Concept of Force and Newton’s

Laws

8.01 W01D2 and W01D3

Getting Started

Please go to https://lms.mitx.mit.edu/ and find Classical Mechanics MIT-8.01. Open up our course website.

Teaching Team Introductions

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

Assignments and Due Dates Complete Reading Assignment for Week One. Complete Online Prepset 1 and submit answers online before due date Week 1 Friday at 8:30 am. Complete Online Prepset 2 and submit answers online before due date Week 2 Monday at 8:30 am. Problem Set 1 due Week 2 Tuesday at 9 pm. Put your problem set solutions in your section box outside room 26-152. Please write your name. section, and table number on your problem set or you may not get credit

Group Learning

Students will form groups of three and work together on problem solving and experiments throughout the semester. We encourage you to form your own groups. On Friday sit with your group, ask your TA which group you are in, and note your section and Table number.

Ground Rules for 8.01

Read Course Information document (sent via email and online on the 8.01 MITx Welcome page where you can find information about: Course websites Online textbook and reading assignments Prep-set and problem sets due dates Collaboration policy Sunday tutoring and office hours In-class concept questions Grading policy Exam dates

Exam Dates Exam 1: Thursday Sept 26, 7:30 pm - 9:30 pm Conflict Exam 1: Friday Sept 27, 8-10 am, 9-11 am

Exam 2: Thursday Nov 7, 7:30 pm - 9:30 pm Conflict Exam 2: Fri Nov 8, 8-10 am, 9-11 am

In-Class Quiz 1: Friday Nov 22 Final Exam: Date/time to be announced by Registrar

Things to Do For Week One

Complete Registration Assignment on MITx: 8.01 Website. By going to https://lms.mitx.mit.edu/ and finding Classical Mechanics MIT-8.01. Find links to Textbook from MITx: 8.01 Website

How Mechanics Fits in With the Rest of Physics

Important Recent Discoveries in Physics Bose-Einstein Condensate (1995) Accelerating expansion of the universe (1997) (Dark Energy?)

Experimental proof that neutrinos have mass (1998) Extrasolar planets Discovery of the Higgs Boson (2012) Topological Insulators (material whose surface contains conducting states but interior is an insulator) with promising applications in spintronic devices and quantum computers) LIGO detection of colliding black holes (2016) and colliding neutron stars (2017)

Colliding Neutron Stars

This neutron star collision produced around 200 Earth masses of pure gold, and maybe 500 Earth masses of platinum.

Simulation: Jet and debris from neutron star collisionhttps://youtu.be/e7LcmWiclOs

The explosive impact created ripples in space-time that reached Earth at 8:41 am ET Aug 17, 2017 and were detected by LIGO

A Few Open Problems in Physics

Dark Matter/ Dark Energy Quantum Gravity Room Temperature Superconductors Extrasolar life What is time? Many more open problems (Spacetime = entanglement; Is the vacuum topologically ordered?; Black hole information/firewalls

We will use Newton’s Laws to explain: Conservation laws: momentum, energy, and angular momentum Non-relativistic collision of particles Translation and rotation of rigid bodies Simple harmonic motion Gyroscopic motion Planetary motion Motion in non-inertial reference frames

Newtonian Mechanics:

8.01 Differs from High School Physics

No memorization of formulas Apply fundamental principles to analyze unfamiliar situations Develop expert problem solving skills starting from a conceptual understanding of laws mechanics, using a set of “tools” to apply the laws of physics, and finally a sophisticated use of mathematics when solving equations of motion.

Demonstration:

Spinning Bicycle Wheel (gyroscopic motion)

Newton’s First Law

If an isolated body is at rest; it remains at rest (v = 0) as long as no force acts on it. If an isolated body is moving at constant velocity then it continues to move at that velocity as long as no force acts on it.

Newton’s Second Law

Non-Isolated object: Force changes motion. For a point-like object This is called the equation of motion. For a system of particles we shall see that only external forces change the momentum of the system

F = m a

Fext =

dpsys

dt

Interaction Pairs

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. Forces always come in interaction pairs.

!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

Group Activity:

Pulling springs Pulling ropes

A B

Group Problem: Pulling a Rope

Activity

Activity: Two people pull on opposite ends of a rope. Each Group use the White Boards to answer the following questions: a)  identify the forces acting on each person and the rope.

Draw three separate force diagrams: one for each person and one for the rope. Represent the forcers by arrow for the direction and a symbol for magnitude

b)  For each force on your free-body diagrams, describe the action-reaction force associated with it.

A B

Pulling a Rope Force Diagrams

FE,Rg mR g

FA,R FB,R

Interaction Pairs: and !FA,R = −

!FR,A

AFR,A

FE,Ag mA gNG,A

b

fG,Ab NG,A

ffG,Af

B

NG,b

fG,Bb

FR,BFE,B mB g

NG,Bf

fG,Bf

g

!FB,R = −

!FR,B

CQ: Car-Truck Collision

A large truck collides head-on with a small car. During the collision 1. the truck exerts a greater force on the car than the car exerts on the truck.

2.  the car exerts a greater force on the truck than the truck exerts on the car.

3.  the truck exerts the same force on the car as the car exerts on the truck. 4. the truck exerts a force on the car but the car does not exert a force on the truck.

CQ: 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.  Both of the above are true. 4.  None of the above are true.

N

mg

Newton’s Second Law: Physics and Mathematics

F = m a

physics ⇔ geometry

cause of motion (why)

⇔description of motion (how)

dynamics ⇔ kinematics

Kinematics and One-Dimensional Motion

Kinema means movement in Greek Mathematical description of motion

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

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

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

Inertial Reference Frames: Newton’s Laws

“It is always possible to find a coordinate system with respect to which isolated bodies (no forces acting on them) move uniformly.” 1 Such a coordinate system is called an inertial frame. Newton’s First Law is part definition, in that in an inertial reference frame isolated bodies by definition move uniformly, and part a statement about the physical world, in that inertial frames exist. Newton’s Second Law only holds in inertial reference frames. Newton’s Third tests whether or not a system is isolated. 1 Kleppner and Kolenkow, An Introduction to Mechanics, p. 56-7.

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

r(t) = x(t)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

CQ: 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

Instantaneous Velocity and Differentiation

For each time interval , calculate the x-component of the average velocity Take limit as generates a sequence of 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

Differentiation Rule for Polynomials

Rule for Polynomials and inverse powers:

x(t) = Atn; n ≠ 0dxdt

= nAtn−1

CQ: One-Dim. Kinematics The graph shows the position as a function of time for two trains A and B running on parallel tracks. For times greater than t = 0, which of the following is true: 1.  At time t2, both trains have the same velocity.

2.  Both trains have the same velocity at some time before t2, .

3.  The velocity of train A is always greater than the velocity of train B for the interval 0 < t < t2

t

train A train B

t = t2

x(t)

Group Problem: Street Scene

Two bicycle riders A and B are a distance d0 apart at time t = 0 and riding towards each other with constant speeds. Rider A has speed vA and rider B has speed vB. At time t = 0, a person is midway between the two riders and walking towards rider A with speed vP. i) At what time t1 does rider A just pass the person? ii) How far is the person from rider B at time t1?

A

vA vBvP

B

Kinematics:

Acceleration and Integration

Instantaneous Acceleration

Average acceleration for time interval Δt: Instantaneous acceleration: limit of a sequence of average accelerations

aave,x (t) = Δvx / Δt

0 +x

ix(t)

ax (t) = lim

Δt→0

Δvx

Δt= lim

Δt→0

vx (t + Δt)− vx (t)Δt

≡dvx

dt

Concept Q.: Acceleration The graph shows the velocity as a function of time for two trains A and B running on parallel tracks. For times greater than t = 0, which of the following is true: 1.  At time t2, both trains have the

same acceleration

2.  Both trains have the same acceleration at some time t1 before t2 , 0< t1 < t2 .

3.  Both trains have the same

velocity at some time t1 before t2 , 0< t1 < t2 .

vx (t)

t

train A

train B

t = t2

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

where a0 and g are positive constants. 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

Change in Velocity: Integral of Acceleration

From calculus The differential is given by

Therefore the change in the x-component of the velocity is the integral of the x-component acceleration Integration is the inverse operation of differentiation

vx (t1)− vx (t0 ) = ax ( ′t )d ′t

′t =t0

′t =t1

∫

vx (t1)− vx (t0 ) = dvx

′t =t0

′t =t1

∫

dvx = axdt

ax (t)

t0+ t

t1

Velocity as a Function of Time

Suppose instead of some difference in velocity we would like to find the velocity as a function of time. Then let t 1= t

vx (t)− vx (t0 ) = ax ( ′t )d ′t

′t =t0

′t =t

∫

ax (t)

t0+ t

t1

Change in Position: Integral of Velocity The integral of the x-component of the velocity vs. time is the displacement

x(t)− x(t0 ) = vx ( ′t )d ′t

′t =t0

′t =t

∫

ax (t)

t0+ t

t1

t0+ t

t1 = t

vx (t)

Integral Formula for Poynomials

′t n d ′t′t =t0

′t =t

∫ = ′t n+1

n +1 ′t =t0

′t =t

= t n+1

n +1− t0

n+1

n +1; n ≠ −1

Worked Ex.: Time-Dependent Acceleration

Acceleration is a non-constant function of time with , , and . Change in velocity: Velocity function: Change in position: Position function: :

vx (t)− v0 = A ′t 2 d ′t

′t =0

′t =t

∫ = A′t 3

3′t =0

′t =t

= At3

3

x(t)− x0 = (v0 +

A ′t 3

3)d ′t ⇒

′t =0

′t =t

∫ x(t) = v0t + A′t 4

12⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎞

⎠⎟0

′t =t

= v0t +A

12t4

ax (t) = At2

v0 ≠ 0 t0 = 0 x0 ≠ 0

x(t) = x0 + v0t +

A12

t4

vx (t) = v0 +

At3

3

Group Problem: Sports Car At t = 0 , a sports car starting with an initial speed v0 moving in the positive x direction at x = 0 accelerates with an acceleration given by

with positive constants A and B . Find expressions for the velocity and position of the sports car as functions of time for

a(t) = At − Bt 3, for 0 < t < (A / B)1/2

0 < t < (A / B)1/2

Special Case: Constant Acceleration

Acceleration: Velocity: Position:

vx (t) = vx ,0 + axt

a = constant

vx (t)− vx ,0 = ax d ′t′t =0

′t =t

∫ ⇒

vx (t) = vx ,0 + axt

x(t)− x0 = (vx ,0 + ax ′t )d ′t′t =0

′t =t

∫ ⇒

x(t) = x0 + vx ,0t +12

axt2