lecture notes for sections 12-1-12-2 (instructor version)

8/12/2019 Lecture Notes for Sections 12-1-12-2 (Instructor Version)

1/18

INTRODUCTION &

RECTILINEAR KINEMATICS: CONTINUOUS MOTION

Todays Objectives:

Students will be able to:

1. Find the kinematic quantities

(position, displacement, velocity,and acceleration) of a particle

traveling along a straight path.

In-Class Activities:

Reading Quiz

Applications

Relations between s(t), v(t),

and a(t) for generalrectilinear motion.

Relations between s(t), v(t),

and a(t) when acceleration is

constant. Concept Quiz

Group Problem Solving

Attention Quiz


2/18

READING QUIZ

1. In dynamics, a particle is assumed to have ____B_____.

A) both translation and rotational motions

B) only a mass

C) a mass but the size and shape cannot be neglected

D) no mass or size or shape, it is just a point

2. The average speed is defined as _____C_____.

A) r/t B) s/t

C) sT/t D) None of the above.


3/18

APPLICATIONS

The motion of large objects,

such as rockets, airplanes, or

cars, can often be analyzed

as if they were particles.

Why? Consider movement of center of mass only,neglecting rotation.

If we measure the altitude

of this rocket as a function

of time, how can wedetermine its velocity and

acceleration? s=f(t), v=ds/dt, a=dv/dt ,Assuming the rocket is moving only vertically.


4/18

APPLICATIONS

(continued)

A sports car travels along a straight road.

Can we treat the car as a particle? Yes, straight line and no rotation.

If the car accelerates at a constant rate, how can we

determine its position and velocity at some instant?

s=so+vot+act

2

v=vo+act


5/18

An Overview of Mechanics

Statics: The study of

bodies in equilibrium.Dynamics:

1. Kinematics concerned with

the geometric aspects of motion2. Kinetics - concerned with

the forces causing the motion

Mechanics: The study of how bodies

react to forces acting on them.


6/18

RECTILINEAR KINEMATICS: CONTINIOUS MOTION

(Section 12.2)

A particle travels along a straight-line path

defined by the coordinate axis s.

The total distance traveledby the particle, sT, is a positive scalar

that represents the total length of the path over which the particletravels.

Theposition 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 rand s are meters (m) or feet (ft).

The displacement of the particle is

defined as its change in position.Vector form: r = r - r Scalar form: s = s - s


7/18

VELOCITY

Velocity is a measure of the rate of change in the position of a particle.It is a vectorquantity (it hasboth magnitude and direction). The

magnitude of the velocity is called speed, with units of m/s or ft/s.

The average velocity of a particle during atime interval t is

vavg= r / t

The instantaneous velocity is the time-derivative of position.

v = dr / dt

Speed is the magnitude of velocity: v = ds / dt

Average speed is the total distance traveled divided by elapsed time:

(vsp)avg = sT / t


8/18

ACCELERATION

Acceleration is the rate of change in the velocity of a particle. It is avectorquantity. 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 (eliminate time)

The instantaneous acceleration is the timederivative of velocity.

Vector form: a = dv / dt

Scalar form: a = dv / dt = d2s / dt2

Acceleration can be positive (speed

increasing) or negative (speed decreasing).


9/18

SUMMARY OF KINEMATIC RELATIONS:

RECTILINEAR MOTION

Differentiateposition 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.

Velocity:

=

t

o

v

vo

dtadv

=

s

s

v

v oo

dsadvvor

=

t

o

s

so

dtvds

Position:


10/18

CONSTANT ACCELERATION

The three kinematic equations can be integrated for the special casewhen 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:

tavv co +=yields=

t

o

c

v

v

dtadvo

2

coo

s

t(1/2) atvss ++=

yields=

t

os

dtvds

o

)s-(s2a)(vv oc2

o

2 +=yields= s

s

c

v

v oo

dsadvv


11/18

EXAMPLE

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.

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.


12/18

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

=> a = 20 m/s2 (or in the direction) when t = 4 s

2) Calculate the distance traveled in 4s by integrating the

velocity using so = 0:

v = ds / dt => ds = v dt =>

=> s so = 2 t2 t3

=> s 0 = 2(4)2 (4)3 => s = 32 m ( or )

(ask question concerning path of particle)

=t

o

s

s

(4 t 3 t2) dtdso


13/18

CONCEPT QUIZ

1. A particle moves along a horizontal path with its velocityvarying with time as shown. The average acceleration of the

particle is ____D_____.

A) 0.4 m/s2 B) 0.4 m/s2

C) 1.6 m/s2 D) 1.6 m/s2

2. A particle has an initial velocity of 30 ft/s to the left. If it

then passes through the same location 5 seconds later with a

velocity of 50 ft/s to the right, the average velocity of the

particle during the 5 s time interval is ___D____.

A) 10 ft/s B) 40 ft/s

C) 16 m/s D) 0 ft/s(Can we calculate average speed from this problem statement? No! We do not know the total path distance)

t = 2 s t = 7 s

3 m/s 5 m/s


14/18

GROUP PROBLEM SOLVING

Given:Ball A is released from restat 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.(Know: For A so, s, vo and ac; For B s, so, and ac)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.


15/18



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) => t = 1.115 s


16/18



2) Now consider ball B. It is thrown 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

=> (vB)o = 31.4 ft/s


17/18

ATTENTION QUIZ

2. A particle is moving with an initial velocity of v = 12 ft/s

and constant acceleration of 3.78 ft/s2 in the same direction

as the velocity. Determine the distance the particle has

traveled when the velocity reaches 30 ft/s. (B)v2 = v0

2 + 2ac(s so); (30)2 + (12)2 + 2(3.78)(s 0); s = 100 ft

A) 50 ft B) 100 ft

C) 150 ft D) 200 ft

1. A particle has an initial velocity of 3 ft/s to the left ats0 = 0 ft. Determine its position when t = 3 s if the

acceleration is 2 ft/s2 to the right. (A)

A) 0.0 ft B) 6.0 ftC) 18.0 ft D) 9.0 ft

s = so + vot + act2 ; s = (0) + (-3)(3) + .5(2)(3)2 = -9 + 9 = 0


18/18

lecture notes for sections 12-1-12-2 (instructor version)

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