Werner Ens 518 Allen Office Hours: Monday 2 - 6 pm (or by appointment)

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Werner Ens 518 Allen Office Hours: Monday 2 - 6 pm (or by appointment). Werner Ens 518 Allen Office Hours: Monday 2 - 6 pm (or by appointment) My PowerPoint lecture notes available on-line www.physics.umanitoba.ca/~ens/16.102lectures.html - PowerPoint PPT Presentation

Transcript

Werner Ens

518 Allen

Office Hours: Monday 2 - 6 pm(or by appointment)

Werner Ens

518 Allen

Office Hours: Monday 2 - 6 pm(or by appointment)

My PowerPoint lecture notes available on-line

www.physics.umanitoba.ca/~ens/16.102lectures.html

(Complete lecture notes prepared last year by Professor Birchall are available at the course web site)

Quiz1. Which arrows correctly represent

acceleration?

(a) (b) (c) (d) (e)

Quiz2. Which is the trajectory after the line is cut?

(a)

(b)

(c)

(d)

Quiz3. What happens to an astronaut if his tether to

an orbiting space station is cut?

(a) Stays with space station

(b) Falls behind in orbit

(c) Falls to earth

(d) Lost in space (moves away)

(e) Remains stationary with respect to the sun

Quiz4. What happens if an astronaut releases a ball

while standing on the moon?

(a) It falls to the ground (moon)

(b) It rises

(c) It stays where it is released

(d) Moves horizontally (moon’s horizon)

Quiz solutions1. Which arrows correctly represent

acceleration?

(a) (b) (c) (d) (e)

Quiz solutions2. Which is the trajectory after the line is cut?

(a)

(b)

(c)

(d)

Quiz solutions3. What happens to an astronaut if his tether to

an orbiting space station is cut?

(a) Stays with space station

(b) Falls behind in orbit

(c) Falls to earth

(d) Lost in space (moves away)

(e) Remains stationary with respect to the sun

Quiz solutions4. What happens if an astronaut releases a ball

while standing on the moon?

(a) It stays where it is

(b) It rises

(c) It falls to the ground (moon)

(d) Moves horizontally (moon’s horizon)

Quiz solutions

1 d, 2 b, 3 a, 4 c

Chapter 1

Introduction

Mathematical Concepts

1) Representation of physical quantities

• Arithmetic is abstract: 10 - 5 = 5• A physical quantity (distance) requires

comparison to a standard: L = 10 is meaningless

m

10 m

• A physical quantity is an algebraic product of a number and a unit: L = 10 m

2) Units combine algebraically; Dimensional Analysis

d3 = d1 + d2

d1 = 5m;

d2 = 8m

d3 = 5m + 8m

= (5 + 8)m

=13m

2) Units combine algebraically; Dimensional Analysis

A = l • w

l = 8m

w = 4m

A = 8m• 4m = 32m2

V = lwh ≠ 32h

V = 32m2h

If h = 2m, then V = 64m3

Example: Vector Addition• Jogger runs 145 m 20.0º East of North

– Displacement vector A

• Then 105 m 35.0º South of East– Displacement vector B

• Find resultant displacement C=A+B

A =145 m; θA =70º

θA

20º

x (E)

y (N) rA

35º rB

rC

θC

θB

B =105 m; θB =360º−35º

→ 325º or −35º

Find C and θC

Step 0: Draw a picture & organize data

Example: Vector Addition

A =145 m; θA =70ºθA20º

x (E)

y (N) rA

• Step 1: Convert to component representation

Ax =AcosθA =(145m)cos70º=49.6m

Ax

Ay

Ay =AsinθA =(145m) sin70º=136m

Example: Vector Addition

x (E)

y (N) 35º rB

θB

B =105 m; θB =360º−35º

→ 325º or −35º

• Step 1: Convert to component representation

Ax =AcosθA =(145m)cos70º=49.6mAy =AsinθA =(145m) sin70º=136m

By

Bx

By =BsinθB =(105m) sin(-35º) =−60.2mBx =BcosθB =(105m) cos(-35º) =86.0m

Example: Vector Addition• Step 2: Sum components

Ax =49.6mAy =136m

By =−60.2mBx =86.0mθA

20º

x (E)

y (N) rA

35º rB

rC

θC

θB

Cx =Ax + Bx =135.6m

Cy =Ay + By =76.0m

Example: Vector Addition• Step 3: Convert result to geometric representation

(magnitude, direction)

Ax =49.6mAy =136m

By =−60.2mBx =86.0m

x (E)

y (N)

rC

θC

Cx =135.6m

Cy =76.0m

C = Cx2 +Cy

2 =155m

Cx

Cy

tanθc =Cy

Cx

=.560 → θC = 29.2º (North of East)

Example: Vector Addition• Step 3: Convert to geometric representation (magnitude, direction)

x (E)

y (N)

rC

θC

Cx =135.6m

Cy =−76.0m

C = Cx2 +Cy

2 =155m

Cx

Cy

tanθc =Cy

Cx

=−.560 → θC = −29.2º

Example: Vector Addition• Step 3: Convert to geometric representation (magnitude, direction)

x (E)

y (N)

rC

θC Cx =−135.6m

Cy =−76.0m

C = Cx2 +Cy

2 =155m

Cx

Cy

tanθc =Cy

Cx

=.560 → θC = 29.2º?

→ θC = 29.2º +180º = 209.2º

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