logging in or signing up Circular Motion fox1100 Download Post to : URL : Related Presentations : Let's Connect Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Copy embed code: Embed: Flash iPad Dynamic Copy Does not support media & animations Automatically changes to Flash or non-Flash embed WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 7255 Category: Education License: All Rights Reserved Like it (5) Dislike it (0) Added: September 19, 2008 This Presentation is Public Favorites: 3 Presentation Description Circular Motion Comments Posting comment... Premium member Presentation Transcript Circular Motion : Circular Motion Examples: Planets, atoms, cars on curves, CD-ROM’s, propellers, etc. Rotational Kinematics : Rotational Kinematics How do we describe an object moving in a circle? Slide 3: Measure angles relative to reference line: Δθ Θ = 00 Θ = 600 Express Angles in Radians : Express Angles in Radians 1 revolution = 3600 3600 = 2π radians 1 radian = 57.30 Position in circular motion is expressed as an angle, preferably in radians. ‘x’ in linear motion, ‘θ’ in circular motion. Example : Example A fan blade, with a diameter of 1 m, is rotating at 1 revolution every second. Through what angle does the tip of the blade turn in 0.3 seconds? 1 rev in 1 second 0.5 rev in 0.5 sec 0.3 rev in 0.3 sec, so 0.3 rev = 1080 = 1.88 radians Speed in Circular Motion : Speed in Circular Motion Since we’re not using ‘x’ to measure position, we can’t define speed as Δx/Δt. But, since ‘θ’ replaces ‘x’, why not define angular speed by Δθ/Δt = (θf – θo)/(tf – to) This is called angular speed and is given the symbol (omega), ω. ω = Δθ/Δt In units of radians/second, i.e. rad/s. Angular Velocity : Angular Velocity ω = Δθ/Δt Connection between Linear and Circular Motion : Connection between Linear and Circular Motion q = x/r = v/r a = a/r Example : Example What is the angular speed of a 33 1/3 rpm record? ω =33.33 rev/min = 33.33 rev/60 sec ω = 33.33 ( 2 π rad)/60 s = 3.49 rad/s Acceleration in Circular Motion : Acceleration in Circular Motion Consider your rotating car tires as you accelerate from 25 mph to 55 mph. What is happening to the rotational speed of the tires? R = 33 cm. V = 11 m/s V=24.5 m/s ωo ωf Linear and Angular Connection : Linear and Angular Connection ω = v/r So, ωo = (11 m/s)/0.33 m = 33 rad/s And ωf = (24.5 m/s)/0.33 m = 73.5 rad/s. Therefore the change in angular speed, ωf – ωo = 40.5 rad/s = Δ ω Angular Acceleration : Angular Acceleration When you have changing angular speeds, this means the object has an angular acceleration, α (alpha), which is calculated by α =Δ ω/Δt In units of radians/second2 = rad/s2 Characterizing Circular Motion : Characterizing Circular Motion Radius, r Angular position, θ Angular displacement, Δθ Angular speed, ω=Δθ/Δt Angular acceleration, α=Δω/Δt Kinematics of Circular Motion : Kinematics of Circular Motion ω=Δθ/Δt α=Δω/Δt ωave=(ωf + ωo)/2 Θ = ωavet ωf= ωo +αt Θ = Θo + ωot + ½ αt2 ωf2 = ωo2 + 2αΔθ Kinematics Example : Kinematics Example A flywheel of a machine is rotating at 12 rev/s. Through what angle will the wheel be displaced from its original position after 5 seconds? Angular speed, ω = 12 rev/s = 75 rad/s Θ = ωavet = 75 rad/s * 5 s = 375 rad = 2148750. 59.6875 revolutions, so .6875 revolutions from start position = 247o. Slide 16: A turntable revolves at 33 1/3 rpm. It is shut off and slow to a stop in 6.3 seconds. What is the angular acceleration? Through what angle did it turn as it slow to a stop? ωf=0, ωo = 33.33 rpm =3.49 rad/s, t = 6.3 s ωf= ωo +αt Θ = Θo + ωot + ½ αt2 Dynamics of Rotation : Dynamics of Rotation Examine circular motion taking Newton’s Laws into consideration. 1st Law- 2nd Law- 3rd Law- Slide 18: Dynamics of Rotation 1st Law Is Moon at rest? Is Moon moving in a straight line? Conclusion MOON EARTH Slide 19: Dynamics of Rotation 1st Law Objects executing circular motion have a net force acting on them…even if you can’t see the agent of the force. What force acts on the Moon? MOON EARTH Slide 20: Dynamics of Rotation 2nd Law FNET = ma a is a vector defined by a = Δv/Δt Δv = vf – vo For circular motion the speeds are the same, but the directions aren’t. MOON EARTH Slide 21: Dynamics of Rotation MOON EARTH -vo vf Δv= vf-vo Let’s visually examine the change in velocity Slide 22: Dynamics of Rotation MOON EARTH vo vf Δv= vf-vo So, a Δv exists because the direction is changing, not the magnitude. How do we find the acceleration? Dv/v = Dr/r Dv/v = vDt/r Dv/Dt = v2/r ac = v2/r Slide 23: Dynamics of Rotation MOON EARTH The acceleration toward the center is called ‘centripetal’ acceleration, ac, given by ac = v2/R In magnitude and directed toward center. R v v Slide 24: Dynamics of Rotation MOON EARTH So, Newton’s 2nd Law for rotation becomes, F =mac = mv2/R In magnitude and directed toward center. R v v Slide 25: Dynamics of Rotation MOON EARTH A physical statement that relates cause and effect. Cause = F, effect = mv2/R F = mv2/R The right side is ‘what you see’, the left side is ‘why’. R v v Slide 26: Dynamics of Rotation MOON,m EARTH,M What is the ‘cause’ for the Moon’s motion? F =GMm/R2 Newton’s Universal Law of Gravity. G = 6.67 x 10-11 Nm2/kg2 R v v Slide 27: ULG-two objects of given masses separated by known distance exert a gravitational force of attraction on each other whose size is determined from F =GMm/R2 You are sitting next to a person whose mass is 55 kg. Your mass is 75 kg. What is the force of attraction between you if you are 0.8 m apart (center to center)? F =(6.67 x 10-11Nm2/kg2)(75kg)(55kg)/(0.8m)2 F=0.00000043 N = 0.43 microNewtons How do we know the mass of the Earth? : How do we know the mass of the Earth? Using the ULG, F =GMm/R2 And 2nd Law, F = mv2/R, Combine, GMm/R2 = mv2/R M = v2R/G M = (ωR)2R/G M = ω2R3/G Slide 29: M = ω2R3/G R = 380,000,000 m ω = 2π rad /27.3 days =2.664 x 10-6 rad/s So, M = (2.664 x 10-6 rad/s)2(380,000,000 m)3/(6.67 x 10-11 Nm2/kg2) M = 5.84 x 1024 kg. True value 5.98 x 1024 kg. Slide 30: Earth and Moon orbit the center of mass of the system. Located 1070 miles below the Earth’s surface or 2880 miles from center of Earth. Problem Solving Strategy for Circular Motion Problems : Problem Solving Strategy for Circular Motion Problems Is it Kinematics or Dynamics Kinematics-You are trying to characterize the motion by its position, speed or acceleration. Click here. Dynamics-You are trying to relate the motion to its causes. Click here. Slide 32: A tire of diameter 26 inches is spinning with a constant angular velocity of 2 rad/s. What is the centripetal acceleration of a point on the rim of the tire? R = 0.33 m ω = 2 rad/s Centripetal Accel. ac = v2/R v = ωR ac = v2/R = ω2 R = (2 rad/s)2*0.33 m ac =1.32 m/s2, directed toward axle. Another example, click here. Slide 33: A dentist’s drill spins at 1800 rpm. If it takes 6 seconds to stop when turned off, what is the angular acceleration of the drill? Initial angular speed, ωo = 1800 rpm=188 rad/s Final angular speed = ωf = 0 rad/s Time, t = 6 s. Angular Accel, α = Δω/Δt=(0-188rad/s)/(6s) = -31 rad/s2 Slide 34: A car moving at 25 m/s rounds a curve of radius 100 m and is just on the verge of slipping. So if that is the fastest that it can round this curve, what is the maximum speed it can travel on a curve of radius 300 m? In both cases the force keeping that car from slipping will be the same, i.e. static friction. So, F = mv2/R is the equation we will apply to each case. v12/R1 = v2 2/R2 v2 = v1(R2/R1)1/2 Next Example Slide 35: A 0.50-kg mass is attached to the end of a 1.0-m string. The system is whirled in a horizontal circular path. If the maximum tension that the string can withstand is 350 N. What is the maximum speed of the mass if the string is not to break. M = 0.5 kg, R = 1.0 m, F(max) = 350 N F = mv2/R V = ( FR/m)1/2 = (350*1.0/0.5)1/2 = 26.5 m/s Next Example Slide 36: A car goes around a flat curve of radius 50 m at a speed of 14 m/s. What must be the minimum coefficient of friction between the tires and the road for the car to make the turn? V = 14 m/s, R = 50 m. F = mv2/R and f = µN = µmg, so µmg = mv2/R µ = v2/gR = (14 m/s)2/(9.8 m/s2*50m) µ =0.4 Next Example Slide 37: The hydrogen atom consists of a proton of mass 1.67X10-27 kg and an orbiting electron of mass 9.11X10-31 kg. In one of its orbits, the electron is 5.3X10-11 m from the proton. What is the mutual gravitational attractive forces between the electron and proton? F = GM1M2/R2 =(6.67e-11*1.67e-27*9.11e-31)/(5.3e-11)2 F =3.6 x 10-47 N Slide 38: 3rd Law: F(earth on moon) = -F(moon on Earth) FEM = -FME FEM -FME Energy of Orbiting Objects : Energy of Orbiting Objects Consider Moon. It has velocity, so it has Kinetic Energy. E = K + U. What is the potential energy of a bound object? Slide 40: K U E K is pos, U is neg, E is neg. You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.