one complete period f, f = 1 1 s Number of cycles per second Revolution Frequency 1Hz (hertz) Connection: The period and the frequency are reciprocal: T = 1 f 18

19 ANALYSIS OF THE DISPLACEMENT Experiment: Mathematical explanation: projection of circular motion By right of the figure: Displacement is a vector: where 19

20 ANALYSIS OF THE VELOCITY By right of the figure: Thus: where Note: v max = Aω and v min = 0 20

21 ANALYSIS OF THE ACCELERATION By right of the figure: Thus: where Acceleration and displacement have opposit directions Note: a max = Aω 2, a min = 0 and a(t) = -ω 2 y(t) 21

22 DYNAMICS OF SIMPLE HARMONIC MOTION Harmonic force: it s proportional to the displacement what it causes, but has opposit direction F h The force of a spring is harmonic Dynamic condition: The forces, which effect on the object, must be harmonic ΣF = ma where a = -Aω 2 sinωt = -ω 2 y(t) and F = Dy(t) By right of the previous informations: 22

23 PERIOD OF SIMPLE HARMONIC MOTION Suspicion: T(m, D) Example: T(m) m f = 1 2π D m 23

24 SIMPLE HARMONIC MOTION SUMMARY Displacement: y t = Asinωt Velocity: v t = Aω cos ωt Acceleration: a t = Aω 2 sin ωt Period of oscillation: T osc = 2π m D 24

25 SIMPLE HARMONIC MOTION PROBLEMS 1) An object on a spring oscillates with an amplitude of 10 cm and a period of 0.8 s. Calculate its displacement 0.1 s after it leaves the equilibrium position! Calculate the maximal velocity! 2) An oscillating object s frequency is 2 1/s and the amplitude of the motion is 0.2 cm. Calculate its displacement and acceleration after s! 3) A force of 20 N elongates a spring with 5 cm. We load this spring with a mass of 4 kg, and make it oscillate. Calculate the frequency! 25

26 ROTATION OF RIGID BODIES Rigid body Definition: any two points of the body have constant distance Types of motion According to the number of fixed points: 0 fixed point: linear and rotational motion 1 fixed point: the points of the body move on the surface of a sphere around the fixed point 2 fixed points: rotational motion around an axis, the points of the body do circular motion 3 fixed points: the body stays still, no motion 26

27 DYNAMICS OF ROTATIONAL MOTION Characterization of the rotating effect: Depends on: Force (F) Distance (k) between the linear of the force and the axis TORQUE (M) M = Fk M = 1Nm ΣM β cause effect 27

28 CONNECTION BETWEEN TORQUE AND ANGULAR ACCELERATION Connection: Direct proportion The friction of the two parameters is constant: M β = const = θ θ: moment of inertia 28

29 BASIC EQUATION OF ROTATIONAL MOTION θ: moment of inertia = rotational inertia θ = m i r i 2 i θ = 1kgm 2 Basic equation: ΣM = θβ Examples: Barrel: θ = 1 2 mr2 Sphere: θ = 2 5 mr2 next=1&list=pl9296a42aedca

30 ANALOGY BETWEEN LINEAR AND ROTATIONAL MOTION SUMMARY Distance: s Velocity: Linear Acceleration: Mass: m Force: F v = s t a = v t Basic equation: ΣF=ma Angle: φ Rotational Angular velocity: Angular acceleration: Rotational inertia: Torque: M ω = φ t Basic equation: ΣM = θβ θ β = ω t Momentum: I=mv Angular momentum: N = θω 30

31 ROTATIONAL MOTION PROBLEMS 1. Calculate the revolution of a sphere which radius is 2 cm, and which mass is 0.3 kg, if 2 Nm torque effects on it for 5 seconds N force effects for 2 s on a circumferential point of a barrel. The mass of the barrel is 1 kg and its radius is 6 cm. Calculate the angular momentum! 31

32 THANK YOU FOR YOUR ATTENTION! 32

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