Sections 12.1 - 12.2

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INTRODUCTION RECTILINEAR KINEMATICS
CONTINUOUS MOTION

• Todays Objectives
• Students will be able to
• Find the kinematic quantities (position,
  displacement, velocity, and acceleration) of a
  particle traveling along a straight path.

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An Overview of Mechanics
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Kinematics is the branch of classical mechanics
that describes the motion of points, bodies
(objects) and systems of bodies (groups of
objects) without consideration of the causes of
motion. The study of kinematics is often referred
to as the geometry of motion. To describe
motion, kinematics studies the trajectories of
points, lines and other geometric objects and
their differential properties such as velocity
and acceleration. In mechanical engineering,
robotics and biomechanics8 to describe the
motion of systems composed of joined parts
(multi-link systems) such as an engine, a robotic
arm or the skeleton of the human body.
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RECTILINEAR KINEMATICS CONTINIOUS
MOTION (Section 12.2)
A particle travels along a straight-line path
defined by the coordinate axis s.
The position of the particle at any instant,
relative to the origin, O, is defined by the
position vector r, or the scalar s. Scalar s can
be positive or negative. Typical units for r
and s are meters (m) or feet (ft).
The total distance traveled by the particle, sT,
is a positive scalar that represents the total
length of the path over which the particle
travels.
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VELOCITY
Velocity is a measure of the rate of change in
the position of a particle. It is a vector
quantity (it has both magnitude and direction).
The magnitude of the velocity is called speed,
with units of m/s or ft/s.
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ACCELERATION
Acceleration is the rate of change in the
velocity of a particle. It is a vector quantity.
Typical units are m/s2 or ft/s2.
As the book indicates, the derivative equations
for velocity and acceleration can be manipulated
to get a ds v dv
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SUMMARY OF KINEMATIC RELATIONS RECTILINEAR MOTION
Differentiate position to get velocity and
acceleration.
v ds/dt a dv/dt or a v dv/ds
Integrate acceleration for velocity and
position.
Note that so and vo represent the initial
position and velocity of the particle at t 0.
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CONSTANT ACCELERATION
The three kinematic equations can be integrated
for the special case when acceleration is
constant (a ac) to obtain very useful
equations. A common example of constant
acceleration is gravity i.e., a body freely
falling toward earth. In this case, ac g
9.81 m/s2 32.2 ft/s2 downward. These equations
are
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Procedure for Analysis Coordinate System.
Establish a position coordinate s along the path
and specify its fixed origin and positive
direction. Since motion is along a straight
line, the vector quantities position, velocity,
and acceleration can be represented as algebraic
scalars. For analytical work the sense of s, v,
and a is then defined by their algebraic signs.
The positive sense for each of these scalars can
be indicated by an arrow shown alongside each
kinematic equation as it is applied. Kinematic
Equations. If a relation is known between any
two of the four variables a, v, s and t, then a
third variable can be obtained by using one of
the kinematic equations, a dv/ dt, v ds/ dt
or a ds v dv, since each equation relates all
three variables. Whenever integration is
performed, it is important that the position and
velocity be known at a given instant in order to
evaluate either the constant of integration if an
indefinite integral is used, or the limits of
integration if a definite integral is used.
Remember that Eqs. 12-4 through 12-6 have only
limited use. These equations apply only when the
acceleration is constant and the initial
conditions are s So and v Vo when t O.
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EXAMPLE
Given A particle travels along a straight line
to the right with a velocity of v ( 4 t 3 t2
) m/s where t is in seconds. Also, s 0 when
t 0.
Find The position and acceleration of the
particle when t 4 s.
Plan Establish the positive coordinate, s, in
the direction the particle is traveling. Since
the velocity is given as a function of time, take
a derivative of it to calculate the acceleration.
Conversely, integrate the velocity function to
calculate the position.
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EXAMPLE (continued)
Solution
1) Take a derivative of the velocity to
determine the acceleration. a dv / dt
d(4 t 3 t2) / dt 4 6 t gt a 20
m/s2 (or in the ? direction) when t 4 s
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GROUP PROBLEM SOLVING
Given Ball A is released from rest at a height
of 40 ft at the same time that ball B is thrown
upward, 5 ft from the ground. The balls pass one
another at a height of 20 ft.
Find The speed at which ball B was thrown upward.
Plan Both balls experience a constant downward
acceleration of 32.2 ft/s2 due to gravity. Apply
the formulas for constant acceleration, with ac
-32.2 ft/s2.
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GROUP PROBLEM SOLVING (continued)
Solution
1) First consider ball A. With the origin
defined at the ground, ball A is released from
rest ((vA)o 0) at a height of 40 ft ((sA )o
40 ft). Calculate the time required for ball A
to drop to 20 ft (sA 20 ft) using a position
equation.
sA (sA )o (vA)o t (1/2) ac t2 So, 20 ft
40 ft (0)(t) (1/2)(-32.2)(t2) gt t
1.115 s
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GROUP PROBLEM SOLVING (continued)
Solution
2) Now consider ball B. It is throw upward from
a height of 5 ft ((sB)o 5 ft). It must reach a
height of 20 ft (sB 20 ft) at the same time
ball A reaches this height (t 1.115 s). Apply
the position equation again to ball B using t
1.115s.
sB (sB)o (vB)ot (1/2) ac t2 So, 20 ft 5
(vB)o(1.115) (1/2)(-32.2)(1.115)2 gt (vB)o
31.4 ft/s
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End of the Lecture Example 12.1/2/3/4/5 textbook