chapter 6 motion in two dimensions
DESCRIPTION
Chapter 6 Motion in Two Dimensions. Merrill Physics: Principles and Problems. A. Motion in two dimensions 1. Will use the kinematic equations: d = vt d = v o t + 1/2 at 2 v f 2 = v o 2 + 2ad v f = v o + at d = {(v f + v o ) / 2}(t). 2. Will use Newton’s Laws - PowerPoint PPT PresentationTRANSCRIPT
Chapter 6Chapter 6Motion in Two Motion in Two DimensionsDimensions
Merrill Physics: Principles and Merrill Physics: Principles and ProblemsProblems
A. Motion in two dimensions A. Motion in two dimensions
1. Will use the kinematic 1. Will use the kinematic equations:equations:
d = vtd = vt
d = vd = voot + 1/2 at t + 1/2 at 22
vvff22 = v = voo
2 2 + 2ad + 2ad
vvff = v = vo o + at + at
d = {(vd = {(vff + v + vo o ) / 2}(t)) / 2}(t)
2. Will use Newton’s Laws2. Will use Newton’s Laws
11stst: An object in motion remains in : An object in motion remains in motion unless acted on by a net motion unless acted on by a net forceforce
22ndnd : :
F = m aF = m a
where F is net forcewhere F is net force
3. And coefficients of friction:3. And coefficients of friction:
ff f f / F/ F N N
Projectile MotionProjectile Motion
Objectives:Objectives: Understand that the vertical and Understand that the vertical and
horizontal components of a projectile horizontal components of a projectile are independentare independent
Determine the range, maximum Determine the range, maximum height and flight time of projectilesheight and flight time of projectiles
*As you throw a ball, or any type of object, *As you throw a ball, or any type of object, what shape does it make in the air?what shape does it make in the air?
Definitions used for Projectile Motion Definitions used for Projectile Motion (Ballistics) (Ballistics)
a.a. Projectile -Projectile - object that only object that only has gravity acting on ithas gravity acting on it
b.b. Trajectory - path a projectile Trajectory - path a projectile followsfollows
c.c. Shape of the trajectory Shape of the trajectory depends on your point of view and depends on your point of view and the object.the object.
starting pointending point
What force accelerates the object in the x-direction?(None once you let go or the bullet is fired) *this assumes no air resistance for simplicity
So velocity is constant in the x-direction
(1) Top looking down on (1) Top looking down on object’s pathobject’s path
(2) Side view of object’s motion(2) Side view of object’s motion
startend
What force accelerates in the y-direction?Gravity!
So VY is a changing value.
4. Three assumptions used in 4. Three assumptions used in analyzing motionanalyzing motion
Assumptions:Assumptions:(1)(1) acceleration due to gravity acceleration due to gravity
is is 9.8 m/s9.8 m/s2 2 downdown(2) air resistance is negligible(2) air resistance is negligible(3)(3) rotation of the earth does rotation of the earth does
not not affect motionaffect motion
This is not rocket science! This is not rocket science! …. …. Otherwise we would include these!Otherwise we would include these!
Independence of Motion in Two Independence of Motion in Two DimensionsDimensions
What forces act on an object after What forces act on an object after it’s initial thrust if you ignore air it’s initial thrust if you ignore air resistance?resistance?
Projectiles launched at an angleProjectiles launched at an angle
Objectives:Objectives: Determine the range, maximum Determine the range, maximum
height and flight time of height and flight time of projectiles launched at an angleprojectiles launched at an angle
Projectiles Launched HorizontallyProjectiles Launched Horizontally
Example: A stone is thrown horizontally off a Example: A stone is thrown horizontally off a 44m cliff at velocity of 15 m/s. How long will it 44m cliff at velocity of 15 m/s. How long will it take to hit the bottom? How far will it go take to hit the bottom? How far will it go horizontally if the air resistance is negligible?horizontally if the air resistance is negligible?
Given:Given:
vvxx = 15 m/s = 15 m/s ddxx = v = vxxt = 15m(t)t = 15m(t)
vvooyy= 0m/s = 0m/s ddyy = v = vooyyt + ½att + ½at22
aayy = -9.8 m/s = -9.8 m/s22 44m = ½(9.8m/s44m = ½(9.8m/s22)t)t22
* Time is the same in the vertical and horizontal * Time is the same in the vertical and horizontal directions!directions!
t2 = 2(44m)/9.8m/ss22
t = 3s and 15m/s(3s) = 45m
Projectiles Launched at an AngleProjectiles Launched at an Angle
b.b.
(1) velocity in x-direction never changes(2) velocity in y-direction changes
vy-f = vy-o + g t
vy-f2 = vy-o
2 + 2 gdy
(3) vR2 = vx
2 + vy2
Projectile Motion Problem Solving Projectile Motion Problem Solving
a. a. Projectile is an object that only has Projectile is an object that only has gravity acting on itgravity acting on itb. b. Break given initial velocity into x and Break given initial velocity into x and y components. y components. vv at an angle at an angle (1)horizontal motion: v(1)horizontal motion: vxx = = vv cos cos(2)vertical motion: v(2)vertical motion: vyy = = vv sin sinc.c. use subscripts to identify x & y use subscripts to identify x & y
components ofcomponents of velocity, displacement, velocity, displacement, accelerationacceleration
d. Use independence of vertical and d. Use independence of vertical and horizontal motion to solve problemshorizontal motion to solve problems
e. Draw diagrams to aid you in e. Draw diagrams to aid you in visualizing what is occurringvisualizing what is occurring
f.f. Write down given information with Write down given information with signssigns
g. Note: ag. Note: axx = 0 = 0 a ayy = - 9.8 m/s = - 9.8 m/s2 2
- - time is the same for both directionstime is the same for both directions
- vertical motion is symmetrical!- vertical motion is symmetrical!
Velocity of the ProjectileVelocity of the Projectile
The velocity of the projectile at any The velocity of the projectile at any point of its motion is the vector sum point of its motion is the vector sum of its x and y components at that of its x and y components at that pointpoint
• Remember to be careful about the Remember to be careful about the angle’s quadrantangle’s quadrant
2 2 1tan yx y
x
vv v v and
v
Example: Object is launched at an angle of 60Example: Object is launched at an angle of 60oo at a velocity of 110 m/s. How far will it go if at a velocity of 110 m/s. How far will it go if the air resistance is negligible?the air resistance is negligible?
Given:Given:
vvx x = 110 cos 60 = 55 m/s= 110 cos 60 = 55 m/s
vvooyy = 110 sin 60 = 95.3 m/s= 110 sin 60 = 95.3 m/s
vvffy y = - 95.3 m/s= - 95.3 m/s
aayy = -9.8 m/s = -9.8 m/s22
* Commonality is with TIME. How long it is in * Commonality is with TIME. How long it is in the air is same for horizontal and vertical the air is same for horizontal and vertical motion. Knowing time, you can find distance motion. Knowing time, you can find distance in xin x
vvyyff = v = v00yy
+ a + ayytt
t = 19.4 sec use travel in the vertical t = 19.4 sec use travel in the vertical direction to determine how long object direction to determine how long object is in flight. Remember, object is is in flight. Remember, object is accelerated in the y-direction.accelerated in the y-direction.
Knowing time then can find out how farKnowing time then can find out how far
vvxx = d = dxx / t since uniform motion in x- / t since uniform motion in x-direct.direct.
ddx x = (55 m/s) (19.4 sec) = 1070 meters= (55 m/s) (19.4 sec) = 1070 meters
i. Example Problem:i. Example Problem:
A punter kicks a football at an A punter kicks a football at an angle of 30angle of 30o o with the horizontal at with the horizontal at an initial speed of 20 m/s. Where an initial speed of 20 m/s. Where should the punt returner position should the punt returner position himself to catch the ball just before himself to catch the ball just before it strikes the ground?it strikes the ground?
Givens:Givens:
v
30
vy = 20 m/s sin 30 = 10 m/s
vx = 20 cos 30 = 17.3 m/s
vy 0 = 10 m/s vyf
=-10m/s
vyf = vy 0
+ at
where a=-9.8m/s2
t = 2.04 sec
dx = vx t =17.3 m/s (2.04s)
= 35.3 m
Projectile Motion Simulation labsProjectile Motion Simulation labs
http://phet.colorado.edu/en/simulatiohttp://phet.colorado.edu/en/simulation/projectile-motionn/projectile-motion
http://phet.colorado.edu/sims/projectihttp://phet.colorado.edu/sims/projectile-motion/projectile-motion_en.htmlle-motion/projectile-motion_en.html
Periodic MotionPeriodic Motion
Explain centripetal acceleration of an Explain centripetal acceleration of an object in circular motionobject in circular motion
Apply Newton’s Laws to circular Apply Newton’s Laws to circular motionmotion
Define simple harmonic motionDefine simple harmonic motion Apply Newton’s Laws to harmonic Apply Newton’s Laws to harmonic
motionmotion
B. Periodic Motion B. Periodic Motion
1.1. Uniform Circular MotionUniform Circular Motion
a. Consider a model airplane a. Consider a model airplane on control strings - the hand pulls on control strings - the hand pulls on cable and keeps an inward on cable and keeps an inward force on the plane so it flies in a force on the plane so it flies in a circle. Let go of the string and it circle. Let go of the string and it flies in a straight lineflies in a straight line
A nA net force, acting perpendicular to a et force, acting perpendicular to a mass moving at constant speed, mass moving at constant speed, changes direction of motion of the changes direction of motion of the mass.mass.
(1) (1) Force is always 90Force is always 90o o to to direction of motion of direction of motion of
the massthe mass
(2)(2) Since it is perpendicular, Since it is perpendicular, cannot cannot change change the magnitude of the magnitude of
velocity only directionvelocity only direction..
(3)(3) Net force is Net force is centripetal force, centripetal force,
FFcc..
Change in direction is a change in Change in direction is a change in velocity (velocity (vvso object is so object is acceleratedaccelerated (1)F(1)Fcc is towards the center, so is towards the center, so
acceleration is towards the acceleration is towards the center.center.
(2) Called (2) Called centripetal acceleration.centripetal acceleration.
Centripetal Acceleration Formulas: Centripetal Acceleration Formulas:
(1) (1) aacc = v = v22/r/r
(2)(2) FFcc = ma = mac c = mv= mv22/r/r
Key: Centripetal force is always Key: Centripetal force is always caused by some other action - caused by some other action - friction, pulling, pushing, etc.friction, pulling, pushing, etc.
We define period of circular motion We define period of circular motion (T) to be the time to go around the (T) to be the time to go around the circumference one timecircumference one time
For uniform circular motion time to For uniform circular motion time to go around circle is constantgo around circle is constant
v = d/t (d = the circumference) v = d/t (d = the circumference)
where d = 2where d = 2r & t = T (the period)r & t = T (the period)
thereforethereforevv22r/T andr/T and
aacc = v = v22/r = 4/r = 4r/r/TT22
FFnetnet = F = Fcc = ma = macc = mv = mv22/r = m(4/r = m(4r/r/TT22))
Example: A person is flying an 80.0g model Example: A person is flying an 80.0g model airplane in a horizontal path at the end of a airplane in a horizontal path at the end of a string 10.0m long. If the string is horizontal string 10.0m long. If the string is horizontal and exerts a 2.65N force on the hand, what and exerts a 2.65N force on the hand, what is the speed of the plane?is the speed of the plane?
Givens:Givens: FormulasFormulasm= 0.080kgm= 0.080kg v = 2v = 2r/T r/T r = 10.0 mr = 10.0 m FFcc = ma = macc = mv = mv22/r/r
V = (Fccr/m)1/2 1/2 = 18.2 m/s
A 200g object is tied on a cord and A 200g object is tied on a cord and whirled in a horizontal circle of radius whirled in a horizontal circle of radius 1.20m at 3 rev/sec. The cord is 1.20m at 3 rev/sec. The cord is horizontal. Determine acceleration horizontal. Determine acceleration and tension in the cord.and tension in the cord.
Givens:Givens: FormulasFormulas
m= 0.200kgm= 0.200kg v = v = 22r/T r/T r = 1.20mr = 1.20m
rev/sec rev/sec (freq.)(freq.)vvm/sm/s
secsec
a. aa. acc = v = v22 / r = 426 m/s / r = 426 m/s22
b. b. FFcc = m a = m ac c = 85 N= 85 N
What is the maximum speedWhat is the maximum speed at which a at which a car can round a curve of 25m radius on a car can round a curve of 25m radius on a level road if level road if static static between the tires and between the tires and
road is 0.80road is 0.80
FFc c = F= Fff = = static static FFN N == static static m(-g)m(-g)
FFc c = ma= mac c = = static static m(-g) = mvm(-g) = mv22/r/r
Solving for v: v = (rSolving for v: v = (rstaticstatic--g)g)1/21/2
Therefore v = (25mx0.80x9.8m/sTherefore v = (25mx0.80x9.8m/s22))1/21/2
v = (196mv = (196m22/s/s22))1/2 1/2 = 14 m/s = 14 m/s * the maximum velocity only depends on the * the maximum velocity only depends on the
radius of the curve and radius of the curve and static , static , not the mass!not the mass!
To make an object turn or To make an object turn or rotate, apply a torque. rotate, apply a torque.
11.1 Torque
Every time you open a door, turn on a water Every time you open a door, turn on a water faucet, or tighten a nut with a wrench, you faucet, or tighten a nut with a wrench, you exert a turning force. exert a turning force.
Torque Torque is produced by this turning force and is produced by this turning force and tends to produce rotational acceleration. tends to produce rotational acceleration.
Torque is different from force.Torque is different from force.• Forces tend to make things accelerate.Forces tend to make things accelerate.• Torques produce rotation.Torques produce rotation.
Torque
A torque produces rotation.A torque produces rotation.
Torque
Changing Circular Motion (Torque)Changing Circular Motion (Torque)
The door is free to rotate about an axis The door is free to rotate about an axis through the hinge at point through the hinge at point OO
There are three factors that determine the There are three factors that determine the effectiveness of the force in opening the effectiveness of the force in opening the door:door:• (1) The (1) The magnitudemagnitude of the force of the force• (2) The distance of the application of the force(2) The distance of the application of the force• (3) The (3) The angleangle at which the force is applied at which the force is applied
TorqueTorque
Magnitude of forceMagnitude of force
Distance from axisDistance from axis
Direction Direction
When a perpendicular force is applied, the When a perpendicular force is applied, the lever arm is the distance between the lever arm is the distance between the doorknob and the edge with the hinges.doorknob and the edge with the hinges.
Torque
When the force is perpendicular, the distance from When the force is perpendicular, the distance from the turning axis to the point of contact is called the the turning axis to the point of contact is called the lever arm. lever arm.
If the force is not at right angle to the lever arm, If the force is not at right angle to the lever arm, then only the perpendicular component of the then only the perpendicular component of the force will contribute to the torque. force will contribute to the torque.
Torque
The same torque can be produced by a The same torque can be produced by a large force with a short lever arm, or a large force with a short lever arm, or a small force with a long lever arm. small force with a long lever arm.
The same force can produce different The same force can produce different amounts of torque. amounts of torque.
Greater torques are produced when both Greater torques are produced when both the force and lever arm are large.the force and lever arm are large.
Torque
TorqueTorque = Frsin= Frsinθθ
Units – NmUnits – Nm
If If θθ = 90 = 90o o
Then sinThen sinθθ = 1 = 1
= Fr or Fd where d = the lever arm = Fr or Fd where d = the lever arm lengthlength
Torque, Torque, , is the tendency of a force to , is the tendency of a force to rotate an object about some axisrotate an object about some axis
= Frsin= Frsin
is the torqueis the torque F is the forceF is the force
symbol is the Greek tausymbol is the Greek tau
rsinrsin is the length of the lever arm acting is the length of the lever arm actingperpendicular to the force where perpendicular to the force where is the angleis the anglebetween F and rbetween F and r
SI unit is NSI unit is N..mm
Although the magnitudes of the applied forces Although the magnitudes of the applied forces are the same in each case, the torques are are the same in each case, the torques are different.different.
Torque
think!think!If you cannot exert enough torque to turn a stubborn If you cannot exert enough torque to turn a stubborn bolt, would more torque be produced if you fastened bolt, would more torque be produced if you fastened a length of rope to the wrench handle as shown?a length of rope to the wrench handle as shown?
Answer: Answer:
No, because the lever arm is the same. To increase No, because the lever arm is the same. To increase the lever arm, a better idea would be to use a pipe the lever arm, a better idea would be to use a pipe that extends upward.that extends upward.
Torque
do the math!do the math!What is the weight of the block hung at the 10-cm What is the weight of the block hung at the 10-cm mark?mark?
Balanced Torques
do the math!do the math!The block of unknown weight tends to rotate the The block of unknown weight tends to rotate the system of blocks and stick system of blocks and stick counterclockwisecounterclockwise, and , and the 20-N block tends to rotate the system the 20-N block tends to rotate the system clockwiseclockwise.. The system is in balance (static equilibrium) when the The system is in balance (static equilibrium) when the two torques are equal:two torques are equal:
counterclockwise torque = clockwise torquecounterclockwise torque = clockwise torque
Balanced Torques
Rearrange the equation to solve for the unknown weight:Rearrange the equation to solve for the unknown weight:
The lever arm for the unknown weight is 40 cm. The lever arm for the unknown weight is 40 cm.
The lever arm for the 20-N block is 30 cm.The lever arm for the 20-N block is 30 cm.
The unknown weight is thus 15 N.The unknown weight is thus 15 N.
Balanced Torques
Torque problemsTorque problems
Complete Gantry Crane Worksheet Complete Gantry Crane Worksheet (Due Thurs 1/7/2014)(Due Thurs 1/7/2014)
Simple Harmonic Motion ObjectivesSimple Harmonic Motion Objectives
Identify objects in simple harmonic Identify objects in simple harmonic motion.motion.
Determine variables that affect the Determine variables that affect the period of a pendulum.period of a pendulum.
Simple Harmonic or Periodic MotionSimple Harmonic or Periodic Motion
Examples – clock pendulum, Examples – clock pendulum, vibrating guitar stringvibrating guitar string
Simple Harmonic MotionSimple Harmonic Motion
Described by two quantities:Described by two quantities:• Period Period • AmplitudeAmplitude
Harmonic Motion in a SpringHarmonic Motion in a Spring
Simple Harmonic MotionSimple Harmonic MotionA weight attached to a spring undergoes simple A weight attached to a spring undergoes simple harmonic motion. harmonic motion. A marking pen attached to the bob traces a sine curve A marking pen attached to the bob traces a sine curve on a sheet of paper that is moving horizontally at on a sheet of paper that is moving horizontally at constant speed.constant speed.A sine curve is a pictorial representation of a wave.A sine curve is a pictorial representation of a wave.A A sine curve sine curve is a pictorial representation of a SHM.is a pictorial representation of a SHM.
A stone suspended at the end of a string is a simple A stone suspended at the end of a string is a simple pendulum.pendulum.
Pendulums swing back and forth with such regularity Pendulums swing back and forth with such regularity that they have long been used to control the motion of that they have long been used to control the motion of clocks. clocks.
The time of a back-and-forth swing of the pendulum is The time of a back-and-forth swing of the pendulum is its its period. period.
Galileo discovered that the period of a pendulum Galileo discovered that the period of a pendulum depends only on its length—its mass has no effect. depends only on its length—its mass has no effect.
Motion of a Pendulum
The period of the pendulum depends only on the The period of the pendulum depends only on the length of a pendulum and the acceleration of gravity. length of a pendulum and the acceleration of gravity.
Motion of a Pendulum
The back-and-forth vibratory motion—called The back-and-forth vibratory motion—called oscillatory motion—of a swinging pendulum is oscillatory motion—of a swinging pendulum is also also simple harmonic motion (SHM).simple harmonic motion (SHM).
glT 2
Two pendulums of the same length have Two pendulums of the same length have the same period regardless of mass.the same period regardless of mass.
Motion of a Pendulum
Example ProblemExample ProblemHow long must a pendulum be on the How long must a pendulum be on the Moon, where g=1.6 m/sMoon, where g=1.6 m/s22, to have a , to have a period of 2.0 s?period of 2.0 s?
T = 2.0s = 2T = 2.0s = 2(l/1.6m/s(l/1.6m/s22))1/21/2
(2.0s/2(2.0s/2))2 2 = l/1.6m/s= l/1.6m/s22
l = 1.6m/sl = 1.6m/s22(4s(4s22/4/422)) = 0.16 m or 16cm= 0.16 m or 16cm
w|| is the part of w that is ______________________motion.w|| _______________ as increases
w|| pulls the bob back towards __________________
w|| is called a ___________________ force.
The weight w can be ______________________________ :
w
w||
w┴
w
w||
w┴
equilibrium
equilibriumincreases
restoring
resolved into components
This componentis ______________to the motion andis ____________ in direction to the____________
parallel to the
perpendicular
tension
opposite
A long pendulum has a longer period than a shorter A long pendulum has a longer period than a shorter pendulum. pendulum.
It swings back and forth more slowly—less frequentlyIt swings back and forth more slowly—less frequently—than a short pendulum. —than a short pendulum.
Just as a long pendulum has a greater period, a Just as a long pendulum has a greater period, a person with long legs tends to have a slower stride person with long legs tends to have a slower stride than a person with short legs. than a person with short legs.
Giraffes and horses run with a slower gait than do Giraffes and horses run with a slower gait than do short-legged animals such as hamsters and mice.short-legged animals such as hamsters and mice.
Motion of a Pendulum