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KINEMATICS KINEMATICS The study of motion The study of motion Translational Rotational Vibrational

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### Transcript of KINEMATICS TTTThe study of motion Translational Rotational Vibrational.

KINEMATICSKINEMATICS The study of motionThe study of motion

Translational

Rotational

Vibrational

DEFINITIONSDEFINITIONS

• Distance, Distance, ΔΔd – change in position.d – change in position.

• Displacement, Δd – Change in position in a particular direction.

NOTE ON DELTA NOTATION : Δ INDICATES THE CHANGE IN A QUANTITY.

d d d or d dfinal initial 1 0

DEFINITIONSDEFINITIONS

v dt

•Speed, v – time rate of change of position.

v dt

•Velocity, v – Speed in a particular direction.

DEFINITIONSDEFINITIONS

Acceleration, a – Time rate of change of a velocity.

a vt

KINEMATIC GRAPHICKINEMATIC GRAPHICINTERPRETATIONINTERPRETATION

16

12

8

4

0

d (m)

0 5 10 15 20 25t(s)

Go To :

v

a

vave

vinst

Drawtan

Δd

Δd

Go tov vs tgraph

Next slide

Velocity is the slope of a Velocity is the slope of a position vs. time curve.position vs. time curve.

Since Since slope = rise/run,slope = rise/run,

Just as ,Just as ,my

x

v dt

QUALITATIVE ANALYSISQUALITATIVE ANALYSISAt what time(s) is the velocity :At what time(s) is the velocity :

a)a) Positive?Positive?

b)b) Negative?Negative?

c)c) Zero?Zero?

d)d) Increasing?Increasing?

e)e) Decreasing?Decreasing?

f)f) Constant?Constant?

{Back to graph}

QUALITATIVE ANALYSISQUALITATIVE ANALYSISAt what time(s) is the acceleration :At what time(s) is the acceleration :

(a)(a) Positive?Positive?

(b)(b) Negative?Negative?

(c)(c) Zero?Zero?

{Back to graph}

(a)When is the velocity increasing?(b)When is the velocity decreasing?(c)When is the velocity constant?

Hints

QUANTITATIVE ANALYSISQUANTITATIVE ANALYSISFind the average velocity betweenFind the average velocity between

t = 5.0 and 18.0 st = 5.0 and 18.0 s

HINT

SOLUTION

vd

tave

vm m

s s

m

sm save

12 0 4 0

18 0 5 0

8 0

13 00 62

. .

. .

.

.. /

vd

td d

t tave

1 0

1 0

QUANTITATIVE ANALYSISQUANTITATIVE ANALYSISFind the instantaneous velocity at Find the instantaneous velocity at

HINT 1 : Draw a tangent to the curve at 20.0 s.

Go to graph

SOLUTION

HINT 2 :Pick 2 convenient points to determine its slope. ( try 13 s and 23s)

vm m

s s

m

sm sinst

58 16 0

230 130

10 2

10 0102

. .

. .

.

.. /

vd

td d

t tinst

1 0

1 0

What is the total displacement What is the total displacement between 11.0 and 18.0 s?between 11.0 and 18.0 s?

HINT : Note that this is a graph of Position vs. Time

SOLUTION :

d d d 1 0

Go to graph

d m m m 12 0 14 0 2 0. . .

What is the distance traveled What is the distance traveled between 11.0 and 18.0 s?between 11.0 and 18.0 s?

HINT : In the time period given, the body moves forwards and backwards.

SOLUTION :

Δd = d1 – d0 forward + d1 – d0 backward

Δd = Δdforward + Δdbackward

Δd = 18.0 m – 14.0m + 12.0m – 18.0m

Δd = 4.0m + 6.0m = 10.0 m

Go to graph

KINEMATIC GRAPHICKINEMATIC GRAPHICINTERPRETATIONINTERPRETATION

+2

+1

0

-1

-2

v (m/s)

0 5 10 15 20 25t(s)

Go To :

a1

a2

Δd

Δd(hint)

Go tod vs tgraph

NextSlide

III

I

II

Δdarea

Δdsoln)

Show

On this graph

Acceleration is the slope of a Acceleration is the slope of a velocity vs. time curve.velocity vs. time curve.

Since Since slope = rise/run,slope = rise/run,

Just as ,Just as ,my

x

a vt

At what time(s) is the acceleration:At what time(s) is the acceleration:

a)a) Positive?Positive?

b)b) Negative?Negative?

c)c) Zero?Zero?

d)d) Increasing?Increasing?

e)e) Decreasing?Decreasing?

f)f) Constant?Constant?

{Back to graph}

Find the average Find the average accelerationacceleration

between 8.0 and 25.0 s.between 8.0 and 25.0 s.

av

t

HINT :

Solution : a v vt t

ms

ms

s s

1 0

1 0

0 0 2 0

250 8 0

. .

. .

( )

amss

ms

2 0

17 0012 2

.

..

Go tov vs tgraph

Displacement is the area Displacement is the area between the curve and the between the curve and the horizontal origin on a v vs. t horizontal origin on a v vs. t graph.graph.

From ,vd

t

If v is constant, area is rectangular : d v tIf a is constant, area is trapezoidal : d v tave

OR d v v t.... ( ) 12 0 1

Find the displacement Find the displacement between 11.0 and 19.0s.between 11.0 and 19.0s.

HINT :

Above the origin, the displacement is positive.

Below the origin, the displacement is negative.

Above the origin, the displacement is positive.

Below the origin, the displacement is negative.

Go tov vs tgraph

SOLUTION : Δdtotal = Δdabove + Δdbelow

Δdtotal = 12 0 1 1( )v v t +

12 1 2 2( )v v t v t2 3+

Δdtotal = ½(+2.0m/s+0)3.0s + ½(0+(-2.0m/s))3.0s +(-2m/s)2.0s

Δdtotal = +3.0m + (-3.0m) + (-4.0m) = -4.0m

Go tov vs tgraph

{Total area = Area I + Area II + Area III}{Total area = Area I + Area II + Area III}

What distance is traveled What distance is traveled between 11.0 and 19.0 s?between 11.0 and 19.0 s?Click here for a

HINT :

Add the absolute values of the areas under the curve.

Go tov vs tgraph

SOLUTION :

Δdtotal= |½(+2.0m/s+0)3.0s| + |½(0+(-2.0m/s))3.0s +(-2m/s)2.0s|

Δdtotal = 3.0m + 3.0m + 4.0m = 10.0m

Δdtotal = Δdabove + Δdbelow

Δdtotal= 12 0 1 1( )v v t +

12 1 2 2( )v v t v t2 3+

{AREA I} {AREA II} {AREA III}

DISTANCEGo tov vs tgraph

SIMPLE KINEMATICS PROBLEMSSIMPLE KINEMATICS PROBLEMS

{1} A car moving south along a level road {1} A car moving south along a level road

increases its velocity uniformly fromincreases its velocity uniformly from

16 m/s to 32 m/s in 10.0 s.16 m/s to 32 m/s in 10.0 s.

(a) What is the car’s acceleration?(a) What is the car’s acceleration?

(b) What is it’s average velocity?(b) What is it’s average velocity?

SOLVING PROCEDURESOLVING PROCEDURE List given and identify the unknown.List given and identify the unknown.Let south Let south ≡ +≡ +

VV00 = + 16 m/s = + 16 m/s V V11 = + 32 m/s = + 32 m/s ΔΔt = 10.0 st = 10.0 s

a = ?a = ? V Vaveave = ? = ?Draw and label a diagram.Draw and label a diagram.

VV00VV11

SOLVING PROCEDURESOLVING PROCEDURE Determine relevant equations relating the Determine relevant equations relating the

known and the unknown.known and the unknown.

av

t

v v vave 12 0 1( )

Derive (if needed) working equations Derive (if needed) working equations for the unknown in terms of the for the unknown in terms of the

known.known.

av

t

v v

t

1 0

v v vave 12 0 1( ) Is already a working equation!

PLUG IN THE KNOWNPLUG IN THE KNOWN

am s m s

s

( / ) ( / )

.

32 16

10 0

v m s m save 12 32 16{( / ) ( / )}

Do the dimensional analysis.Do the dimensional analysis.

msms

s

mss

m s

2

msmsms

ARE THE UNITS CONSISTANT WITH WHAT YOU ARE SOLVING FOR?ARE THE UNITS CONSISTANT WITH WHAT YOU ARE SOLVING FOR?

Do the math.Do the math.

ms2

ms

a

( ) ( )

..

32 16

10 016

vave 12 32 16 24{( ) ( )}

And tag on the units.

Acceleration due to gravity.Acceleration due to gravity.[Bodies in freefall][Bodies in freefall]

At the earth’s surface, all falling At the earth’s surface, all falling bodies accelerate at the same rate bodies accelerate at the same rate in a vacuum.in a vacuum.

g = 9.80 m/s2g = 9.80 m/s2

SAMPLE PROBLEMSAMPLE PROBLEM A ball is dropped from a bridge A ball is dropped from a bridge

and strikes the water after 8.0 s.and strikes the water after 8.0 s.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

a) At what speed does the ball strike the water?

b) How high is the bridge?b) How high is the bridge?

V0 = 0 m/sΔt = 8.0sg = -9.80 m/s2

Diagram :

Δd g

v1

V0 = 0