chapter 3 kinematics in two or three dimensions; vectors · 2017. 9. 26. · • vectors and...
TRANSCRIPT
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Lecture PowerPoints
Chapter 3
Physics for Scientists &Engineers, with Modern
Physics, 4th edition
Giancoli
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Chapter 3
Kinematics in Two or ThreeDimensions; Vectors
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Opening Question
A box is dropped from a moving helicopter atpoint A as it flies in a horizontal direction. Whichpath in the drawing below best describes the pathof the box (ignore air resistance) as seen by aperson standing on the ground?
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Units of Chapter 3
• Vectors and Scalars
• Addition of Vectors—Graphical Methods
• Subtraction of Vectors, and Multiplication of aVector by a Scalar
• Adding Vectors by Components
• Unit Vectors
• Vector Kinematics
• Projectile Motion
• Solving Problems Involving Projectile Motion
• Relative Velocity
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3-1 Vectors and Scalars
A vector has magnitude aswell as direction.
Some vector quantities:displacement, velocity, force,momentum
A scalar has only a magnitude.
Some scalar quantities: mass,time, temperature
The GREEN arrows representthe velocity vector at eachpoint.
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Representing Vectors
Vectors can be represented in a variety of ways by
using Bold type and/or lines or arrows, etc. aboveor below the vector symbol. Eg.
value.averageormeanwith the
confusedbemaytheyasVandavoidbest toisIt
VV~
VV~
V
VVVVV
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3-2 Addition of Vectors—Graphical Methods
For vectors in onedimension, simpleaddition and subtractionare all that is needed.
You do need to be carefulabout the signs, as thefigure indicates.
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3-2 Addition of Vectors—Graphical MethodsIf the motion is in two dimensions, the situation is somewhatmore complicated.
Here, the actual travel paths are at right angles to one another; wecan find the displacement by using the Pythagorean Theorem.
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3-2 Addition of Vectors—Graphical Methods
Adding the vectors in the opposite order gives thesame result:
EofN27atkm11.212 DDDR
1221 VVVV
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3-2 Addition of Vectors—Graphical Methods
Even if the vectors are not at rightangles, they can be added graphically byusing the tail-to-tip method.
Note that (as in the previous example) the orderof adding the vectors is not important.
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3-2 Addition of Vectors—Graphical Methods
Where there are two vectors the parallelogrammethod may also be used.
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3-3 Subtraction of Vectors, andMultiplication of a Vector by a Scalar
In order to subtract vectors, wedefine the negative of a vector, whichhas the same magnitude but pointsin the opposite direction.
Then we add the negative vector.
1212 VVVV
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3-3 Subtraction of Vectors, andMultiplication of a Vector by a Scalar
A vector can be multiplied by a scalarc; the result is a vector c that has thesame direction but a magnitude cV. If c isnegative, the resultant vector points inthe opposite direction.
V
V
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3-4 Adding Vectors by Components
Any vector can be expressed as the sumof two other vectors, which are called itscomponents. Usually the other vectors arechosen so that they are perpendicular toeach other.
x
y
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V
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vectorofcomponentstheareand
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VVV
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3-4 Adding Vectors by Components
If the components areperpendicular, they can befound using trigonometricfunctions.
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3-4 Adding Vectors by Components
The components are effectively one-dimensional,so they can be added arithmetically.
yyyxxx 2121 VVVandVVV
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3-4 Adding Vectors by Components
Adding vectors:
1. Draw a diagram; add the vectors graphically.
2. Choose x and y axes.
3. Resolve each vector into x and y components.
4. Calculate each component using sines and cosines.
5. Add the components in each direction.
6. To find the length and direction of the vector, use:
and .
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3-4 Adding Vectors byComponents
Example 3-2: Mail carrier’sdisplacement.
A rural mail carrier leaves thepost office and drives 22.0 kmin a northerly direction. Shethen drives in a direction 60.0°south of east for 47.0 km. Whatis her displacement from thepost office?
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This image cannot currently be displayed.
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3-4 Adding Vectors byComponents
Example 3-3: Three short trips.
An airplane trip involves threelegs, with two stopovers. Thefirst leg is due east for 620 km;the second leg is southeast(45°) for 440 km; and the thirdleg is at 53° south of west, for550 km, as shown. What is theplane’s total displacement?
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S50.4Eatkm980ntDisplacemeTotal
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3-5 Unit Vectors
Unit vectors have magnitude 1.
Using unit vectors, any vectorcan be written in terms of itscomponents:
V
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Eg. Using Unit Vectors to Solve the Mail CourierProblem – (Example 3-2)
kmˆ18.7ˆ23.5
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3-6 Vector Kinematics
In two or threedimensions, thedisplacement is avector:
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3-6 Vector Kinematics
As Δt and Δr becomesmaller and smaller, theaverage velocityapproaches theinstantaneous velocity.
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3-6 Vector Kinematics
v
v
The instantaneousacceleration is in thedirection of Δ = 2 – 1,and is given by:
v
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3-6 Vector Kinematics
Using unit vectors,
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3-6 Vector Kinematics
Generalizing the one-dimensional equationsfor constant acceleration:
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3-7 Projectile Motion
A projectile is anobject moving in twodimensions under theinfluence of Earth'sgravity; its path is aparabola.
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It can be understoodby analyzing thehorizontal and verticalmotions separately.
3-7 Projectile Motion
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3-7 Projectile Motion
The speed in the x-directionis constant; in the y-direction the object moveswith constant acceleration g.
This photograph shows two ballsthat start to fall at the same time.The one on the right has an initialspeed in the x-direction. It can beseen that vertical positions of thetwo balls are identical at identicaltimes, while the horizontalposition of the yellow ballincreases linearly.
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3-7 Projectile Motion
If an object is launched at an initial angle of θ0
with the horizontal, the analysis is similar exceptthat the initial velocity has a vertical component.
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3-8 Solving Problems InvolvingProjectile Motion
Projectile motion is motion with constantacceleration in two dimensions, where theacceleration is g and is down.
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3-8 Solving Problems InvolvingProjectile Motion
1. Read the problem carefully, and choose theobject(s) you are going to analyze.
2. Draw a diagram.
3. Choose an origin and a coordinate system.
4. Decide on the time interval; this is the same inboth directions, and includes only the time theobject is moving with constant acceleration g.
5. Examine the x and y motions separately.
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3-8 Solving Problems InvolvingProjectile Motion
6. List known and unknown quantities.Remember that vx never changes, and thatvy = 0 at the highest point.
7. Plan how you will proceed. Use theappropriate equations; you may have tocombine some of them.
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3-8 Solving Problems InvolvingProjectile Motion
Example 3-6: Driving off acliff.
A movie stunt driver on amotorcycle speedshorizontally off a 50.0 m highcliff. How fast must themotorcycle leave the cliff topto land on level ground below,90.0 m from the base of thecliff where the cameras are?Ignore air resistance.
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3-8 Solving Problems InvolvingProjectile Motion
Example 3-7: A kicked football.
A football is kicked at an angle θ0 = 37.0° with a velocityof 20.0 m/s, as shown. Calculate (a) the maximumheight, (b) the time of travel before the football hits theground, (c) how far away it hits the ground, (d) thevelocity vector at the maximum height, and (e) theacceleration vector at maximum height. Assume theball leaves the foot at ground level, and ignore airresistance and rotation of the ball.
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m7.359.802
12.0y
0y9.80212.00
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(a) Maximum Height
Only need to consider vertical motion andcalculate time to maximum height.
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(b) Time for ball to hit ground
Consider vertical motion and calculate time tomaximum height.
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9.80t12.00
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avv 0yy
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(c) How far away it hits the ground
Horizontal motion at constant velocity and time offlight = 2.45 s.
m39.22.4516.0t
m/s 16.037.020.0θ0
x
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vx
vv
coscos
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(d) The velocity vector at the maximum height
m/sˆ16.0horizontalisMotion ivv 0x
(e) The acceleration vector at the maximum height
2m/sˆ-9.80
pointsallat
ja
ga
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3-8 Solving Problems InvolvingProjectile Motion
A child sits upright in a wagonwhich is moving to the right atconstant speed as shown. Thechild extends her hand and throwsan apple straight upward (from herown point of view), while thewagon continues to travel forwardat constant speed. If air resistanceis neglected, will the apple land (a)behind the wagon, (b) in thewagon, or (c) in front of thewagon?
Conceptual Example 3-8: Where does theapple land?
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3-8 Solving Problems InvolvingProjectile Motion
Conceptual Example 3-9: The wrong strategy.
A boy on a small hill aims his water-balloon slingshothorizontally, straight at a second boy hanging from atree branch a distance d away. At the instant the waterballoon is released, the second boy lets go and fallsfrom the tree, hoping to avoid being hit. Show that hemade the wrong move. (He hadn’t studied physicsyet.) Ignore air resistance.
Both water-balloonand second boy havethe same vertical(vertical) acceleration
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3-8 Solving Problems InvolvingProjectile Motion
Example 3-10: Level horizontal range.
(a) Derive a formula for thehorizontal range R of a projectile interms of its initial speed v0 and angleθ0. The horizontal range is defined asthe horizontal distance the projectiletravels before returning to itsoriginal height (which is typically theground); that is, y(final) = y0. (b)Suppose one of Napoleon’s cannonshad a muzzle speed, v0, of 60.0 m/s.At what angle should it have beenaimed (ignore air resistance) tostrike a target 320 m away?
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:MotionHorizontal(a)
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θθ22θθ2
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rangesamethegivewillanglesBoth
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3-8 Solving Problems InvolvingProjectile Motion
Example 3-11: A punt.
Suppose the football in Example 3–7 waspunted and left the punter’s foot at a height of1.00 m above the ground. How far did thefootball travel before hitting the ground? Setx0 = 0, y0 = 0.
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NB: Cannot use horizontal range
v0 = 20.0 m/s0 = 37.0y0 = 0y = -1 m
v0x = 20.0 cos 37.0 = 16.0 m/sv0y = 20.0 sin 37.0 = 12.0 m/sa = -g = -9.8 m/s2
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3-8 Solving Problems Involving Projectile Motion
Example 3-12: Rescue helicopter drops supplies.A rescue helicopter wants to drop a package of supplies to isolatedmountain climbers on a rocky ridge 200 m below. If the helicopter istraveling horizontally with a speed of 70 m/s (250 km/h), (a) how far inadvance of the recipients (horizontal distance) must the package bedropped? (b) Suppose, instead, that the helicopter releases the packagea horizontal distance of 400 m in advance of the mountain climbers.What vertical velocity should the package be given (up or down) so thatit arrives precisely at the climbers’ position? (c) With what speed doesthe package land in the latter case?
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Example 3-12: Rescue helicopter drops supplies.
Try this one yourself
Answer: (a) 447 m(b) 7 m/s downwards(c) 94 m/s
Worked solution in text book.
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3-8 Solving Problems InvolvingProjectile Motion
Projectile motion is parabolic:
Taking the equations for x and y as a functionof time, and combining them to eliminate t, wefind y as a function of x:
This is the equation for a parabola.
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3-8 Solving Problems InvolvingProjectile Motion
Examples of projectilemotion. Notice theeffects of air resistance.
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3-9 Relative Velocity
We have already considered relative speed inone dimension; it is similar in two dimensionsexcept that we must add and subtract velocitiesas vectors.
Each velocity is labeled first with the object,and second with the reference frame in whichit has this velocity.
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Here, vWS is the velocityof the water in the shoreframe, vBS is the velocityof the boat in the shoreframe, and vBW is thevelocity of the boat in thewater frame.
3-9 Relative Velocity
The relationship betweenthe three velocities is:
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3-9 Relative Velocity
Example 3-14: Heading upstream.
A boat’s speed in stillwater is vBW = 1.85 m/s. Ifthe boat is to traveldirectly across a riverwhose current has speedvWS = 1.20 m/s, at whatupstream angle must theboat head?
upstream40.41.85
1.20θ 11
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BW
WS
v
v
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3-9 Relative Velocity
Example 3-15: Heading across the river.
The same boat (vBW = 1.85 m/s) nowheads directly across the riverwhose current is still vWS = 1.20 m/s.
(a) What is the velocity (vBS)(magnitude and direction) of theboat relative to the shore?
(b) If the river is 110 m wide, howlong will it take to cross and howfar downstream will the boat bethen?
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vBW = 1.85 m/s, vWS =1.20 m/s.
(a) What is the velocity (vBS)(magnitude and direction) ofthe boat relative to the shore?
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(b) If the river is 110 m wide, howlong will it take to cross and howfar downstream will the boat bethen?
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3-9 Relative Velocity
Example 3-16: Car velocities at 90°.
Two automobiles approach a street corner atright angles to each other with the same speedof 40.0 km/h (= 11.1 m/s), as shown. What is therelative velocity of one car with respect to theother? That is, determine the velocity of car 1 asseen by car 2.
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The diagram at left shows the situation from a referenceframe fixed to the Earth.
We need to view from a reference frame in which car 2 is atrest which is shown in the diagram at right.
Car 2 sees both the Earth and Car 1 moving towards it.
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2Cartowards45atkm/h56.64040v
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Summary of Chapter 3
• A quantity with magnitude and direction is avector.
• A quantity with magnitude but no direction isa scalar.
• Vector addition can be done either graphicallyor by using components.
• The sum is called the resultant vector.
• Projectile motion is the motion of an objectnear the Earth’s surface under the influence ofgravity.