03_physics_lecture_15 mar 2014

MS101: PhysicsDr. Ahmed Amin Hussein

[email protected]

May 3, 2023 Prepared By: Dr. Ahmed Amin 1

mailto:[email protected]

Chapter 3

Acceleration and Newton’s Second Law of Motion

May 3, 2023 2Prepared By: Dr. Ahmed Amin

Position & Displacement Velocity Newton’s Second Law of Motion Applying Newton’s Second Law Relative Velocity

§3.1 Position & Displacement


The position (r) of an object describes its location relative to some origin or other reference point.

The displacement is the change in an object’s position. It depends only on the beginning and ending positions.

if rrr

Example


Margaret walks to the store using the following path: 0.500 miles west, 0.200 miles north, 0.300 miles east. What is her total displacement? Give the magnitude and direction.

x

y

r3

r2

r1

r

Take north to be in the +y direction and east to be along +x.

Example continued


The displacement is r = rf ri. The initial position is the origin; what is rf?

The final position will be rf = r1 + r2 + r3. The components are rfx = r1 + r3 = 0.2 miles and rfy = +r2 = +0.2 miles.

miles 283.022 yx rrr

45 and 1tan

x

y

r

r

Using the figure, the magnitude and direction of the displacement are

x

y

rry

rx

N of W.

§3.2 Velocity


Velocity is a vector that measures how fast and in what direction something moves.

Speed is the magnitude of the velocity. It is a scalar.


Path of a particle

Start

finish

r

vav is the constant speed that results in the same displacement in a given time interval.

tripof time traveleddistancespeed Average

t

rvav velocityAverage

txv x,av :be wouldcomponent - xThe


y

x

ri rf

t

rvav Points in the direction of r

r

vi

The instantaneous velocity points tangent to the path.vf

A particle moves along the blue path as shown. At time t1 its position is ri and at time t2 its position is rf.


On a graph of position versus time, the average velocity is represented by the slope of a chord.

x (m)

t (sec)t1 t2

x1

x2

12

12,av velocityAverage

ttxxv x


tt

rv0

lim velocityousInstantane

x (m)

t (sec)

This is represented by the slope of a line tangent to the curve on the graph of an object’s position versus time.


The area under a velocity versus time graph (between the curve and the time

axis) gives the displacement in a given interval of time.

v(m/s)

t (sec)

Example


Consider Margaret’s walk to the store in the example on slides 3 and 4. If

the first leg of her walk takes 10 minutes, the second takes 8 minutes, and

the third 7 minutes, compute her average velocity and average speed during

each leg and for the overall trip.

t

rvav velocityAverage

tripof time traveleddistancespeed Average

Use the definitions:

Margaret walks to the store using the following path: 0.500 miles west, 0.200 miles north, 0.300 miles east.


x

y

r3

r2

r1

r

Legt

(hours)vav

(miles/hour)Average speed

(miles/hour)

1 (0.5 miles)

0.167(10 mins)

3.00 (west) 3.00

2 (0.2 miles)

0.133(8 mins)

1.50 (north) 1.50

3 (0.3 miles)

0.117(7 mins)

2.56 (east) 2.56

Total trip 0.417(25 mins)

0.679(45 N of W)

2.40miles 283.022 yx rrr

45 and 1tan

x

y

r

r

Example


Speedometer readings are obtained and graphed as a car comes to a stop along a straight-line path. How far does the car move between t = 0 and t = 16 seconds?

Since there is not a reversal of direction, the area between the curve and the time axis will represent the distance traveled.

The rectangular portion has an area of Lw = (20 m/s)(4 s) = 80 m. The triangular portion has an area of ½bh = ½(8 s) (20 m/s) = 80 m. Thus, the total area is 160 m. This is the distance traveled by the car.

§3.3 Newton’s Second Law of Motion


t

vaavonaccelerati Average

A nonzero acceleration changes an object’s state of motion.

tt

va0

limonaccelerati ousInstantane

These have interpretations similar to vav and v.


y

x

vi

ri rf

vf

v

Points in the direction of v.t

vaav

The instantaneous acceleration can point in any direction.

A particle moves along the blue path as shown. At time t1 its position is ri and at time t2 its position is rf.

Example


If a car traveling at 28 m/s is brought to a full stop 4.0 s after the brakes are applied, find the average acceleration during braking.

2av m/s 0.7

s 0.4m/s 280

tva

Given: vi = +28 m/s, vf = 0 m/s, and t = 4.0 s.

Example


At the beginning of a 3 hour trip you are traveling due north at 192 km/hour. At the end, you are traveling 240 km/hour at 45 west of north.

(a) Draw the initial and final velocity vectors.

x (east)

y (north)

vivf


(b) Find v.

km/hr 3.2245cos

km/hr 170045sin

ifiyfyy

fixfxx

vvvvv

vvvv

The components are

km/hr 17122 yx vvv

5.71312.0tan1312.0tan 1

x

y

v

vSouth of west


(c) What is aav during the trip?

t

vaav

2av,

2av,

km/hr 43.7hr 3km/hr 3.22

km/hr 7.56hr 3km/hr 170

tv

a

tva

yy

xx

The magnitude and direction are:

5.7)1310.0(tan1310.0tan

km/hr 2.57

1

,av

,av

22av,

2av,av

x

y

yx

a

a

aaa

South of west

Newton’s 2nd Law


The acceleration of a body is directly proportional to the net force acting on the body and inversely proportional to the body’s mass.

Mathematically: aFFa mm

netnet or

An object’s mass is a measure of its inertia. The more mass, the more force is required to obtain a given acceleration.

The net force is just the vector sum of all of the forces acting on the body, often written as F.

If a = 0, then F = 0. This body can have:Speed = 0 which is called static equilibrium, orspeed 0, but constant, which is called dynamic equilibrium.

§3.4 Applying Newton’s Second Law


aF m

Force units: 1 N = 1 kg m/s2.

Example


Find the tension in the cord connecting the two blocks as shown. A force of 10.0 N is applied to the right on block 1. Assume a frictionless surface. The masses are m1 = 3.00 kg and m2 = 1.00 kg.

F block 2

block 1

Assume that the rope stays taut so that both blocks have the same acceleration.

FBD for block 2:

TF

w1

N1

x

y

x

T

w2

N2

y

FBD for block 1:

011

1

wNF

amTFF

y

x

022

2

wNF

amTF

y

x

Apply Newton’s 2nd Law to each block:

amTF 1

amT 2

These two equations contain the unknowns: a and T.

To solve for T, a must be eliminated. Solve for a in (2) and substitute in (1).

(1)

(2)

N 5.2

kg 1kg 31

N 10

1

1

2

1

2

1

21

211

mmFT

TmmT

mTmF

mTmamTF

Example continued:

TMA Exercises


Problems :

Page # 7, 11, 17, 35, 49

Additional questions up to slide # 14

Chapter 3, Questions 8, 10, 11, 12, 13, and 14

03_physics_lecture_15 mar 2014

Documents