1

Test One:Review Topics

8.01W05D2

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Dimensions, Units, and Problem Solving Strategies

8.01W01D3

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Dimensions in Mechanics :Base Quantities

ampereIIElectric current

kelvinΘTThermodynamic Temperature

moleNnAmount of substance

candelaJIVLuminous intensity

M

T

L

Symbol for dimension

secondtTime

meterlLength

kilogrammMass

SI base unit

Symbol for quantity

Name

4

SI Base Units:Second: The second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom.

Meter: The meter (m) is now defined as the distance traveled by a light wave in vacuum in 1/299,792,458 seconds.

Mass: The SI standard of mass is a platinum-iridium cylinder assigned a mass of one 1 kg

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Speed of LightIn 1983 the General Conference on Weights and Measures defined the speed of light to be the best measured value at that time:

This had the effect that length became a derived quantity, but the meter was kept around for practicality

However, it is impossible to measure the speed of light these days!

c = 299,792, 458 meters/second

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Problem Solving Strategy: Dimensional Analysis

When trying to find a dimensional correct formula for a quantity from a set of given quantities, an answer that is dimensionally correct will scale properly and is generally off by a constant of order unity

Since:

[desired quantity] = Mα Lβ Tγ where α β, and γ are known

Combine the given quantities correctly so that:

[desired quantity] = Mα Lβ Tγ = (given1)X (given2)Y (given3)Z

- solve for X, Y, Z to match correct dimensions of desired quantity

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Problem Solving: Four Stages of Attack

1. Understand the Problem and Models2. Plan your Approach – Models and

Schema3. Execute your plan (does it work?)4. Review - does answer make sense?

- return to plan if necessary

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Understand: Get it in Your Head

What concepts are involved? Represent problem - Draw pictures, graphs, storylines…Similarity to previous problems?What known models/physical principles are involved?e.g. motion with constant acceleartion; two bodiesFind special features, constraintse.g. different acceleration before and after some instant in time

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Plan your Approach

Model: Real life contains great complexity, so in physics you actually solve a model problem that contains the essential elements of the real problem.

Build on familiar models

Lessons from previous similar problems

Select your system, Pick coordinates to your advantageAre there constraints, given conditions?

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Execute the PlanFrom general model to specific equations

Examine equations of the models, constraints

Is all/enough understanding embodied in Eqs.?

count equations and unknowns

simplest way through the algebra

Solve analytically (numbers later)

Keep notation simple (substitute later)

Keep track of where you are in your plan

Check that intermediate results make sense

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Stuck?Represent the Problem in New Way

– Graphical– Pictures with descriptions– Pure verbal– Equations

Could You Solve it if…– the problem were simplified?– you knew some other fact/relationship?– You could solve any part of problem, even a

simple one?

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Review

a. Does the solution make sense?Check units, special cases, scaling. Is your answer reasonably close to a simple estimation?

b. If it seems wrong, review the whole process.c. If it seems right,

review the pattern and models used,note the approximations,tricky/helpful math steps.

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Problem Solving Strategy: Estimation

Identify a set of quantities that can be estimated or calculated.

What type of quantity is being estimated?

How is that quantity related to other quantities, which can be estimated more accurately?

Establish an approximate or exact relationship between these quantities and the quantity to be estimated in the problem

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Kinematics and One Dimensional Motion

8.01W02D1

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Coordinate System in One Dimension

Used to describe the position of a point in space

A coordinate system consists of:

1. An origin at a particular point in space2. A set of coordinate axes with scales and labels3. Choice of positive direction for each axis: unit

vectors 4. Choice of type: Cartesian or Polar or Spherical

Example: Cartesian One-Dimensional Coordinate System

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One-Dimensional Kinematics: Summary

Position:

Displacement:

Average velocity:

Instantaneous Velocity

Average Acceleration

Instantaneous Acceleration

ˆ( ) ( )t x t=x i

ˆ( )x tΔ ≡ Δr i

ˆ ˆ( ) ( )xxt v tt

Δ≡ =

Δv i i

0( ) limx t

x dxv tt dtΔ →

Δ≡ ≡

Δ

ˆ ˆxx

v at

Δ≡ =

Δa i i

0ˆ ˆ( ) lim x x

t

v dvtt dtΔ →

Δ= ≡

Δa i i

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Instantaneous Velocityx-component of the velocity is equal to the slope of the tangent line of the graph of x-component of position vs. time at time t

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Instantaneous Acceleration

( ) ( )0 0 0

ˆ ˆ ˆ ˆ ˆ( ) ( ) lim lim lim x xx xx xt t t

v t t v tv dvt a t at t dtΔ → Δ → Δ →

+ Δ −Δ= ≡ = = ≡

Δ Δa i i i i i

The x-component of acceleration is equal to the slope of the tangent line of the graph of the x-component of the velocity vs. time at time t

ax (t) =dvx

dt

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Summary: Constant

AccelerationAcceleration

Velocity

Position

vx(t) = vx,0 + axt

20 ,0

1( )2x xx t x v t a t= + +

ax = constant

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Problem Solving Strategies: One-Dimensional Kinematics

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I. Understand – get a conceptual grasp of the problem

Question 1: How many objects are involved in the problem?

Question 2: How many different stages of motion occur?

Question 3: For each object, how many independent directions are needed to describe the motion of that object?

Question 4: What choice of coordinate system best suits the problem?

Question 5: What information can you infer from the problem?

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II. Devise a PlanSketch the problem

Choose a coordinate system

Write down the complete set of equations for the position and velocity functions; identify any specified quantities; clean up the equations.

Finding the “missing links”: count the number of independent equations and the number of unknowns.

You can solve a system of n independent equations if you have exactly n unknowns.

Look for constraint conditions

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III. SolveDesign a strategy for solving a system of equations.

Check your algebra and dimensions.

Substitute in numbers.

Check your results and units.

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IV. ReviewCheck your results, do they make sense (think!)

Check limits of an algebraic expression (be creative)

Think about how to extend model to cover more general cases (thinking outside the box)

Solved problems act as models for thinking about new problems. (Mechanics provides a foundation of of solved problems.)

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Non-Uniform Acceleration,Vectors,

Kinematics in Two-Dimensions

8.01 W02D2

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Summary: Time Dependent Acceleration

Acceleration is a non-constant function of time

Change in velocity

Change in position

,00

( ) ( )t t

x x xt

v t v a t dt′=

′=

′ ′− = ∫

( )xa t

00

( ) ( )t t

xt

x t x v t dt′=

′=

′ ′− = ∫

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VectorsA vector is a quantity that has both direction and magnitude. Let a vector be denoted by the symbol The magnitude of is denoted by

(1) Vectors can exist at any point Pin space.

(2) Vectors have direction and magnitude.

(3) Vector Equality: Any two vectors that have the same direction and magnitude are equal no matter where in space they are located.

AA

| A≡A |

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Vector Addition

A BLet and be two vectors. Define a new vector ,the “vector addition” of

and by the geometric construction shown in either figure

= +C A BA B

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Vector DecompositionChoose a coordinate system with an origin and axes. We can decompose a vector into component vectors along each coordinate axis, for example along the x,y, and z-axes of a Cartesian coordinate system. A vector at P can be decomposed into the vector sum,

x y z= + +A A A A

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Unit Vectors and ComponentsThe idea of multiplication by real numbers allows us to define a set of unit vectors

at each point in space with

Vector components:

( )ˆ ˆ ˆ, ,i j k

1, 1, 1ˆ ˆ ˆ| | | | | |= = =i j k

x x y y z zˆ ˆ ˆA , A , A= = =A i A j A k

x y zˆ ˆ ˆA A A= + +A i j k

( )x y zA , A , A=A

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Coordinate System

1. An origin as the reference point

2. A set of coordinate axes with scales and labels

3. Choice of positive direction for each axis

4. Choice of unit vectors at each point in space

Coordinate system: used to describe the position of a point in space and consists of

Cartesian Coordinate System

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Vector Decomposition in Two Dimensions

Consider a vector

x- and y components:

Magnitude:

Direction:

( 0)x yA , A ,=A

Ax = Acos(θ), Ay = Asin(θ)

A = Ax2 + Ay

2

Ay

Ax

=Asin(θ)Acos(θ)

= tan(θ) θ = tan−1( Ay / Ax )

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Vector Addition

Vector Sum:Components

A = Acos(θA ) i + Asin(θA ) j

B = Bcos(θB ) i + Bsin(θB ) j

= +C A B

Cx = Ax + Bx , Cy = Ay + By

Cx = C cos(θC ) = Acos(θA )+ Bcos(θB )Cy = C sin(θC ) = Asin(θA ) + Bsin(θB )

ˆ ˆ ˆ ˆ( ) ( ) cos( ) sin( )x x y y C CA B A B C Cθ θ= + + + = +C i j i j

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Vector Description of MotionPosition

Displacement

Velocity

Acceleration

ˆ ˆ( ) ( ) ( )t x t y t= +r i j

( ) ( )ˆ ˆ ˆ ˆ( ) ( ) ( )x ydx t dy tt v t v t

dt dt= + ≡ +v i j i j

( )( ) ˆ ˆ ˆ ˆ( ) ( ) ( )yxx y

dv tdv tt a t a tdt dt

= + ≡ +a i j i j

ˆ ˆ( ) ( ) ( )t x t y tΔ = Δ + Δr i j

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Constant Acceleration

0, 0, ,x x x y y yv v a t v v a t= + = +

x = x0 + v x 0t +

12

a xt 2 , y = y0 + v y 0t +12

a yt 2

2ax (x − x0 ) = vx2 − vx ,0

2

• Components of Velocity:

• Components of Position:

• Eliminating t:

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Two-Dimensional Motion

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Projectile Motion

totalintotal

in totalin

x x

y y

F m am

F m a

⎧ =⎪= ⇒ ⎨=⎪⎩

F a

A projectile is fired from a height y0 with an initial speed v0

at an angle θ above the horizontal. Ignore air resistance

Gravitational Force Law

Newton’s Second Lawgrav grav grav, grav

ˆym g F m g= − ⇒ = −F j

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Kinematic Equations

Acceleration:

Velocity:

Position:

ya g= −

20 ,0

1( )2yy t y v t gt= + −

,0( )y yv t v gt= −

0xa =

,0( )x xv t v=

0 ,0( ) xx t x v t= +

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Initial ConditionsInitial position:

Initial velocity:

Velocity components:

Initial speed:

Direction:

0 0 0ˆ ˆx y= +r i j

0 ,0 ,0ˆ ˆ( ) x yt v v= +v i j

,0 0 0 ,0 0 0cos , sinx yv v v vθ θ= =

( )1/ 22 20 0 ,0 ,0| | x yv v v= = +v

,010

,0

tan y

x

vv

θ − ⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠

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Concept of Force and Newton’s Laws of Motion

8.01 W03D1

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Concept of Force Force is a vector quantity

The magnitude of the total force is defined to beF = (mass) x (magnitude of the acceleration)

The direction of the total force on a body is the same as the direction of the acceleration.

The SI units for force are newtons (N):1 N = 1kg . m/s2

total m≡F a

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ForcesGravitation

Electric and magnetic forces

Elastic forces (Hooke’s Law)

Frictional forces: static and kinetic friction, fluid resistance

Contact forces: normal forces and static friction

Tension and compression

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Superposition Principle Apply two forces and on a body,

the total force is the vector sum of the two forces:

Notation: The force acting on body 1 due to the interaction between body 1 and body 2 is denoted by

1F 2F

total1 2= +F F F

3 31 32total = +F F F

Example: The total force exerted on m3 by m1 and m2 is:

12F

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Free Body Diagram Represent each force that is acting on the object by an arrow on a free body

force diagram that indicates the direction of the force

Choose set of independent unit vectors and draw them on free body diagram.

Decompose each force in terms of vector components.

Add vector components to find vector decomposition of the total force

iF1 2

T = + +⋅⋅⋅F F F

1, 2,T T T

x x xF F F= + + ⋅⋅⋅

, , ,ˆ ˆ ˆ

i i x i y i zF F F= + +F i j k

1, 2,T T T

y y yF F F= + + ⋅⋅⋅

1, 2,T T T

z z zF F F= + + ⋅⋅⋅

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Newton’s First Law

Every body continues in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed upon it.

Use coordinate system as a ‘reference frame’to describe the position, velocity, and acceleration of objects.

constanti = ⇒ =∑F 0 v

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Relatively Inertial Reference Frames

Two reference frames.

Origins need not coincide.

One moving object has differentposition vectors in different frames

Relative velocity between the two reference frames

is constant since the relative acceleration is zero

1 2= +r R r

d dt=V R

d dt= =A V 0

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Law of Addition of VelocitiesSuppose the object is moving; then, observers in

different reference frames will measure different velocities

Velocity of the object in Frame 1:

Velocity of the object in Frame 2:

Velocity of an object in two different reference frames

1 1d dt=v r

2 2d dt=v r

1 2ddt dt dt

= +r rR

1 2= +v V v

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Newton’s Second LawThe change of motion is proportional to the motive

force impresses, and is made in the direction of the right line in which that force is impressed,

, , ,1 1 1

, , . N N N

x i x y i y z i zi i i

F ma F ma F ma= = =

= = =∑ ∑ ∑

When multiple forces are acting,

In Cartesian coordinates:

1.

N

ii

m=

=∑F a

.m=F a

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Newton’s Third Law

To every action there is always opposed an equal reaction: or, the mutual action of two bodies upon each other are always equal, and directed to contrary parts.

Action-reaction pair of forces cannot act on same body; they act on different bodies.

1,2 2,1= −F F

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Force Law: Newtonian Induction

Definition of force has no predictive content.

Need to measure the acceleration and the mass in order to define the force.

Force Law: Discover experimental relation between force exerted on object and change in properties of object.

Induction: Extend force law from finite measurements to all cases within some range creating a model.

Second Law can now be used to predict motion!

If prediction disagrees with measurement adjust model.

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Force Law: Gravitational Force near the Surface of the EarthNear the surface of the earth, the gravitational interaction between a body and the earth is mutually attractive and has a magnitude of

where is the gravitational mass of the body and g is a positive constant.

grav gravm g=F

mgrav

g = 9.80665 m ⋅ s−2

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Empirical Force Law: Hooke’sLaw

Consider a mass m attached to a spring

Stretch or compress spring by different amounts produces different accelerations

Hooke’s law:

Direction: restoring spring to equilibrium

Hooke’s law holds within some reasonable range of extension or compression

| | k l= ΔF

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Force Laws: Contact Forces Between Surfaces

The contact force between two surfaces is denoted by the vector

Normal Force: Component of the contact force perpendicular to surface and is denoted by

Friction Force: Component of the contact force tangent to the surface and is denoted by

Therefore the contact force can be modeled as a vector sum

,normalsurface hand ≡F N

,tangentsurface hand ≡F f ≡ +C N f

,totalsurface hand ≡F C

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Kinetic FrictionThe kinetic frictional force fk is proportional to the normal force, but independent of surface area of contact and the velocity.

The magnitude of fk is

where µk is the coefficients of friction.

Direction of fk: opposes motion

fk = μk N

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Static FrictionVaries in direction and magnitude depending on applied forces:

Static friction is equal to it’s maximum value

0 ≤ fs ≤ fs,max = μsN

fs,max = μs N

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Tension in a Rope

The tension in a rope at a distance x from one end of the rope is the magnitude of the action-reaction pair of forces acting at that point ,

left,right right,left( ) ( ) ( )T x x x= =F F

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Application of Newton’s Second Law

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Concept of System: ReductionModeling complicated interaction of objects by isolated a

subset (possible one object) of the objects as the system

Treat each object in the system as a point-like object

Identify all forces that act on that object

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Model: Newton’s Laws of MotionSystem: Point mass with applied force

Description of System: – Objects: Point Mass– State Variables: m, a(t), r(t)– Agents: real forces on object

Multiple Representations;– Words, Force Diagrams, Equations

Interactions:– Force Laws: contact, spring, universal gravity,

uniform gravity, drag.

Law of Motion:∑F = ma– Origin and Type of forces, Vectors

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Methodology for Newton’s 2nd LawUnderstand – get a conceptual grasp of the problem

Sketch the system at some time when the system is in motion.

Draw free body diagrams for each body or composite bodies:

Each force is represented by an arrow indicating the direction of the force

Choose an appropriate symbol for the force

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II. Devise a Plan Choose a coordinate system:

Identify the position function of all objects and unit vectors.

Include the set of unit vectors on free body force diagram.

Apply vector decomposition to each force in the free body diagram:

Apply superposition principle to find total force in each direction:

ˆ ˆ ˆ( ) ( ) ( )i x i y i z iF F F= + +F i j k

( ) ( )( ) ( )( ) ( )

total1 2

total

1 2

total1 2

ˆ :ˆ :

ˆ :

x x x

y y y

z z z

F F F

F F F

F F F

= + +

= + +

= + +

i

j

k

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II. Devise a Plan: Equations of Motion

Application of Newton’s Second Law

This is a vector equality; the two sides are equal in magnitude and direction.

total1 2 .m= + + ⋅⋅⋅ =F F F a

( ) ( )( ) ( )( ) ( )

1 2

1 2

1 2

ˆ :ˆ :

ˆ :

x x x

y y y

z z z

F F ma

F F ma

F F ma

+ + =

+ + =

+ + =

i

j

k

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II. Devise a Plan (cont’d)

Analyze whether you can solve the system of equations

Common problems and missing conditions.

Constraint conditions between the components of the acceleration.

Action-reaction pairs.

Different bodies are not distinguished.

Design a strategy for solving the system of equations.

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III. Carry Out your Plan

Hints:

Use all your equations. Avoid thinking that one equation alone will contain your answer!

Solve your equations for the components of the individual forces.

65

IV. Look BackCheck your algebra

Substitute in numbers

Check your result

Think about the result: Solved problems become models for thinking about new problems.

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Worked Example: Pulley and Inclined Plane

A block of mass m1, constrained to move along a plane inclined at angle ϕ to the horizontal, is connected via a massless inextensible rope that passes over a massless pulley to a second block of mass m2. Assume the block is sliding up the inclined plane. The coefficient of kinetic friction is μk. Assume the gravitational constant is g. Calculate the acceleration of the blocks.

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Solution: Pulley and Inclined Plane

Coordinate system

Free body force diagrams

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Solution: Pulley and Inclined Plane

Constraint:Object on inclined plane:

Suspended Object: Solution: T − m1g sinφ − μkm1g cosφ = m1a

1 1 k 1ˆ : sinT m g f m aφ− − =i

1 1ˆ : cos 0N m g φ− =j

fk = μk N = μkm1g cosφ

2 2 2ˆ : m g T m a− =j

a ≡ ay ,2 = ax ,1

T = m2g − m2a

m2g − m2a − m1g sinφ − μkm1g cosφ = m1a

a =

(m2 − m1(sinφ + μk cosφ))m1 + m2

g

69

Circular Motion

70

Position and Displacement: position vector of an object moving in a

circular orbit of radius R

: change in position between time t and time t + ∆t

Position vector is changing in direction not in magnitude.

The magnitude of the displacement is the length of the chord of the circle:

( )tr

( )tΔr

2 sin( / 2)R θΔ = Δr

( ) ( )t t tΔ = + Δ −r r r

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Direction of Velocity

Sequence of chord directions approach direction of velocity as ∆tapproaches zero.

The direction of velocity is perpendicular to the direction of the position and tangent to the circular orbit.

Direction of velocity is constantly changing.

rΔ

72

Small Angle ApproximationWhen the angle is small:

Power series expansion

Using the small angle approximation with , the magnitude of the displacement is

sinφ ≈ φ, cosφ ≈ 1

sinφ = φ −φ3

3!+

φ5

5!− ...

cosφ = 1−φ 2

2!+

φ 4

4!− ...

2 sin( / 2)R Rθ θΔ = Δ ≈ Δr

φ = Δθ / 2

73

Speed and Angular SpeedThe speed of the object undergoing circular motion is proportional to the rate of change of the angle with time:

Angular speed:

0 0 0lim lim lim

t t t

R dv R R Rt t t dt

θ θ θ ωΔ → Δ → Δ →

Δ Δ Δ≡ = = = = =

Δ Δ Δr

v

ω =dθdt

(units: rad ⋅ s-1)

74

Circular Motion: Constant Speed, Period, and Frequency

In one period the object travels a distance equal to the circumference:

Period: the amount of time to complete one circular orbit of radius R

Frequency is the inverse of the period:

s = 2π R = vT

2 2 2R RTv Rπ π π

ω ω= = =

11 (units: s or Hz)2

fT

ωπ

−= =

75

Acceleration and Circular Motion

When an object moves in a circular orbit, the direction of the velocity changes and the speed may change as well.

For circular motion, the acceleration will always have a radial component (ar) due to the change in direction of velocity

The acceleration may have a tangential component if the speed changes (at). When at =0, the speed of the object remains constant

76

Direction of Radial Acceleration: Uniform Circular

MotionSequence of chord directions approaches direction of radial acceleration as Δt approaches zero

Perpendicular to the velocity vector

Points radially inward

Δv

77

Change in Magnitude of Velocity:Uniform Circular MotionChange in velocity:

Magnitude of change in velocity:

Using small angle approximation

( ) ( )t t tΔ = + Δ −v v v

2 sin ( / 2)v θΔ = Δv

v θΔ ≅ Δv

78

Radial Acceleration: Constant Speed Circular Motion Any object traveling in a circular orbit with a

constant speed is always accelerating towards the center.

Direction of velocity is constantly changing.

Radial component of ar (minus sign indicates direction of acceleration points towards center)

ar = − lim

Δt→0

ΔvΔt

= − limΔt→0

vΔθΔt

= −v limΔt→0

ΔθΔt

= −vdθdt

= −vω = −v2

R

ar = −v2

R= − ω 2 R

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Tangential AccelerationThe tangential acceleration is the rate of change of the magnitude of the velocity

Angular acceleration: rate of change of angular velocity with time

2

20 0lim limt t t

v d da R R R Rt t dt dt

ω ω θ αΔ → Δ →

Δ Δ= = = = =

Δ Δ

α =

dωdt

=d 2θdt2 (units: rad ⋅ s-2 )

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Alternative forms of Magnitude of Radial Acceleration

Parameters: speed v, angular speed ω, angular frequency f, period T

2 22 2

2

4(2 )rv Ra R R fR T

πω π= = = =

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Summary: Kinematics of Circular Motion

Arc length

Tangential velocity

Tangential acceleration

Centripetal2

2r

va v RR

ω ω= = =

2

2tdv da R Rdt dt

θ α= = =

v =dsdt

= Rdθdt

= Rω

s = Rθ

82

Cylindrical Coordinates: Circular Motion

Coordinates

Unit vectors

Position

Velocity

Acceleration

ˆ( ) ( )t r t=r r

(r,θ , z)

ˆˆ ˆ( , , )θr z

ˆ ˆ( ) ( ) ( )dt r t r tdtθ ω= =v θ θ

ˆ( ) ( )t r t=r r

ˆˆr ta a= +a r θ2 2( / ),r ta r v r a rω α= − = − =

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Modeling Problems: Circular Motion

Always has a component of acceleration pointing radiallyinward

May or may not have tangential component of acceleration

Draw Free Body Diagram for all forces

mv2/r is not a force but mass times acceleration and does not appear on force diagram

Choose a sign convention for radial direction and check that signs for forces and acceleration are consistent

84

Newton’s Second Law:Equations of Motion for Circular

Motion

85

Table Problem: Lecture Demo Centripetal Force

A wheel is connected via a pulley to a motor. A thread is knotted and placed through a hole in a ping pong ball of mass m. A wing nut secures the thread holding the ball a distance r from the center of the wheel. The wheel is set in motion. When a satisfactory angular speed w is reached, the string is cut and the ball comes off at a tangent to the spinning wheel, traveling vertically upward with a vertical displacement ∆y = h. Find the angular velocity of the wheel and the tension in the string just before it was cut.

86

Table Problem: Experiment 2One end of a spring is attached to the central axis of a motor. The axis of the motor is in the vertical direction. A small ball of mass m2 is then attached to the other end of the spring. The motor rotates at a constant frequency f . Neglect the gravitational force exerted on the ball. Assume that the ball and spring rotate in a horizontal plane. The spring constant is k. Let r0 denote the unstretched length of the spring.

(i) How long does it take the ball to complete one rotation?(ii) What is the angular frequency of the ball in radians per sec?(iii) What is the radius of the circular motion of the ball?