Download - Kinematics II

Transcript
Page 1: Kinematics II

Kinematics IIKinematics IIWeek 2Week 2

Page 2: Kinematics II

ContentsContents RadiansRadians Uniform Circular MotionUniform Circular Motion Projectile MotionProjectile Motion Summary of Kinematics I & IISummary of Kinematics I & II

Page 3: Kinematics II

RadiansRadiansA radian is a measure of an angle.A radian is a measure of an angle.

Radius = 1Radius = 1 ӨӨ = = arcarc length length

360˚ = 2360˚ = 2π radiansπ radians 180180˚̊ = π radians = π radians

Unit Circle; radius = 1

ӨӨ

arcarc

Page 4: Kinematics II

Quiz 1Quiz 1Which of the following radian Which of the following radian measures is equivalent to 90measures is equivalent to 90˚?˚?

1.1. 2π2π2.2. ππ3.3. π/2π/24.4. π/3π/35.5. π/4π/4

Page 5: Kinematics II

Quiz 1Quiz 1Which of the following radian Which of the following radian measures is equivalent to 90measures is equivalent to 90˚?˚?

180180˚̊ = π radians = π radians 9090˚̊ = π/2 radians = π/2 radians

ANSWER: π/2ANSWER: π/2

Page 6: Kinematics II

Arc LengthArc Length When radius = 1When radius = 1

arcarc = = ӨӨ When radius = rWhen radius = r

arcarc = r = rӨӨ

* * ӨӨ must be in radians must be in radians

ӨӨ

arcarcrr

Page 7: Kinematics II

Uniform Circular MotionUniform Circular Motion ““Uniform” implies Uniform” implies

constant speed constant speed while moving in a while moving in a circlecircle

Angular Angular FrequencyFrequency

ω = ω = ӨӨ/sec/sec

ӨӨ

Page 8: Kinematics II

Quiz 2Quiz 2An object in circular motion moves An object in circular motion moves 180180˚ per second has an angular ˚ per second has an angular frequency of what?frequency of what?

1.1. 2π2π2.2. 3π/23π/23.3. π/2π/24.4. π/3π/35.5. π/4π/4

Page 9: Kinematics II

Quiz 2Quiz 2An object in circular motion moves An object in circular motion moves 180180˚ per second has an angular ˚ per second has an angular frequency of what?frequency of what?

ANSWER: ANSWER: ω = π/2 radians/secω = π/2 radians/sec

Page 10: Kinematics II

Angular SpeedAngular Speedarc = rarc = rӨӨω = ω = ӨӨ/sec/sec

Angular speed:Angular speed:vv = arc/sec= arc/sec

= (r= (rӨӨ)/sec)/sec= r(= r(ӨӨ/sec)/sec)= rω= rω

ӨӨ

ӨӨ

Page 11: Kinematics II

SummarySummary ӨӨ = arc length (r=1) = arc length (r=1) arc = rarc = rӨӨ ω = ω = ӨӨ/sec/sec v = arc/sec = rωv = arc/sec = rω

ӨӨ

rr

arcarc

Page 12: Kinematics II

Quiz 3Quiz 3An object in circular motion travels An object in circular motion travels at 3 radians/sec in a circle of radius at 3 radians/sec in a circle of radius 5m. What is its angular speed?5m. What is its angular speed?

1.1. 3/5 m/sec3/5 m/sec2.2. 5/3 m/sec5/3 m/sec3.3. 2 m/sec2 m/sec4.4. 8 m/sec8 m/sec5.5. 15 m/sec15 m/sec

Page 13: Kinematics II

Quiz 3Quiz 3An object in circular motion travels at 3 An object in circular motion travels at 3 radians/sec in a circle of radius 5m. What radians/sec in a circle of radius 5m. What is its angular speed?is its angular speed?

ω = 3 radians/secω = 3 radians/secv = rωv = rωv = (5)(3) = 15 m/secv = (5)(3) = 15 m/sec

ANSWER: 15 m/secANSWER: 15 m/sec

Page 14: Kinematics II

Centripital ForceCentripital Force An object An object

travelling in travelling in circular motion circular motion experiences a experiences a centripetal force centripetal force FF

The induced The induced acceleration isacceleration isa = va = v22/r/r

F

Page 15: Kinematics II
Page 16: Kinematics II

Quiz 4Quiz 4Imagine that the centripetal force applied Imagine that the centripetal force applied on an object moving in circular motion is on an object moving in circular motion is suddenly shut off. What would happen?suddenly shut off. What would happen?

1.1. The object spirals outwardsThe object spirals outwards2.2. The object spirals inwardsThe object spirals inwards3.3. The object continues to move in uniform circular The object continues to move in uniform circular

motionmotion4.4. The object decelerates while continuing to move in The object decelerates while continuing to move in

a circlea circle5.5. The object moves away in a straight lineThe object moves away in a straight line

Page 17: Kinematics II

Quiz 4Quiz 4Imagine that the centripetal force Imagine that the centripetal force applied on an object moving in applied on an object moving in circular motion is suddenly shut off. circular motion is suddenly shut off. What would happen?What would happen?

ANSWER: The object moves away in ANSWER: The object moves away in a straight line (Newton’s first law)a straight line (Newton’s first law)

Page 18: Kinematics II

Projectile MotionProjectile Motion An object An object

experiences experiences gravitational gravitational acceleration acceleration g=9.80m/sg=9.80m/s22 downwards as it downwards as it movesmoves

Page 19: Kinematics II

Projectile MotionProjectile Motion Projectile motion is Projectile motion is

really two really two independent 1-D independent 1-D motionsmotions

One in the x One in the x directiondirection

One in the y One in the y directiondirection

Page 20: Kinematics II

x-directionx-directionThe x-direction experiences NO The x-direction experiences NO acceleration, thusacceleration, thus

x = vx = vxxtt

(It’s that simple!)(It’s that simple!)

Page 21: Kinematics II

y-directiony-directionThe y-direction experiences an The y-direction experiences an acceleration downwards, thusacceleration downwards, thus

y = vy = vyyt – ½gtt – ½gt22

Page 22: Kinematics II

Velocity ComponentsVelocity Componentsx = vx = vxxtty = vy = vyyt – ½gtt – ½gt22

vv

vvxx

vvyy

ӨӨ

vvxx = v cos = v cosӨӨvvyy = v sin= v sinӨӨ

Page 23: Kinematics II

Quiz 5Quiz 5At what angle should a cannon be At what angle should a cannon be positioned so that the projectile positioned so that the projectile travels the farthest?travels the farthest?

1.1. 90˚90˚2.2. 60˚60˚3.3. 45˚45˚4.4. 30˚30˚5.5. 0˚0˚

Page 24: Kinematics II

Quiz 5Quiz 5At what angle should a cannon be At what angle should a cannon be positioned so that the projectile positioned so that the projectile travels the farthest?travels the farthest?

ANSWER: 45˚ANSWER: 45˚

Page 25: Kinematics II

Proof of Quiz 5Proof of Quiz 5x = vx = vxxtty = vy = vyyt – ½gtt – ½gt22

vvxx = v cos = v cosӨӨvvyy = v sin= v sinӨӨ

Let y = 0Let y = 0 t = 2vt = 2vyy/g/g x = 2vx = 2vxxvvyy/g/g x = (2vx = (2v2 2 sinsinӨӨ cos cosӨӨ)/g)/g x = (vx = (v2 2 sin2sin2ӨӨ)/g)/g xxmaxmax = v = v22/g when 2/g when 2ӨӨ = 90˚ = 90˚ ӨӨ = 45˚ = 45˚

y = 0 y = 0

* Trig. Identity:* Trig. Identity:sin2sin2ӨӨ = 2sin = 2sinӨӨ cos cosӨӨ

Page 26: Kinematics II

ReviewReview Linear Motion under Linear Motion under

Uniform AccelerationUniform Accelerationx = xx = x00 + v + v00t + ½att + ½at22

v = v = vv00 + at + atΔΔx = vx = vaveavet = [(vt = [(v00 + + v)/2]tv)/2]tvv22 - v - v00

22 = 2a = 2aΔΔxx

Uniform Circular MotionUniform Circular MotionӨӨ = arc length (r=1) = arc length (r=1)arc = rarc = rӨӨω = ω = ӨӨ/sec/secv = arc/sec = rωv = arc/sec = rωa = va = v22/r/r

Projectile MotionProjectile Motionx = vx = vxxtty = vy = vyyt – ½gtt – ½gt22

vvxx = v cos = v cosӨӨvvyy = v sin= v sinӨӨ

Page 27: Kinematics II

Next WeekNext Week Choice 1Choice 1

– Review kinematicsReview kinematics Choice 2Choice 2

– Newton’s lawsNewton’s laws

Page 28: Kinematics II

The EndThe End