D R M A R T A S T A S I A K
D E P A R T M E N T O F C Y T O B I O L O G Y A N D P R O T E O M I C S
KINEMATICS
lecture based on © 2016 Pearson Education, Ltd.
• Vectors
• Reference Frames
• Displacement
• Average Velocity
• Instantaneous Velocity
• Acceleration
• Motion at Constant Acceleration
• Solving Problems
• Falling Objects
• Graphical Analysis of Linear Motion
• Projectile Motion
• Linear Momentum
VECTORS
• A vector has magnitude as well as direction.
• Examples: displacement, velocity, acceleration,
force, momentum
• A scalar has only magnitude
• Examples: time, mass, temperature, energy
VECTORS
• Magnitude
• Direction – the line of action (line segment) and
sense (orientation).
• Origin (tail) of the vector - point of application,
initial point
A
B
Origin
line segment sense
D
VECTOR ADDITION – ONE DIMENSION
A person walks 8 km East
and then 6 km East.
Displacement =14 km East
A person walks 8 km East
and then 6 km West.
Displacement = 2 km East
VECTOR ADDITION
• A person walks 10 km East
and 5.0 km North
• Order doesn’t matter
21 DDDR
2
2
2
1 DDDR
kmkmkmDR 2.11)5()10( 22
RD
D2sin
0121 5.26)2.11
5(sin)(sin
km
km
D
D
R
GRAPHICAL METHOD OF VECTOR ADDITION TAIL TO TIP METHOD
1V
2V
3V
RV
GRAPHICAL METHOD OF VECTOR ADDITION TAIL TO TIP METHOD
1V
2V
3V
RV
1V
2V
3V
PARALLELOGRAM METHOD
SUBTRACTION OF VECTORS
• Negative of vector has
same magnitude but points
in the opposite direction
• For subtraction, we add the
negative vector.
MULTIPLICATION BY A SCALAR
• A vector V can be multiplied by a scalar c;
the result is a vector c•V that has the same direction
but a magnitude c•V.
• If c is negative, the resultant vector points in the
opposite direction.
ADDING VECTORS BY COMPONENTS
• Any vector can be expressed as the sum of two
other vectors, which are called its components.
Usually the other vectors are chosen so that they
are perpendicular to each other.
TRIGONOMETRY REVIEW
Opp
osi
te
Adjacent
Hypotenuse
Hypotenuse
Oppositesin
Hypotenuse
Adjacentcos
cos
sin
Adjacent
Oppositetan
ADDING VECTORS BY COMPONENTS
• If the components are perpendicular, they can be
found using trigonometric functions.
sinVVy
cosVVx
V
Vy
Hypotenuse
Oppositesin
V
VxHypotenuse
Adjacentcos
cos
sin
Adj
Opptan
ADDING VECTORS BY COMPONENTS
Adding vectors:
1. Draw a diagram; add the vectors graphically.
2. Choose x and y axes.
3. Resolve each vector into x and y components.
4. Calculate each component using sines and
cosines.
5. Add the components in each direction.
6. To find the length and direction of the vector, use:
V
Vysin
REFERENCE FRAMES
Any measurement of position, distance, or speed
must be made with respect to a reference frame.
REFERENCE FRAMES
Coordinate axes
REFERENCE FRAMES
COORDINATE SYSTEM
DISPLACEMENT
• distance
• displacement
• The displacement is written:
• Displacement is positive
• Δx=30m-10m=20m
• Displacement is negative
• Δx=10m-30m=-20m
DISPLACEMENT
VELOCITY
• Speed: how far an object travels in a given time
interval
• Velocity includes directional information:
VELOCITY AS A VECTOR
𝑣 =∆𝑥
∆𝑡
AVERAGE VELOCITY
𝑣 =𝑥2 − 𝑥1
𝑡2 − 𝑡1=
Δ𝑥
Δ𝑡
INSTANTANEOUS VELOCITY
constant velocity varying velocity
𝑣 = lim∆𝑡→0
Δ𝑥
Δ𝑡=
𝛿𝑥
𝛿𝑡
ACCELERATION
ACCELERATION
VECTOR
ACCELERATION AS A VECTOR
𝑎 =∆𝑣
∆𝑡
ACCELERATION
There is a difference between negative acceleration
and deceleration
ACCELERATION
Negative acceleration is acceleration in the negative
direction as defined by the coordinate system.
ACCELERATION
Deceleration occurs when the acceleration is
opposite in direction to the velocity.
ACCELERATION
The instantaneous acceleration is the average
acceleration, in the limit as the time interval becomes
infinitesimally short.
𝑎 = lim∆𝑡→0
Δ𝑣
Δ𝑡=
𝛿𝑣
𝛿𝑡=
𝛿2𝑥
𝛿𝑡
MOTION AT CONSTANT ACCELERATION
• The average velocity of an object during a time
interval t is
• The acceleration, assumed constant, is
MOTION AT CONSTANT ACCELERATION
• In addition, as the velocity is increasing at a
constant rate, we know that
• Combining these last three equations, we find:
MOTION AT CONSTANT ACCELERATION
• We can also combine these equations so as to
eliminate t:
• We now have all the equations we need to solve
constant-acceleration problems.
FALLING OBJECTS
The same acceleration
FALLING OBJECTS
FALLING OBJECTS
9.80 m/s2
FREE FALL—HOW FAST?
• The velocity acquired by an object
starting from rest is
• So, under free fall, when
acceleration is 10 m/s2, the speed is
• 10 m/s after 1 s.
• 20 m/s after 2 s.
• 30 m/s after 3 s.
And so on.
© 2015 Pearson Education, Inc.
Velocity = acceleration ´ time
GRAPHICAL ANALYSIS OF LINEAR MOTION
constant velocity
GRAPHICAL ANALYSIS OF LINEAR MOTION
VARYING VELOCITY
GRAPHICAL ANALYSIS OF LINEAR MOTION
Displacement
Displacement
SUMMARY
• Kinematics is the description of how objects move
with respect to a defined reference frame.
• Displacement is the change in position of an object.
• Average speed is the distance traveled divided by
the time it took; average velocity is the
displacement divided by the time.
• Instantaneous velocity is the limit as the time
becomes infinitesimally short.
SUMMARY
• Average acceleration is the change in velocity
divided by the time.
• Instantaneous acceleration is the limit as the time
interval becomes infinitesimally small.
• The equations of motion for constant acceleration
are given in the text; there are four, each one of
which requires a different set of quantities.
• Objects falling (or having been projected) near the
surface of the Earth experience a gravitational
acceleration of 9.80 m/s2.
D R M A R T A S T A S I A K
D E P A R T M E N T O F C Y T O B I O L O G Y A N D P R O T E O M I C S
KINEMATICS IN TWO DIMENSION
lecture based on © 2016 Pearson Education, Ltd.
PROJECTILE MOTION
• two dimensions
• parabola
PROJECTILE MOTION
ay=o
tgrounded=tdropped vertically
EQUATIONS FOR PROJECTILE MOTION
• Horizontal X Vertical Y
• ax=0 ay = - g
• vx= constant
00 xvv tgvv yy 0
2
002
1tgtvyy y tvxx x00
)(2 0
2
0
2 yygvv yy
INITIAL VELOCITY
• For y=0 𝑣0 = 𝑣
• v = vx0 = constans
sin00 vvy
cos00 vvx
sin00 vvy sin00 vvy
PROBLEM SOLVING—A GENERAL APPROACH
• Read the problem carefully; then read it again.
• Draw a sketch, and then a free-body diagram.
• Choose a convenient coordinate system.
• List the known and unknown quantities
• Find relationships between the knowns and the unknowns.
• Estimate the answer.
• Solve the problem without putting in any numbers
(algebraically); once you are satisfied, put the numbers in.
• Keep track of dimensions.
• Make sure your answer is reasonable.
EXAMPLE
• A football is kicked at an angle of 50˚ above the
horizontal with a velocity of 18 m/s.
• Calculate the maximum height and the range
as well as how long it is in the air
• Assume that the ball was kicked at ground level
and lands at ground level.
Θ=50˚
V0=18m/s
Hmax, R , t=?
EXAMPLE • A football is kicked at an angle of 50˚ above the horizontal with a
velocity of 18.0 m / s. Calculate the maximum height. Assume that the ball was kicked at ground level and lands at ground level.
cos0 vvx sin0 vvy
tgvv yy 0 0
g
vt
y
up
0
280.9
8.13
sms
m
s41.12
0maxmax2
1gttvyyH yo
2
0max
sin
2
1sin0
g
vg
g
vvH y
mH 7.9max
at top:
sms
m /6.11)50)(cos18(
sms
m /8.13)50)(sin18(
g
v sin
22max )41.1)(8.9(
2
1)41.1)(8.13(0 s
sms
smH
LEVEL HORIZONTAL RANGE
• Range is determined by time it takes for ball to return to
ground level or perhaps some other vertical value.
• If ball hits something a fixed distance away, then time is
determined by x motion
• If the motion is on a level field, when it hits: y = 0
• Solving we find
• We can substitute this in the x equation to find the range R
22
002
100
2
1tgtvtgtvyy yoy
g
vt
y02
g
v
g
vv
g
vvtvxR
yoxy
xox00
2
000
0
cossin22)
2(
LEVEL HORIZONTAL RANGE
• We can use a trig identity
• Greatest range: θ= 450
• θ = 300 and 600 have same
range.
2sincossin2
g
vR
2sin2
0)1545( 00
• Caution– the range formula has limited usefulness. It is
only valid when the projectile returns to the same
vertical position.
EXAMPLE • A football is kicked at an angle of 50˚ above the horizontal with a
velocity of 18.0 m / s. Calculate the maximum height. Assume that the ball was kicked at ground level and lands at ground level.
• Assume time down = time up
• For Range:
• Could also use range formula
)41.1)(2( st s82.2
tvxxR x00 )82.2()6.11(0 ss
m m33
g
vR
2sin2
0
msm
sm33
/8.9
)50()2(sin)/18(2
02
EXAMPLE-VERTICAL PROJECTION A rescue plane wants to drop supplies to isolated mountain climbers
on a rocky ridge 235 m below. If the plane is traveling horizontally with a speed of 250 km/h (69.4 m/s) how far in advance of the recipients
(horizontal distance) must the goods be dropped?
Coordinate system is 235 m below plane
22
2
1
2
10235 tgtgmy
tvxx xo 0
g
yt
)()2(
mx
ssm
481
)93.6()/4.69(0
s93.6
2/8.9
)235()2(
sm
m
smvv
my
x /4.69
235
0
?x 00 yv
PROJECTILE MOTION IS PARABOLIC
• In order to demonstrate that
projectile motion is parabolic, the
book derives y as a function of x.
When we do, we find that it has
the form:
• This is the equation for a parabola
D R M A R T A S T A S I A K
D E P A R T M E N T O F C Y T O B I O L O G Y A N D P R O T E O M I C S
LINEAR MOMENTUM
lecture based on © 2016 Pearson Education, Ltd.
• Momentum and Its Relation to Force
• Conservation of Momentum
• Collisions and Impulse
• Conservation of Energy and Momentum in Collisions
• Elastic Collisions in One Dimension
• Inelastic Collisions
• Collisions in Two or Three Dimensions
• Center of Mass (CM)
• CM for the Human Body
• Center of Mass and Translational Motion
MOMENTUM AND ITS RELATION TO FORCE
• Momentum is a vector symbolized by the symbol p,
and is defined as
• The rate of change of momentum is equal to the
net force:
• This can be shown using Newton’s second law.
CONSERVATION OF MOMENTUM
• During a collision, measurements show that the
TOTAL MOMENTUM DOES NOT CHANGE:
CONSERVATION OF MOMENTUM
More formally, the law of conservation of momentum
states:
• The total momentum of an isolated system of
objects remains constant.
CONSERVATION OF MOMENTUM EXPERIMENT
Momentum conservation works for a rocket as long
as we consider the rocket and its fuel to be one
system, and account for the mass loss of the rocket.
CONSERVATION OF MOMENTUM EXPERIMENT
A Wad of Clay Hits Unsuspecting Sled
1 kg clay ball strikes 5 kg sled at 12 m/s and sticks
Momentum before collision:
(1 kg)(12 m/s) + (5 kg)(0 m/s)
Momentum after
= 12 kg·m/s → (6 kg)·(2 m/s)
ELASTIC COLLISION: BILLIARD BALLS
• Whack stationary ball with identical ball moving at velocity vcue
• To conserve both energy and momentum, cue ball stops
dead, and 8-ball takes off with vcue • Momentum conservation: mvcue = mvcue, after + mv8-ball • Energy conservation: ½mv2
cue = ½mv2cue, after + ½mv2
8-ball
• The only way v0 = v1 + v2 and v20 = v2
1 + v22 is if either v1 or
v2 is 0. • Since cue ball can’t move through 8-ball, cue ball gets
stopped.
8
8
DESK TOY PHYSICS
• The same principle applies to the suspended-ball
desk toy, which eerily “knows” how many balls you
let go…
• Only way to simultaneously satisfy energy and
momentum conservation
• Relies on balls to all have same mass
INELASTIC COLLISION
• Energy not conserved (absorbed into other paths)
• Non-bouncy: hacky sack, velcro ball, ball of clay
Momentum before = m1vinitial Momentum after = (m1 + m2)vfinal = m1vinitial (because conserved)
Energy before = ½m1v2
initial Energy after = ½ (m1 + m2)v
2final + heat energy
COLLISIONS AND IMPULSE
• During a collision, objects are deformed due to the
large forces involved.
• Since , we can
• Write
• The definition of impulse:
COLLISIONS AND IMPULSE
• The impulse tells us that we
can get the same change
in momentum with a large
force acting for a short time,
or a small force acting for a
longer time.
• This is why you should bend
your knees when you land;
why airbags work; and why
landing on a pillow hurts less
than landing on concrete.
CONSERVATION OF ENERGY AND MOMENTUM IN COLLISIONS
• Momentum is conserved
in all collisions.
• Collisions in which kinetic
energy is conserved as
well are called elastic
collisions,
and those in which it is not
are called inelastic.
ELASTIC COLLISIONS IN ONE DIMENSION
• Here we have two objects
colliding elastically. We know
the masses and the initial
speeds.
• Since both momentum and
kinetic energy are conserved,
we can write two equations.
This allows us to solve for the
two unknown final speeds.
INELASTIC COLLISIONS
• With inelastic collisions, some of the
initial kinetic energy is lost to
thermal or potential energy. It may
also be gained during explosions,
as there is the addition of chemical
or nuclear energy.
• A completely inelastic collision is
one where the objects stick
together afterwards, so there is only
one final velocity.
COLLISIONS IN TWO OR THREE DIMENSIONS
• Conservation of energy and momentum can also be used to analyze collisions in two or three dimensions, but unless the situation is very simple, the math quickly becomes unwieldy.
• Here, a moving object collides with an object initially at rest. Knowing the masses and initial velocities is not enough; we need to know the angles as well in order to find the final velocities
COLLISIONS IN TWO OR THREE DIMENSIONS PROBLEM SOLVING
• Choose the system. If it is complex, subsystems may be chosen where one or more conservation laws apply.
• Is there an external force? If so, is the collision time short enough that you can ignore it?
• Draw diagrams of the initial and final situations, with momentum vectors labeled.
• Choose a coordinate system.
• Apply momentum conservation; there will be one equation for each dimension.
• If the collision is elastic, apply conservation of kinetic energy as well.
• Solve.
• Check units and magnitudes of result.
SUMMARY
• Momentum of an object:
• Newton’s second law:
• Total momentum of an isolated system of objects is
conserved.
• During a collision, the colliding objects can be
considered to be an isolated system even if external
forces exist, as long as they are not too large.
• Momentum will therefore be conserved during
collisions.
SUMMARY
• In an elastic collision, total kinetic energy is also
conserved.
• In an inelastic collision, some kinetic energy is lost.
• In a completely inelastic collision, the two objects
stick together after the collision.
•