2-Dimensional Motion2-Dimensional Motion

Part I: Projectile MotionPart I: Projectile MotionPart II: Circular MotionPart II: Circular Motion

First of All, What is 2-D Motion?First of All, What is 2-D Motion?

Before, we talked about motion in one Before, we talked about motion in one dimension (just the x axis)dimension (just the x axis)

Now we are investigating things that can Now we are investigating things that can move it 2 dimensions (x and y)!move it 2 dimensions (x and y)!

y

x

2D Motion

x

1D Motion

2-D Motion Part I:2-D Motion Part I:Projectile MotionProjectile Motion

A “projectile” is an object on which the A “projectile” is an object on which the only force acting is gravity.only force acting is gravity.– These are things that fly through the These are things that fly through the

air!air!(Examples: footballs, baseballs, cars (Examples: footballs, baseballs, cars flying off cliffs, bullets, cannon flying off cliffs, bullets, cannon shells, and batman)shells, and batman)

– Usually they are thrown/fired/etc. but Usually they are thrown/fired/etc. but are are notnot things like rockets, which things like rockets, which still have something “pushing” them still have something “pushing” them as they fly through the air!as they fly through the air!

As always, gravity will accelerate the As always, gravity will accelerate the object object downdown..

The Dimensions are The Dimensions are IndependentIndependent

Since 2D motion is Since 2D motion is complicated, it is helpful complicated, it is helpful to think of the x and y to think of the x and y dimensions separately dimensions separately when we analyze motion.when we analyze motion.This is because the x and This is because the x and y dimensions are y dimensions are completely completely independentindependent..This lets us use all of our This lets us use all of our favorite 1D equations favorite 1D equations when solving problems!when solving problems!

Notice that after this idiot Notice that after this idiot jumps off the cliff his jumps off the cliff his velocity in the x dimension velocity in the x dimension does not change. does not change.

BUT……His velocity in the BUT……His velocity in the y dimension is increasing y dimension is increasing due to gravity.due to gravity.

The same is true for this The same is true for this cannonball:cannonball:

GravityGravity

What causes projectiles to What causes projectiles to accelerate?accelerate?– Gravity!Gravity!

In which dimension does In which dimension does gravity exist?gravity exist?– The vertical (y)!The vertical (y)!

The acceleration is always The acceleration is always 9.81 m/s9.81 m/s22 downwarddownward. This is . This is called “acceleration due to called “acceleration due to gravity” or “g”gravity” or “g”Since acceleration Since acceleration onlyonly exists in the exists in the y dimension y dimension this means it is always this means it is always zerozero for the x dimension.for the x dimension.

Horizontal VelocityHorizontal Velocity

Since we have established that acceleration only exists Since we have established that acceleration only exists in the y dimension and the acceleration in the x in the y dimension and the acceleration in the x dimension is zero, what does this mean about the dimension is zero, what does this mean about the velocity of a projectile in the x dimension?velocity of a projectile in the x dimension?– It is always It is always constantconstant!!

This needs to be This needs to be rememberedremembered (by YOU)! (by YOU)!

Solving ProblemsSolving Problems

We are going to take some time to go over We are going to take some time to go over the problem solving process for these the problem solving process for these projectile motion problems.projectile motion problems.It is a “modified” version of GUESS.It is a “modified” version of GUESS.The best part is that you don’t have to The best part is that you don’t have to learn any complicated equations… we will learn any complicated equations… we will use the ones you use the ones you already know from already know from the last unit! It’s easy!the last unit! It’s easy!

GivensGivensOf course, the first step to solving Of course, the first step to solving any problem is to list your givens.any problem is to list your givens.

Always begin by drawing a Always begin by drawing a diagram representing the motion. diagram representing the motion. And remember to include your And remember to include your arrows for initial velocity, arrows for initial velocity, acceleration, displacement, and acceleration, displacement, and the positive direction.the positive direction.

Remember, we are separating the Remember, we are separating the variables into the two dimensions.variables into the two dimensions.

You will write out the same givens You will write out the same givens table as before, but this time it will table as before, but this time it will be a different set for each be a different set for each dimension (x or y)!dimension (x or y)!

xx yy

vvoo = =

v = v =

aaxx = =

t = t =

x = x =

vvoo = =

v = v =

aayy = =

t = t =

y = y =

+

a

v

d

+

v

d

EquationsEquations

The equations you will use are the same The equations you will use are the same ones.ones.

You will just be using the variables from You will just be using the variables from one dimension (x or y) at a time.one dimension (x or y) at a time.

210 0 2x x v t at

210 0 2x x v t at

2 20 02 ( ) fa x x v v 2 20 02 ( ) fa x x v v

0fv v at 0fv v at

TimeTime

In projectile motion, we In projectile motion, we will treat the x and y will treat the x and y variables as completely variables as completely different different setssets..

This means the givens This means the givens we use for one set we use for one set cannotcannot be used with the other!be used with the other!

However, However, TimeTime is a is a variable that is variable that is alwaysalways the the samesame for both x and for both x and y.y.

xx yy

vvoo = 25 m/s = 25 m/s

v = 25 m/sv = 25 m/s

a = 0 m/sa = 0 m/s22

t = 4.0 st = 4.0 s

x = 190 mx = 190 m

vvoo = 0 m/s = 0 m/s

v = 48 m/sv = 48 m/s

aayy = 10.0m/s = 10.0m/s22

t = 4.0 st = 4.0 s

y = 22 my = 22 m

Sample ProblemSample ProblemA cannon is placed at the edge of a cliff so the barrel is 60 meters above A cannon is placed at the edge of a cliff so the barrel is 60 meters above

the ground. The cannon fires a projectile horizontally with a velocity of the ground. The cannon fires a projectile horizontally with a velocity of 95 m/s. How far from the base of the cliff does the projectile land?95 m/s. How far from the base of the cliff does the projectile land?

As always, we will start by drawing a diagram of the situation. Remember to label any givens and include a direction vector for v, d, a, and +

y =60 m

x = ?? m

vy

Vx=95m/s

ay

+

+

After you have drawn your picture, fill in your Givens table.

xx yy

vvoo = =

v = v =

a = a =

t = t =

x = x =

vvoo = =

v = v =

aayy = =

t = t =

y = y =

95m/s

95m/s0m/s2

?

?

0m/s?

10.0m/s2

?

60 m

And of course our unknown variable is x. Always circle the unknown in the table.

dy=60 m

dx = ?? m

vy

Vx=95m/s

ay

+

+

xx yy

vvoo = =

v = v =

a = a =

t = t =

x = x =

vvoo = =

v = v =

aayy = =

t = t =

y = y =

Sample ProblemSample ProblemA cannon is placed at the edge of a cliff so the barrel is 60 meters above A cannon is placed at the edge of a cliff so the barrel is 60 meters above

the ground. The cannon fires a projectile horizontally with a velocity of the ground. The cannon fires a projectile horizontally with a velocity of 95 m/s. How far from the base of the cliff does the projectile land?95 m/s. How far from the base of the cliff does the projectile land?

?

95m/s

95m/s0m/s2

?

0m/s?

10.0m/s2

?

60 m

Now determine the equations you will need to solve the problem.Remember, we will keep the sets of variables completely separate (except time)!!!

x = vot + ½ at2

x = (95)(3.5) + ½ (0)(3.5)2

x = 332.5 m

y = vot + ½ at2

t = √(2y / a)t = √[(2)(60) / (10.0)] = 3.5 s

This is the only equation we can use to solve for x. But we are missing time! We need to use the variables in the y-dimension to find time first.3.5 s 3.5 s

Thelma and Louise drive their car off the edge of the grand Thelma and Louise drive their car off the edge of the grand canyon at a speed of 30 m/s and die 3.4 seconds later.canyon at a speed of 30 m/s and die 3.4 seconds later.(a) how tall is the canyon at that spot?(a) how tall is the canyon at that spot?Answer:Answer:57.8 m57.8 m(b) how far away from the bottom of the canyon do they (b) how far away from the bottom of the canyon do they land?land?Answer:Answer:102 m102 m

Practice ProblemPractice Problem

Which of the following does not Which of the following does not effect the hang time of a effect the hang time of a

projectile?projectile?

A.A. Angled firedAngled fired

B.B. Vertical velocityVertical velocity

C.C. Horizontal velocityHorizontal velocity

D.D. Height it was fired Height it was fired at.at.

Angled Projectile MotionAngled Projectile Motion

Occasionally, you will Occasionally, you will encounter problems encounter problems where the projectile is where the projectile is initially fired at an angle.initially fired at an angle.– This means that the initial This means that the initial

velocity will have a velocity will have a horizontalhorizontal andand a a verticalvertical component.component.

You will need to calculate You will need to calculate these components in these components in order to fill in your givens order to fill in your givens table on each problem.table on each problem.

Vector ComponentsVector Components

In order to find the In order to find the components of a components of a vector (like velocity) vector (like velocity) you will need to use you will need to use those timeless those timeless Trigonometric Trigonometric FunctionsFunctions..

Vector ComponentsVector Components

So we have a So we have a projectile that is fired projectile that is fired at an angle of 30at an angle of 30oo with respect to the with respect to the horizontal at a horizontal at a velocity of 50 m/s.velocity of 50 m/s.We need to think of We need to think of it like the projectile it like the projectile is fired vertically and is fired vertically and horizontally at the horizontally at the same time, giving it same time, giving it both components.both components.

Θ=30o

V=50m/svy

vx

Vector ComponentsVector Components

In order to calculate the In order to calculate the components, we need components, we need to shift vto shift vyy to make a to make a right triangle.right triangle.Then we can use trig Then we can use trig functions to solve for vfunctions to solve for vyy and vand vxx like they are like they are sides of a right triangle.sides of a right triangle.

To solve for vTo solve for vxx, we will , we will use cosine because it is use cosine because it is adjacent and we have adjacent and we have the hypotenuse.the hypotenuse.

Θ=30o

V=50m/svy

vx

Vector ComponentsVector ComponentsTo solve for vTo solve for vxx, we will use cosine , we will use cosine

because it is the adjacent side and because it is the adjacent side and we have the hypotenuse.we have the hypotenuse.

To solve for vTo solve for vyy, use the same , use the same

process but with sine.process but with sine.

Θ=30o

V=50m/s vy

vx

θVvV

vθ

hyp

oppθ

y

y

sin

sin

sin

cos

cos

cos

VvV

v

hyp

adj

x

x

Sample ProblemSample ProblemA soccer ball is kicked at an angle of 70A soccer ball is kicked at an angle of 70oo from the horizontal with a velocity of 20m/s. from the horizontal with a velocity of 20m/s.

What is the What is the rangerange of the soccer ball? (how far away does it land) of the soccer ball? (how far away does it land)

V=20m

/s

vx

vy

Θ=70o

+ xx yy

vvii = =

vvff = =

aaxx = =

t = t =

ddxx = =

vvii = =

vvff = =

aayy = =

t = t =

ddyy = =

+

a

dx

-10.0 m/s2

? ?

0 m?

0 m/s2

As always, start by drawing a diagram, including all vectors.

Begin to fill in what you can on your givens table

?

? ?

?

We need to use trig functions to solve for vix and viy

vix = VcosΘvix=(20)cos(70)

vix=6.84 m/s viy = VsinΘ

viy=(20)sin(70)

viy=18.8 m/s

+18.8m/s

6.84m/s

6.84m/s

-18.8m/sAnd as always circle your

unknown variable.

Sample ProblemSample ProblemA soccer ball is kicked at an angle of 70A soccer ball is kicked at an angle of 70oo from the horizontal with a velocity of from the horizontal with a velocity of

20.0m/s. What is the 20.0m/s. What is the rangerange of the soccer ball? (how far away does it land) of the soccer ball? (how far away does it land)

xx yy

vvii = 6.84 m/s = 6.84 m/s

vvff = 6.84 m/s = 6.84 m/s

aaxx = 0 m/s = 0 m/s22

t = ?t = ?

ddxx = ? = ?

vvii = +18.8 m/s = +18.8 m/s

vvff = -18.8 m/s = -18.8 m/s

aayy = -10.0m/s = -10.0m/s22

t = ?t = ?

ddyy = 0 m = 0 m

V=20m

/s

vx

vy

Θ=70o

+

+

a

dx

dx = vixt + ½ at2

dx = (6.84)(t) + ½ (0)t2

dx = (6.84)(3.76) + 0

dx = 25.7 m

vf = vi + att = (vf – vi) / at = (-18.8 – 18.8) / (-10.0)t = 3.76 s

A cannon shoots a cannon ball at 50° from the A cannon shoots a cannon ball at 50° from the horizontal at a velocity of 300 m/s. horizontal at a velocity of 300 m/s.

What is the range (dWhat is the range (dxx) of the cannon ball? ) of the cannon ball?

Answer:Answer:

~8,900 m~8,900 m

Practice ProblemPractice Problem

Projectile Motion FormulasProjectile Motion Formulas

advv if 222 cosvvix

atvv if

2

21 attvd i

sinvviy

2-D Motion Part II:2-D Motion Part II:Circular MotionCircular Motion

You spin me right round….You spin me right round….

Circular MotionCircular Motion

Circular motion is Circular motion is exactly what is sounds exactly what is sounds like: objects moving in like: objects moving in a a circular pathcircular path..We are going to We are going to investigate how investigate how objects rotate and objects rotate and revolve, and why they revolve, and why they do this…do this…

Rotation vs. RevolutionRotation vs. Revolution

Before we get started lets make a Before we get started lets make a quick differentiation.quick differentiation.When we say “When we say “RotationRotation,,” we will ” we will be talking about something be talking about something spinning.spinning.– The axis of rotation is The axis of rotation is insideinside the the

object.object.

When we say “When we say “RevolutionRevolution” we ” we will be talking about something will be talking about something going around something else.going around something else.– The axis of rotation is The axis of rotation is outsideoutside the the

object.object.

Angular VelocityAngular VelocityFor an object in circular For an object in circular motion, it makes sense to motion, it makes sense to describe its velocity in angular describe its velocity in angular terms. terms. Meaning: we will describe the Meaning: we will describe the velocity of an object in motion velocity of an object in motion by how many degrees (or by how many degrees (or radians) it covers in a certain radians) it covers in a certain amount of time.amount of time.The symbol for angular The symbol for angular velocity is a lower-case velocity is a lower-case omega: omega: ωωAlso, the units for Also, the units for ΘΘ must be in must be in radiansradians..

t

ΔΘω

The equation for angular velocity:The equation for angular velocity:

Radians?!?Radians?!?If you’re in Pre-Calculus (which If you’re in Pre-Calculus (which you should be) then you have you should be) then you have already been introduced to the already been introduced to the idea of radians. But here is the idea of radians. But here is the basic idea:basic idea:– One full revolution of a circle One full revolution of a circle

(360(360oo) is equal to 2) is equal to 2ππ Radians. Radians.– So if you have degrees and So if you have degrees and

want to convert to radians the want to convert to radians the equation is:equation is:

For any problem where you For any problem where you need to find the need to find the angular angular velocityvelocity, be sure that your units , be sure that your units are in are in radiansradians!!

360

2deg

rad

Sample ProblemSample Problem

A child on a merry-go-round completes A child on a merry-go-round completes one rotation in 60 seconds. What is one rotation in 60 seconds. What is the child’s angular velocity?the child’s angular velocity?

ωω = ∆ = ∆ΘΘ / t / t

ωω = (2 = (2ππ) / (60) ) / (60)

ωω = 0.10 rad/s = 0.10 rad/s∆∆ΘΘ is 2 is 2ππ because he because he completes completes one rotationone rotation

Tangential VelocityTangential Velocity

Suppose we have 2 horses on a Suppose we have 2 horses on a carosel. The black horse is 1 meter carosel. The black horse is 1 meter from the center and the white horse from the center and the white horse is 2 meters from the center.is 2 meters from the center.– Which horse has a greater Which horse has a greater angular angular

velocityvelocity??They have the same! They will each They have the same! They will each cover a full rotation (360cover a full rotation (360oo) in the same ) in the same amount of time.amount of time.

– Which horse “feels” like they are going Which horse “feels” like they are going “faster”?“faster”?

The white one!The white one!– The white horse is going faster The white horse is going faster

because it has a greater because it has a greater Tangential Tangential VelocityVelocity..

It covers a greater distance It covers a greater distance (circumference) in the same amount of (circumference) in the same amount of time.time.

Tangential VelocityTangential VelocityHopefully you learned in math Hopefully you learned in math what a “Tangent” line is.what a “Tangent” line is.– (Basically it’s a straight line that (Basically it’s a straight line that

touches one point on a curve.)touches one point on a curve.)

When we talk about the When we talk about the tangential velocity of an object tangential velocity of an object in motion we mean its actual in motion we mean its actual velocity (distance/time).velocity (distance/time).Tangential velocity can be Tangential velocity can be found using the angular found using the angular velocity and radius.velocity and radius.It can also be found if you It can also be found if you know the distance know the distance (circumference) traveled and (circumference) traveled and time.time.The effect of tangential velocity The effect of tangential velocity can hurt!can hurt!

rvT

Sample ProblemSample Problem

A student twirls a rock on a string above their head. The rock A student twirls a rock on a string above their head. The rock completes 2 revolutions in 4.50 seconds and the length of the string completes 2 revolutions in 4.50 seconds and the length of the string is 0.750 meters. What is the tangential velocity of the rock?is 0.750 meters. What is the tangential velocity of the rock?

As always, start by listing Givens and Unknowns:t = 4.50 s∆Θ = 2 revs = 4π radiansr = 0.750 meters.vT = ?

The equation we will need to use is:vT = rωBut there is a problem.. We don’t know ω!So we must use ω = ∆Θ/t

vT = rωvT = (0.75)(?)vT = (0.75)(2.79)vT = 2.09 m/s

ω = ∆Θ/tω = (4π)/(4.5)ω = 2.79 rad/s

Question…Question…

We know that it is possible for an We know that it is possible for an object to rotate in a circle with a object to rotate in a circle with a constant constant speedspeed, but can an object , but can an object rotate with a constant rotate with a constant velocityvelocity??– The answer is The answer is NONO! But why?! But why?– Because the Because the directiondirection is constantly is constantly

changingchanging. Remember: velocity is a . Remember: velocity is a vectorvector and has a magnitude and and has a magnitude and directiondirection!!

What is it called when something What is it called when something has a change in velocity?has a change in velocity?– Acceleration! In this case we refer to Acceleration! In this case we refer to

it as it as Centripetal AccelerationCentripetal Acceleration..

Centripetal AccelerationCentripetal AccelerationCentripetal Acceleration is what causes Centripetal Acceleration is what causes objects to travel in a circle.objects to travel in a circle.““Centripetal” means “center pointing”Centripetal” means “center pointing”As the object is moving, the As the object is moving, the acceleration pushes toward the axis of acceleration pushes toward the axis of rotation (center) and causes it to rotation (center) and causes it to change direction and travel in a circle.change direction and travel in a circle.The equation for Centripetal The equation for Centripetal Acceleration is:Acceleration is:

r

va Tc

2

Sample ProblemSample ProblemA car is doing donuts out in the parking lot and is A car is doing donuts out in the parking lot and is

traveling in a perfect circle with a radius of 15 traveling in a perfect circle with a radius of 15 meters at a constant speed of 18 m/s. What is meters at a constant speed of 18 m/s. What is the magnitude of the centripetal acceleration the magnitude of the centripetal acceleration acting on the car?acting on the car?

Givens:Givens:

vvTT = 18 m/s = 18 m/s

r = 15 mr = 15 m

aacc = ? = ?

aac c = v= vTT22/r/r

aacc = (18) = (18)22 / 15 / 15

aacc = 21.6 m/s = 21.6 m/s22

““g” Forcesg” Forces

““g” Forces are used to describe g” Forces are used to describe acceleration in terms of how it compares acceleration in terms of how it compares to acceleration due to gravity (g) which is to acceleration due to gravity (g) which is 10m/s10m/s22..

When undergoing centripetal acceleration, When undergoing centripetal acceleration, people can undergo many g’s!people can undergo many g’s!

UCM FormulasUCM Formulas

360

2deg

radt

ΔΘω

rvT r

va Tc

2