physics 207: lecture 4, pg 1 physics 207, lecture 4, sept. 15 l goals for chapts. 3 & 4 perform...

Physics 207: Lecture 4, Pg 1

Physics 207, Physics 207, Lecture 4, Sept. 15Lecture 4, Sept. 15 Goals for Chapts. 3 & 4Goals for Chapts. 3 & 4

Perform vector algebraPerform vector algebra (addition & subtraction) graphically or by addition & subtraction) graphically or by x,y & z components x,y & z components

Interconvert between Interconvert between CartesianCartesian and and Polar Polar coordinatescoordinates

Distinguish Distinguish position-timeposition-time graphs from graphs from particle trajectoryparticle trajectory plots plots

Obtain particle velocities and accelerations from trajectory plotsObtain particle velocities and accelerations from trajectory plots Determine both Determine both magnitudemagnitude and and accelerationacceleration parallel and parallel and perpendicular to the trajectory path perpendicular to the trajectory path

Solve problems with multiple accelerations in 2-dimensions Solve problems with multiple accelerations in 2-dimensions (including linear, projectile and circular motion)(including linear, projectile and circular motion)

Discern different reference frames and understand how they relate Discern different reference frames and understand how they relate to particle motion in stationary and moving framesto particle motion in stationary and moving frames

Physics 207: Lecture 4, Pg 2

Physics 207, Physics 207, Lecture 4, Sept. 15Lecture 4, Sept. 15

Assignment: Read Chapter 5 (sections 1- 4 Assignment: Read Chapter 5 (sections 1- 4 carefully)carefully)

MP Problem Set 2 MP Problem Set 2 due Wednesdaydue Wednesday (should have (should have started)started)

MP Problem Set 3, Chapters 4 and 5 (available MP Problem Set 3, Chapters 4 and 5 (available today)today)

Physics 207: Lecture 4, Pg 3

Vector additionVector addition

The sum of two vectors is another vector.

A = B + CB

C A

B

C

Physics 207: Lecture 4, Pg 4

Vector subtractionVector subtraction

Vector subtraction can be defined in terms of addition.

B - C

B

C

= B + (-1)C

B

-C

B - C

A Different direction and magnitude !A = B + C

Physics 207: Lecture 4, Pg 5

Unit VectorsUnit Vectors

A Unit Vector Unit Vector is a vector having length 1 and no units

It is used to specify a direction. Unit vector uu points in the direction of

UUOften denoted with a “hat”: uu = û

U = |U| U = |U| ûû

û û

x

y

z

ii

jj

kk

Useful examples are the cartesian unit vectors [ i, j, ki, j, k ]

Point in the direction of the x, y and z axes.

R = rx i + ry j + rz k

Physics 207: Lecture 4, Pg 6

Vector addition using components:Vector addition using components:

Consider, in 2D, CC = AA + BB.

(a) CC = (Ax ii + Ay jj ) + (Bx i i + By jj ) = (Ax + Bx )ii + (Ay + By )

(b) CC = (Cx ii + Cy jj )

Comparing components of (a) and (b):

Cx = Ax + Bx

Cy = Ay + By

|C| =[ (Cx)2+ (Cy)2 ]1/2

CC

BxAA

ByBB

Ax

Ay

Physics 207: Lecture 4, Pg 7

Exercise 1Exercise 1Vector AdditionVector Addition

A.A. {3,-4,2}{3,-4,2}

B.B. {4,-2,5}{4,-2,5}

C.C. {5,-2,4}{5,-2,4}

D. None of the above

Vector A = {0,2,1} Vector B = {3,0,2} Vector C = {1,-4,2}

What is the resultant vector, D, from adding A+B+C?

Physics 207: Lecture 4, Pg 8

Converting Coordinate SystemsConverting Coordinate Systems In polar coordinates the vector R = (r,)

In Cartesian the vector R = (rx,ry) = (x,y)

We can convert between the two as follows:

• In 3D cylindrical coordinates (r,z), r is the same as the magnitude of the vector in the x-y plane [sqrt(x2 +y2)]

tan-1 ( y / x )

22 yxr

y

x

(x,y)

rrry

rx

j i

cos

cos

yxR

ryr

rxr

y

x

Physics 207: Lecture 4, Pg 9

Exercise: Frictionless inclined planeExercise: Frictionless inclined plane

A block of mass m slides down a frictionless ramp that makes angle with respect to horizontal. What is its acceleration a ?

ma

Physics 207: Lecture 4, Pg 10

Resolving vectors, little g & the inclined plane

g (bold face, vector) can be resolved into its x,y or x’,y’ components g = - g j g = - g cos j’ + g sin i’

The bigger the tilt the faster the acceleration…..

along the incline

x’

y’

x

yg

Physics 207: Lecture 4, Pg 11

Dynamics II: Motion along a line but with a twist(2D dimensional motion, magnitude and directions)

Particle motions involve a path or trajectory

Recall instantaneous velocity and acceleration

These are vector expressions reflecting x, y & z motion

rr = rr(t) vv = drr / dt aa = d2rr / dt2

Physics 207: Lecture 4, Pg 12

Instantaneous VelocityInstantaneous Velocity But how we think about requires knowledge of the path. The direction of the instantaneous velocity is along a line that is

tangent to the path of the particle’s direction of motion.

v

The magnitude of the instantaneous velocity vector is the speed, s. (Knight uses v)

s = (vx2 + vy

2 + vz )1/2

Physics 207: Lecture 4, Pg 13

Average AccelerationAverage Acceleration

The average acceleration of particle motion reflects changes in the instantaneous velocity vector (divided by the time interval during which that change occurs).

a The average

acceleration is a vector quantity directed along ∆v

( a vector! )

Physics 207: Lecture 4, Pg 14

Instantaneous AccelerationInstantaneous Acceleration

The instantaneous acceleration is the limit of the average acceleration as ∆v/∆t approaches zero

The instantaneous acceleration is a vector with components parallel (tangential) and/or perpendicular (radial) to the tangent of the path

Changes in a particle’s path may produce an acceleration The magnitude of the velocity vector may change The direction of the velocity vector may change

(Even if the magnitude remains constant) Both may change simultaneously (depends: path vs time)

Physics 207: Lecture 4, Pg 15

Motion along a path Motion along a path ( displacement, velocity, acceleration )( displacement, velocity, acceleration )

2 or 3D Kinematics : vector equations:

rr = rr(t) vv = drr / dt aa = d2rr / dt2

Velocity :

v = drr / dt

vav = rr / t

Acceleration :

a = dvv / dt

aav = vv / t

vv

y

x

path21 vvv

1v

2v

1- v

Physics 207: Lecture 4, Pg 16

Generalized motion with non-zero acceleration:Generalized motion with non-zero acceleration:

need both path & time

Two possible options:

Change in the magnitude of v

Change in the direction of v

aa = 0

aa = 0

aaaaaa= +

Animation

22 with 0 tr aaaa

aar

||aat

Q1: What is the time sequence in this particle’s motion?

Q2: Is the particle speeding up or slowing down?

a

v

Physics 207: Lecture 4, Pg 17

Examples of motion: ChemotaxisExamples of motion: Chemotaxis

Q1: How do single cell animals locate their “food” ?

Possible mechanism: Tumble and run paradigm.

Q2: How does a fly find its meal?

Possible mechanism: If you smell it fly into the wind, if you

don’t fly across the wind.

An example…..

Physics 207: Lecture 4, Pg 18

KinematicsKinematics The position, velocity, and acceleration of a particle in

3-dimensions can be expressed as:

rr = x ii + y jj + z kk

vv = vx ii + vy jj + vz kk (ii , jj , kk unit vectors )

aa = ax ii + ay jj + az kk

All this complexity is hidden away in

rr = rr(t) vv = drr / dt aa = d2rr / dt2

2

2

dtxd

ax 2

2

dtyd

ay 2

2

dtzd

az

dt

dxvx

dt

dyvy

dtdz

vz

)( txx )( tyy )( tzz

221

0 )( e.g. accel.,constant if with,0

tatvxtx xx

Physics 207: Lecture 4, Pg 19

Special CaseSpecial Case

Throwing an object with x along the horizontal and y along the vertical.

x and y motion both coexist and t is common to both

Let g act in the –y direction, v0x= v0 and v0y= 0

y

t0 4 x

t = 0

4y

t

x

0 4

x vs t y vs t x vs y

Physics 207: Lecture 4, Pg 20

Another trajectoryAnother trajectory

x vs yt = 0

t =10

Can you identify the dynamics in this picture?

How many distinct regimes are there?

Are vx or vy = 0 ? Is vx >,< or = vy ?

y

x

Physics 207: Lecture 4, Pg 21

Another trajectoryAnother trajectory

x vs yt = 0

t =10

Can you identify the dynamics in this picture?

How many distinct regimes are there?

0 < t < 3 3 < t < 7 7 < t < 10

I. vx = constant = v0 ; vy = 0

II. vx = vy = v0

III. vx = 0 ; vy = constant < v0

y

x

What can you say about the acceleration?

Physics 207: Lecture 4, Pg 22

Exercise 2 & 3Exercise 2 & 3Trajectories with accelerationTrajectories with acceleration

A rocket is drifting sideways (from left to right) in deep space, with its engine off, from A to B. It is not near any stars or planets or other outside forces.

Its “constant thrust” engine (i.e., acceleration is constant) is fired at point B and left on for 2 seconds in which time the rocket travels from point B to some point C Sketch the shape of the path

from B to C. At point C the engine is turned off.

Sketch the shape of the path

after point C

Physics 207: Lecture 4, Pg 23

Exercise 2Exercise 2Trajectories with accelerationTrajectories with acceleration

A. A

B. B

C. C

D. D

E. None of these

B

C

B

C

B

C

B

C

A

C

B

D

From B to C ?

Physics 207: Lecture 4, Pg 24

Exercise 3Exercise 3Trajectories with accelerationTrajectories with acceleration

A. A

B. B

C. C

D. D

E. None of these

C

C

C

C

A

C

B

D

After C ?

Physics 207: Lecture 4, Pg 25

Trajectory with constant Trajectory with constant acceleration along the vertical acceleration along the vertical

How does the trajectory appear to

different observers ?

What if the observer is moving with the

same x velocity (i.e. running in parallel)?

x

t = 0

4y

x vs y

t

x

0 4

x vs t

Physics 207: Lecture 4, Pg 26

Trajectory with constant Trajectory with constant acceleration along the vertical acceleration along the vertical

This observer will only see the y motion

x

t = 0

4y

x vs y

x

t = 0

4y

y vs t

In an inertial reference frame all see the same acceleration

Physics 207: Lecture 4, Pg 27

Trajectory with constant Trajectory with constant acceleration along the vertical acceleration along the vertical

What do the velocity and acceleration

vectors look like?

Velocity vector is always tangent to the curve!

Acceleration may or may not be!

y

t = 0

4x

x vs y

Physics 207: Lecture 4, Pg 28

Home Exercise Home Exercise The PendulumThe Pendulum

1= 30°

A) vr = 0 ar = 0

vT = 0 aT 0

B) vr = 0 ar 0

vT = 0 aT = 0

C) vr = 0 ar 0

vT = 0 aT 0

Which statement best describes the motion of the pendulum bob at the instant of time drawn ? i. when the bob is at the top of its swing.ii. which quantities are non-zero ?

Physics 207: Lecture 4, Pg 29

Home Exercise Home Exercise The Pendulum SolutionThe Pendulum Solution

aT

NOT uniform circular motion :If circular motion then ar not zero,If speed is increasing so aT not zero = 30°

a

O

However, at the top of the swing the bob temporarily comes to rest, so v = 0 and the net tangential force is mg sin

C) vr = 0 ar = 0

vT = 0 aT 0

mg

T

Everywhere else the bob has a non-zero velocity and so then

(except at the bottom of the swing)

vr = 0 ar 0

vT 0 aT 0

Physics 207: Lecture 4, Pg 30

Physics 207, Physics 207, Lecture 4, Sept. 15Lecture 4, Sept. 15

Assignment: Read Chapter 5 (sections 1- 4 Assignment: Read Chapter 5 (sections 1- 4 carefully)carefully)

MP Problem Set 2 MP Problem Set 2 due Wednesdaydue Wednesday (should have (should have started)started)

MP Problem Set 3, Chapters 4 and 5 (available MP Problem Set 3, Chapters 4 and 5 (available today)today)