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Concept of Force and Newton’s

Laws

8.01 W01D2

Teaching Staff Introductions

Instructor: Peter Dourmashkin Graduate Teaching Assistant: Undergraduate Teaching Assistants: Technical Instructor:

Things to Do For Week One Complete Registration Assignment on MITx: 8.01 Website https://lms.mitx.mit.edu/courses/MITx/8.01/2015_Fall/about Buy at MIT Coop or Download Textbook from Website Buy Clicker at MIT Coop, bring clicker to class, and register at https://lms.mitx.mit.edu/courses/MITx/8.01/2015_Fall/

courseware/Intro/about:Clickers/ Complete Reading Assignment for Week One. Complete Prepset 1 and submit online before due date Friday Sept 11 Friday 8:30 am

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Introduction to 8.01

Force

Circular Motion

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7

Momentum  

8

Impulse  

Energy

http://sohowww.nascom.nasa.gov/data/LATEST/current_eit_284.gif

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10

Collision  Theory  

Torque

Angular Momentum

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Central Force Motion

Stars orbiting black hole at center of Milky Way galaxy http://www.galacticcenter.astro.ucla.edu/videos/ghezGC_comp3-18_H264_864_VP8.webm

8.01 Differs from High School Physics

Demonstration: Spinning Bicycle Wheel

How Mechanics Fits in With the Rest of Physics

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Newton’s Laws enable us to explain: Conservation laws: momentum, energy, and angular momentum Translation and rotation of rigid bodies Simple harmonic motion Gyroscopic motion Planetary motion

Newtonian Mechanics:

Newtonian Mechanics

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

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Newton’s Second Law

Non-Isolated Object: Force changes motion The 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. This is called the equation of motion.

F = m a = dp

dt

Interaction Pair

We shall refer to objects that interact as an interaction pair. Notation: Denote as the force on object 2 due to the interaction between objects 1 and 2. Similarly, denote as the force on object 1 due to the interaction between objects 1 and 2.

F1,2

F2,1

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.

F2,1 = −

F1,2

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Concept Question: Car-Earth Interaction

Consider a car at rest. We can conclude that the downward gravitational pull of Earth on the car and the upward contact force of Earth on it are equal and opposite because

1.  the two forces form a third law interaction pair. 2.  the net force on the car is zero. 3.  neither of the above.

Concept Question: Interaction Pair

A large truck collides head-on with a small car. During the collision a)  the truck exerts a greater force on the car than the car

exerts on the truck.

b) the car exerts a greater force on the truck than the truck exerts on the car. c) the truck exerts the same force on the car as the car exerts on the truck. d) the truck exerts a force on the car but the car does not exert a force on the truck.

Group Activity: String Theory

1.  At each table, form pairs, and each pair pull on the opposite ends of the provided very light string.

2.  Draw three force diagrams on the board, one for person A, one for person B, and one for the rope.

3.  On each diagram draw all the forces acting on the object in your diagram.

4.  Identify the action-reaction pairs of forces.

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Tug of War Force Diagrams

For the rope we have neglected the gravitational force because we are assuming the rope is very light

Kinematics and One-Dimensional Motion

Kinema means movement in Greek Mathematical description of motion

1)  Position 2)  Displacement 3)  Velocity 4)  Acceleration

Coordinate System Used to describe the position of a point in space and vectors at any point

A coordinate system consists of:

1.  An origin at a particular point in space 2.  A set of coordinate axes with scales and labels 3.  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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Position and Displacement Position vector points from origin to body.

x(t) is called the coordinate position function Change in position vector of the object during the time interval Displacement vector

ˆ( ) ( )t x t=x i

0 +x

ix(t)

Δt = t2 − t1

Δr ≡ [x(t2 )− x(t1)] i

≡ Δx(t)i

0 +x

ix(t)

x(t + t)

x

Concept Question: Displacement An object goes from one point in space to another. After the object arrives at its destination, the magnitude of its displacement is:

1) either greater than or equal to the distance traveled. 2) always greater than the distance traveled.

3) always equal to the distance traveled.

4) either smaller than or equal to the distance traveled.

5) always smaller than the distance traveled.

6) either smaller or larger than the distance traveled.

Average Velocity

The average velocity, , is the displacement divided by the time interval The x-component of the average velocity is given by

vave(t)

vave ≡

ΔrΔt

=ΔxΔt

i = vave,x (t)i

vave,x =

ΔxΔt

Δr

Δt

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Instantaneous Velocity and Differentiation

For each time interval , calculate the x-component of the average velocity Take limit as generates a sequence x-components of average velocity The limiting value of this sequence is x-component of the instantaneous velocity at time t.

tΔ

0tΔ → vave,x (t) = Δx / Δt

limΔt→0

ΔxΔt

= limΔt→0

x(t + Δt) − x(t)Δt

≡dxdt

vx (t) = dx / dt

Instantaneous Velocity

x-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

vx (t) =

dxdt

Worked Example: Differentiation

x(t) = At 2

x(t + Δt) = A(t + Δt)2 = At 2 + 2AtΔt + AΔt 2

x(t + Δt)− x(t)Δt

= 2At + AΔt

dxdt

= limΔt→0

x(t + Δt)− x(t)Δt

= limΔt→0

(2At + AΔt) = 2At

Generalization for Polynomials:

x(t) = Atn

dxdt

= nAtn−1

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Concept Q.: One-Dimensional. The graph shows the position as a function of time for two trains running on parallel tracks. For times greater than t = 0, which of the following is true: 1.  At time tB, both trains have the

same velocity.

2.  Both trains speed up all the time.

3.  Both trains have the same velocity at some time before tB. 4.  Somewhere on the graph, both

trains have the same acceleration.

Average Acceleration

Change in instantaneous velocity divided by the time interval

The x-component of the average acceleration aave ≡

ΔvΔt

=Δvx

Δti =

(vx ,2 − vx ,1)Δt

i =Δvx

Δti = aave,x i

aave,x =

Δvx

Δt

Δt = t2 − t1

Instantaneous Acceleration and Differentiation

For each time interval , calculate the x-component of the average acceleration Take limit as sequence of the x-component average accelerations The limiting value of this sequence is x-component of the instantaneous acceleration at the time t.

Δt → 0 aave,x (t) = Δvx / Δt

limΔt→0

Δvx

Δt= lim

Δt→0

vx (t + Δt) − vx (t)Δt

≡dvx

dt

ax (t) = dvx / dt

Δt

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Instantaneous Acceleration

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

Group Problem: Model Rocket

A person launches a home-built model rocket straight up into the air at y = 0 from rest at time t = 0 . (The positive y-direction is upwards). The fuel burns out at t = t0. The position of the rocket is given by

with a0 and g are positive. Find the y-components of the velocity and acceleration of the rocket as a function of time. Graph ay vs t for 0 < t < t0.

y = 1

2(a0 − g)t2 −

a0

30t6 / t0

4; 0 < t < t0

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