1-mar-07dr. m. s. kariapper1 in this chapter we will learn about the kinematics (displacement,...
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![Page 1: 1-Mar-07Dr. M. S. Kariapper1 In this chapter we will learn about the kinematics (displacement, velocity, acceleration) of a particle in two or three dimensions](https://reader033.vdocument.in/reader033/viewer/2022051401/56649d6c5503460f94a4b398/html5/thumbnails/1.jpg)
1-Mar-07 Dr. M. S. Kariapper 1
• In this chapter we will learn about the kinematics (displacement, velocity, acceleration) of a particle in two or three dimensions.
• Projectile motion, Superposition principle
• Uniform Circular Motion
• Relative Motion
Chapter 4:
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1-Mar-07 Dr. M. S. Kariapper 2
Displacement in a plane
The displacement vector r:
f ir r r
Displacement is the straight line between the final and initial position of the particle.
That is the vector difference between the final and initial position.
ˆ ˆ ˆThe vector r is given by , ,r xi yj zk x y z
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1-Mar-07 Dr. M. S. Kariapper 3
Average Velocity
Average velocity v:
avg
rv
t
Average velocity: Displacement of a particle, r, divided by time interval t.
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1-Mar-07 Dr. M. S. Kariapper 4
Instantaneous Velocity
0lim , ,t
r dr dx dy dzv
t dt dt dt dt
Instantaneous velocity : Limit of the average velocity as t approaches zero. The direction v is always tangent to the particles path.
The instantaneous velocity equals the derivative of the position vector with respect to time.
The magnitude of the instantaneous velocity vector is called the speed (scalar) vv
v
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1-Mar-07 Dr. M. S. Kariapper 5
Checkpoint 2
The figure shows a circular path taken by a particle. If the instantaneous velocity of the particle is , through which quadrant is the particle moving when it is traveling (a) clockwise and (b) counterclockwise around the circle?
ˆ ˆ(2 / ) (2 / )v m s i m s j
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1-Mar-07 Dr. M. S. Kariapper 6
Average Acceleration
Average acceleration:
f iavg
f i
v v va
t t t
Average acceleration: Change in the velocity v divided by the time t during which the change occurred.
Change can occur in direction and magnitude!
Acceleration points along change in velocity v!
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1-Mar-07 Dr. M. S. Kariapper 7
Instantaneous Acceleration
Instantaneous acceleration: limiting value of the ratio
as t goes to zero.
Instantaneous acceleration equals the derivative of the velocity vector with respect to time.
tv
0lim , ,yx z
t
dvdv dvv dva
t dt dt dt dt
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1-Mar-07 Dr. M. S. Kariapper 8
Kinetic Quantities in 1-D and 2-D
Quantities 1 Dimension 2 Dimension
Displacement
Average Velocity
Inst. Velocity
Average Acc.
Inst. Acc.
ox x x or r r
oavg
x x xv
t t
o
avg
r r rv
t t
dt
rd
t
rv
t
0
lim0
limt
x dxv
t dt
o
avg
v v va
t t
o
avg
v v va
t t
2
2
0lim
dt
rd
dt
vd
t
va
t
2
20lim
t
v dv d xa
t dt dt
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1-Mar-07 Dr. M. S. Kariapper 9
Two- (or three)-dimensional motion with constant acceleration a
Trick 1:
The equations of motion we derived before (e.g. kinematic equations) are still valid, but are now in vector form.
Trick 2 (Superposition principle):
Vector equations can be broken down into their x- and y- components. Then calculated independently.
jvivv yx
jyixr
Position vector: Velocity vector:
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1-Mar-07 Dr. M. S. Kariapper 10
Two-dimensional motion with constant acceleration
ov v at
Velocity as function of time
21
2o or r v t at
Position as function of time:
x ox x
y oy y
v v a t
v v a t
2
2
1
21
2
o ox x
o oy y
x x v t a t
y y v t a t
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1-Mar-07 Dr. M. S. Kariapper 11
A particle with velocity vv=(-2.0ii+4.0jj)m/s at t=0 undergoes a constatnt acceleration a of magnitude a =3.0m/s2 at an angle = 130° from the positive direction of the x-axis. What is the particle’s velocity v at t = 5.0s, in unit vector notation and as a magnitude and angle.
x ox xv v a t
What is ax and ay?
2219 /x yspeed v v v m s
y oy yv v a t
2 2cos (3.0 / )(cos130) 1.93 /xa a m s m s 2 2sin (3.0 / )(sin130) 2.30 /ya a m s m s
2.0 ( 1.93)(5) 11.65 /x ox xv v a t m s
Sample Problem 4-5
4.0 (2.30)(5) 15.50 /y oy yv v a t m s
ˆ ˆ: ( 12 / ) (16 / )Answer v m s i m s j
1tan 127 180 53 127y
x
v
v
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1-Mar-07 Dr. M. S. Kariapper 12
Projectile motion
Two assumptions:
1. Free-fall acceleration g is constant.
2. Air resistance is negligible.
- The path of a projectile is a parabola (derivation: see book).
- Projectile leaves origin with an initial velocity of vo.
- Projectile is launched at an angle o
- Velocity vector changes in magnitude and direction.
- Acceleration in y-direction (vertical) is -g.
- Acceleration in x-direction (horizontal) is 0.
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1-Mar-07 Dr. M. S. Kariapper 13
y yov v gt
21
2o yoy y v t gt
x xov v
o xox x v t
Acceleration in x-direction is 0. Acceleration in y-direction is -g.
(Constant velocity) (Constant acceleration)
Projectile motion
Superposition of motion in x-direction and motion in y-direction
The horizontal motion and vertical motion are independent of each other; that is, neither motion affects the other.
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1-Mar-07 Dr. M. S. Kariapper 14
Simultaneous fall demo
Which ball will hit the ground first?
• Straight drop• Straight out• Both at the same
time
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1-Mar-07 Dr. M. S. Kariapper 15
A battleship simultaneously fires two shells at enemy ships.
If the shells follow the parabolic trajectories shown, which ship gets hit first?
A.
B.
C. Both hit at the same time.
D. Need more information.
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1-Mar-07 Dr. M. S. Kariapper 16
Hitting the bull’s eye. How’s that?
Demo. Explanation using Simulation
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1-Mar-07 Dr. M. S. Kariapper 17
Example for a Projectile Motion• A stone was thrown upward from the top of a cliff at an angle
of 37o to horizontal with initial speed of 65.0m/s. If the height of the cliff is 125.0m, how long is it before the stone hits the ground?
cos 65.0 cos37 51.9 /xo ov v m s
21125.0
2yoy v t gt
02502.7880.92502.78 22 tttgt
80.92
)250(80.942.782.78 2
t
stst 4.10or 43.2 st 4.10
sin 65.0 sin 37 39.1 /yo o ov v m s
Since negative time does not exist.
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1-Mar-07 Dr. M. S. Kariapper 18
Example cont’d• What is the speed of the stone just before it hits the
ground?
y yov v gt
22 2 251.9 62.8 81.5 /x yv v v m s
x xov v
• What are the maximum height and the maximum range of the stone?
sin 39.1 9.80 10.4 62.8 /o ov gt m s
cos 65.0 cos37 51.9 /o ov m s
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1-Mar-07 Dr. M. S. Kariapper 19
Uniform Circular Motion
Motion in a circular path at constant speed.
• Velocity is changing, thus there is an acceleration!!
• Acceleration is perpendicular to velocity
• Centripetal acceleration is towards the center of the circle
• Magnitude of acceleration is
• r is radius of circler
var
2
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1-Mar-07 Dr. M. S. Kariapper 20
Relative MotionResults of Physical measurements in different reference frames could be different
PA PB BAr r v t
PA PBBA
d r d rv
dt dt
PA PB BAv v v
Consider that you are driving a car (reference frame B) with (constant) velocity relative to stationary frame A. To you (B), an object (P) in the car does not move while to the person (A) outside the car P is moving in the same speed and direction as your car is.
BAv
O
Frame A
PAr
BAv t
O’
Frame BP
PBr
BAv
PA PBa a
Since we consider only the case where is constant:BAv
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1-Mar-07 Dr. M. S. Kariapper 21
Relative Motion
Moving frame of reference
A boat heading due north crosses a river with a speed of 10.0 km/h. The water in the river has a speed of 5.0 km/h due east.
(a) Determine the velocity of the boat.
(b) If the river is 3.0 km wide how long does it take to cross it?
be br re v v v
In general we have PA PB BAv v v
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1-Mar-07 Dr. M. S. Kariapper 22
Performance Objective
1. Express the position of an object using vector notation.
2. Express the velocity of an object using vector notation.
3. Express the acceleration of an object using vector notation.
4. Analyze projectile motion using x and y components.
5. Explain what is meant by uniform circular motion.
6. Explain what is meant by centripetal acceleration and force..
7. Discuss relative motion in one dimension.
8. Discuss relative motion in two dimensions.
9. Solve word problems involving two and three-dimensional motion.