KINEMATICS TTTThe 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
Go to graphClick here for a
HINT
Click here for the
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
20.0s20.0sClick here for
HINT 1 : Draw a tangent to the curve at 20.0 s.
Go to graph
Click here for the
SOLUTION
Click here for
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?
Click here for a
HINT : Note that this is a graph of Position vs. Time
Click here for the
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?
Click here for a
HINT : In the time period given, the body moves forwards and backwards.
Click here for the
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
Click here for a
HINT :
Click here for the
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.
Click here for a
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
Click here for the
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
Click here for the
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
Click here for the
Diagram :
Δd g
v1
V0 = 0
Click here for the
Given:
Click here for the
Solution :a)
Click here for the
Solution : b)
Move on to KINEMATIC DERIVATIONS