Homework: 49, 51, 70 (p. 134-137)

49. In the figure below, a car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill, the normal force on the driver from the car seat is 0. The driver’s mass is 80.0 kg. What is the magnitude of the normal on the driver from the seat when the car passes through the bottom of the valley?

R

vmFFF

2

Nglcentripeta

At the top of the hill:

R

vmF0F

2

gN

At the bottom of the valley:

g

2

N

2

gNlcentripeta FR

vmF

R

vmFFF

(N)15689.88022mg2FF gN

gF

NF NF

gF

51. An airplane is flying in a horizontal circle at a speed of 600 km/h. If its wings are tilted at angle =400 to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an “aerodynamic lift” that is perpendicular to the wing surface.

Animation

gyl, FF

Fl

Fg

Fl,x

Fl,y

•According to the Bernoulli’s principle, the aerodynamic lift appears due to the air-stream velocity over the top of the airplane greater than that at the bottom.

(1) mgcosFl

R

vmF

2

xl, Fl,x is the centripetal force:

)2(R

vmsinF

2

l

(km) 3.38or (m) 7933)40tan(8.9

7.166

tang

vR (2) and (1)

22

m/s7.166km/h600 v

70. The figure below shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. The cord sweeps out a cone as the bob rotates. The bob has a mass of 0.050 kg, the string has length L=0.90 m and negligible mass, and the bob follows a circular path of circumference 0.94 m. What are (a) the tension in the string and (b) the period of the motion?

mgTsin0F-T gy

T

Fg

Tx

Ty

Tx: the centripetal force

: the angle between the cord and the horizontal circle.

R

vmcosT

R

vmT

22

x

m 51.014.32

(m) 0.94

2π

CR;

L

Rcos

04.80L

Rarccos

(a) (N) .50sin

mgT

(b) (m/s); 5.0

m

RcosθTv

(s) 88.1

v

CP

Animation

Part B Laws of Conservation

Chapter 3 Work and Mechanical Energy

3.1. Kinetic Energy and Work. Power

3.2. Work-Kinetic Energy Theorem

3.3. Work and Potential Energy

3.4. Conservative and Non-conservative Forces. Conservative Forces

and Potential Energy

3.5. Conservation of Mechanical Energy

3.6. Work Done on a System by an External Force. Conservation of

Energy

What is energy?

Energy is the capacity of a system to do work. There are many forms of energy:

• Mechanical energy: •Potential energy, stored in a system •Kinetic energy, from the movement of matter

• Radiant energy (solar energy) • Thermal energy • Chemical energy (chemical bonds of molecules) • Electrical energy (movement of electrons) • Electromagnetic energy (from X-rays to radio waves) • Nuclear energy

Energy can be transformed from one type to another and transferred from one object to another, but the total amount is always the same (the principle of energy conservation).

3.1. Kinetic Energy and Work. Power

2mv2

1K

Kinetic energy is energy associated with the state of motion of an object.

1 joule = 1 J = 1 kg. m2/s2

Work is energy transferred to or from an object by means of a force acting on the object. Energy transferred to the object is positive work and energy transferred from the object is negative work.

Unit: joule

3.1.1. Kinetic Energy and Work

3.1.2. Work done by a force

•To establish an expression for work, we consider a constant force F that accelerates a bead along a wire:

xx maF

d2avv x

2

0

2

dFmv2

1mv

2

1x

2

0

2

Therefore, the work W done on the bead by F is:

cos FddFW x

force)constant aby done(work d FW

A. Work done by a constant force:

B. Work done by a general variable force:

•Choose x small enough, work done by the force in the jth interval:

xFΔW avgj,j

One-dimensional analysis:

xFWW avgj,j

•The total work:

xFlimW avgj,0x

f

i

x

xforce) variableaby done(work F(x)dx W

Three-dimensional analysis:

dzFdyFdxFrdFdW zyx

f

i

f

i

f

i

f

i

y

y

z

zzy

x

xx

r

rdzFdy Fdx FdWW

3.1.3. Power

Power is the rate at which work is done.

t

WPavg

Average power:

Instantaneous power: dt

dWP

Unit: watt (W) 1J/s W 1 watt1

W746ft.lb/s 550 hp 1horsepower 1

MJ 3.6J103.6W)(3600s)(101kW.hhour-kilowatt 1 63

dt

dxFcos

dt

dxFcos

dt

dWP

FvcosP

F=constant:

Instantaneous power: vFP

3.2. Work-Kinetic Energy Theorem

Let K be the change in the kinetic energy of the bead.

WKKΔK if

This can be read as follows:

object the

on donenet work

objectan ofenergy

kinetic in the change

WKK if or

donework

net the

net work thebefore

energy kinetic

done isnet work the

after energy kinetic

Examples:

E1. Work done by the gravitational force:

cos FddFW x

0KK if

For a rising object, =1800:

mgdW

For a falling object, =00:

mgdW

Work done in lifting and lowering an object

gaif WWKKΔK where Wa is the work done by the applied force; Wg is the work done by the gravitational force. If initial and final velocities are zero:

ga WW0ΔK

cos mgdWa

Gravity and an applied force acting on the object:

0KK if

E2. Work done by a spring force:

The spring force is computed by:

law) s(Hooke' dkFs

k: the spring constant (or force constant)

Is an x axis is parallel to the length of the spring:

law) s(Hooke'kx Fx

A spring force is a variable force F=F(x)

To find the work done by the spring force, We have to make assumptions: (1) the spring is massless; (2) It is an ideal spring (it exactly obeys Hooke’s law).

ΔxF-limcosΔxFlimxΔFlimW xj0

xj0

xj0

sxxx

Note: = 1800 and Fxj is the magnitude of the spring force

2

sfi kx2

1 W:x x0, xIf

Work done by an applied force:

saif WWKKΔK

If the block is stationary before and after the displacement, K=0:

sa WW

2

f

2

is kx2

1kx

2

1W

f

i

f

i

xx

xx xs kxdxdxFW

Homework: 1, 2, 8, 15, 24, 26, 29, 36, 43 (p. 159-163)

Review

Chapter 1:

To describe motion, we need to measure: + Displacement: x = xt – x0 (measured in m or cm) + Time interval: t = t – t0 (measured in s)

Δt

distancetotalsavg

dt

dx(t)

Δt

Δx(t)limv(t)

0Δt

Average velocity:

Average speed:

Instantaneous velocity:

12

12avg

tt

vv

Δt

Δva

Average acceleration:

Instantaneous acceleration: 2

2

dt

xd

dt

dv(t)a(t)

Motion in one dimension:

(All sections of Chapter 1, 2)

Two basic equations for constant acceleration:

For freely falling objects: )(m/s 9.8ga 2

Motion in two dimensions:

Projectile motion:

tcosθvxx 000

• Ox: Horizontal motion (no acceleration):

• Oy: Vertical motion (free fall)

gt-sinθvv 00y

2000 gt

2

1-tsinθvyy

•Horizontal range:

0

20 sin2θ

g

vR

constantcosθvv 00x

v

r2T

(T: period)

Where r is the radius of the circle v the speed of the particle

Uniform Circular Motion: The particle is accelerating with a centripetal acceleration:

r

va

2

Relative Velocity and Relative Acceleration:

BAPBPA vvv BAPBPA vvv

Chapter 2:

Newton’s Laws 0For 0Fn

1ii

amFnet

CBBC FF

Some particular forces: gravitational, normal, tension and frictional forces

Friction and Properties of Friction:

Nsmaxs, Fμf sμ is the coefficient of static friction

Nkk Fμf kμ is the coefficient of kinetic friction

Uniform Circular Motion and Non-uniform Circular Motion:

• Uniform circular motion:

• Non-uniform circular motion: