# Circular Motion

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Category: Education

Circular Motion

## 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. 