PE2113-Chapter 12 - Dynamics - Draft1

1

Chapter 12 – Kinematics of a Particle

2

Classification of Mechanics

3

Kinetics

4

12.1 Dynamics

• Dynamics deals with the accelerated motion

of a body.

• Dynamics is divided into 2 parts: Kinematics

and Kinetics – Kinematics deals with the geometric aspects

of the motion of a body

– Kinetics deals with analysis of the forces

casing the motion.

• Dynamics of a particle and then dynamics of

a rigid body.

5

Newton’s Contribution

6

Other Great Masters

7

9

12.2 Rectilinear Kinematics – Continuous Motion

• Rectilinear motion means motion on a straight line

path.

• Recall a particle: has mass but negligible size and

shape.

RECTILINEAR KINEMATICS: CONTINIOUS MOTION

(Section 12.2)

A particle travels along a straight-line path

defined by the coordinate axis s.

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.

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 displacement of the particle is

defined as its change in position.

Vector form: r = r’ - r Scalar form: s = s’ - s

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.

The average velocity of a particle during a

time 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

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

The instantaneous acceleration is the time

derivative of velocity.

Vector form: a = dv / dt

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

Acceleration can be positive (speed

increasing) or negative (speed decreasing).

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.

Velocity:

= t

o

v

v o

dt a dv = s

s

v

v o o

ds a dv v or = t

o

s

s o

dt v ds

Position:

Derivative of Acceleration

14

1. acceleration build-up, with

maximum positive jerk

2. constant maximum

acceleration (zero jerk)

3. acceleration ramp-down,

approaching the desired

maximum velocity, with

maximum negative jerk

4. constant maximum speed

(zero jerk, zero acceleration)

5. deceleration build-up,

approaching the desired

deceleration, with maximum

negative jerk

6. constant maximum

deceleration (zero jerk)

7. deceleration ramp-down,

approaching the desired

position at zero velocity, with

maximum positive jerk

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:

t a v v c o + = yields = t

o

c

v

v

dt a dv o

2 c o o

s

t (1/2) a t v s s + + = yields = t

o s

dt v ds o

) s - (s 2a ) (v v o c

2

o

2 + = yields = s

s

c

v

v o o

ds a dv v

16

17

12.3 Erratic Motion: Graphs

The approach builds on the facts that slope and differentiation

are linked and that integration can be thought of as finding the

area under a curve.

Graphing provides a good way to

handle complex motions that

would be difficult to describe

with formulas.

Graphs also provide a visual

description of motion and

reinforce the calculus concepts of

differentiation and integration as

used in dynamics.

S-T GRAPH

Plots of position vs. time can be

used to find velocity vs. time

curves. Finding the slope of the

line tangent to the motion curve at

any point is the velocity at that

point (or v = ds/dt).

Therefore, the v-t graph can be

constructed by finding the slope at

various points along the s-t graph.

V-T GRAPH

Also, the distance moved

(displacement) of the particle is the

area under the v-t graph during time t.

Plots of velocity vs. time can be used to

find acceleration vs. time curves.

Finding the slope of the line tangent to

the velocity curve at any point is the

acceleration at that point (or a = dv/dt).

Therefore, the acceleration vs. time (or

a-t) graph can be constructed by

finding the slope at various points

along the v-t graph.

A-T GRAPH

Given the acceleration vs. time

or a-t curve, the change in

velocity (v) during a time

period is the area under the a-t

curve.

So we can construct a v-t graph

from an a-t graph if we know the

initial velocity of the particle.

12.4 GENERAL CURVILINEAR MOTION

A particle moving along a curved path undergoes curvilinear motion.

Since the motion is often three-dimensional, vectors are used to

describe the motion.

The position of the particle at any instant is designated by the vector

r = r(t). Both the magnitude and direction of r may vary with time.

A particle moves along a curve

defined by the path function, s.

If the particle moves a distance s along the

curve during time interval t, the

displacement is determined by vector

subtraction: r = r’ - r

VELOCITY

Velocity represents the rate of change in the position of a

particle.

The average velocity of the particle

during the time increment t is

vavg = r/t .

The instantaneous velocity is the

time-derivative of position

v = dr/dt .

The velocity vector, v, is always

tangent to the path of motion.

The magnitude of v is called the speed. Since the arc length s

approaches the magnitude of r as t→0, the speed can be

obtained by differentiating the path function (v = ds/dt). Note

that this is not a vector!

ACCELERATION

Acceleration represents the rate of change in the

velocity of a particle.

If a particle’s velocity changes from v to v’ over a

time increment t, the average acceleration during

that increment is:

aavg = v/t = (v - v’)/t

The instantaneous acceleration is the time-

derivative of velocity:

a = dv/dt = d2r/dt2

A plot of the locus of points defined by the arrowhead

of the velocity vector is called a hodograph. The

acceleration vector is tangent to the hodograph, but

not, in general, tangent to the path function.

12.5 CURVILINEAR MOTION: RECTANGULAR COMPONENTS

It is often convenient to describe the motion of a particle in

terms of its x, y, z or rectangular components, relative to a fixed

frame of reference.

The magnitude of the position vector is: r = (x2 + y2 + z2)0.5

The direction of r is defined by the unit vector: ur = (1/r)r

The position of the particle can be

defined at any instant by the

position vector

r = x i + y j + z k .

The x, y, z components may all be

functions of time, i.e.,

x = x(t), y = y(t), and z = z(t) .

RECTANGULAR COMPONENTS: VELOCITY

The magnitude of the velocity

vector is

v = [(vx)2 + (vy)

2 + (vz)2]0.5

The direction of v is tangent

to the path of motion.

The velocity vector is the time derivative of the position vector:

v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt

Since the unit vectors i, j, k are constant in magnitude and

direction, this equation reduces to v = vx i + vy j + vz k

where vx = = dx/dt, vy = = dy/dt, vz = = dz/dt x y z • • •

RECTANGULAR COMPONENTS: ACCELERATION

The direction of a is usually

not tangent to the path of the

particle.

The acceleration vector is the time derivative of the

velocity vector (second derivative of the position vector):

a = dv/dt = d2r/dt2 = ax i + ay j + az k

where ax = = = dvx /dt, ay = = = dvy /dt,

az = = = dvz /dt

vx x vy y

vz z

• •• ••

•• •

•

The magnitude of the acceleration vector is

a = [(ax)2 + (ay)

2 + (az)2 ]0.5

EXAMPLE

Given: The motion of two particles (A and B) is described by

the position vectors

rA = [3t i + 9t(2 – t) j] m and

rB = [3(t2 –2t +2) i + 3(t – 2) j] m.

Find: The point at which the particles collide and their

speeds just before the collision.

Plan: 1) The particles will collide when their position

vectors are equal, or rA = rB .

2) Their speeds can be determined by differentiating

the position vectors.

EXAMPLE (continued)

1) The point of collision requires that rA = rB,

so xA = xB and yA = yB .

Solution:

Set the x-components equal: 3t = 3(t2 – 2t + 2)

Simplifying: t2 – 3t + 2 = 0

Solving: t = {3 [32 – 4(1)(2)]0.5}/2(1)

=> t = 2 or 1 s

Set the y-components equal: 9t(2 – t) = 3(t – 2)

Simplifying: 3t2 – 5t – 2 = 0

Solving: t = {5 [52 – 4(3)(–2)]0.5}/2(3)

=> t = 2 or – 1/3 s

So, the particles collide when t = 2 s (only common

time). Substituting this value into rA or rB yields

xA = xB = 6 m and yA = yB = 0

EXAMPLE (continued)

2) Differentiate rA and rB to get the velocity vectors.

Speed is the magnitude of the velocity vector.

vA = (32 + 182) 0.5 = 18.2 m/s

vB = (62 + 32) 0.5 = 6.71 m/s

vB = drB/dt = xB i + yB j = [ (6t – 6) i + 3 j ] m/s

At t = 2 s: vB = [ 6 i + 3 j ] m/s • •

vA = drA/dt = = [ 3 i + (18 – 18t) j ] m/s

At t = 2 s: vA = [ 3i – 18 j ] m/s

j yA i xA . + • •

12.6 MOTION OF A PROJECTILE

Projectile motion can be treated as two rectilinear motions, one in

the horizontal direction experiencing zero acceleration and the other

in the vertical direction experiencing constant acceleration (i.e.,

from gravity).

For illustration, consider the two balls on the

left. The red ball falls from rest, whereas the

yellow ball is given a horizontal velocity. Each

picture in this sequence is taken after the same

time interval. Notice both balls are subjected to

the same downward acceleration since they

remain at the same elevation at any instant.

Also, note that the horizontal distance between

successive photos of the yellow ball is constant

since the velocity in the horizontal direction is

constant.

KINEMATIC EQUATIONS: HORIZONTAL MOTION

Since ax = 0, the velocity in the horizontal direction remains

constant (vx = vox) and the position in the x direction can be

determined by:

x = xo + (vox) t Why is ax equal to zero (assuming movement through the air)?

KINEMATIC EQUATIONS: VERTICAL MOTION

Since the positive y-axis is directed upward, ay = – g.

Application of the constant acceleration equations yields:

vy = voy – g t

y = yo + (voy) t – ½ g t2

vy2 = voy

2 – 2 g (y – yo)

EXAMPLE II

Given: Projectile is fired with vA=150 m/s

at point A.

Find: The horizontal distance it travels (R) and

the time in the air.

Plan:

Establish a fixed x, y coordinate system (in this solution,

the origin of the coordinate system is placed at A).

Apply the kinematic relations in x- and y-directions.

EXAMPLE II (continued)

Solving for tAB first, tAB = 19.89 s.

Then, R = 120 tAB = 120 (19.89) = 2387 m

Solution: 1) Place the coordinate system at point A.

Then, write the equation for horizontal motion.

+ xB = xA + vAx tAB

where xB = R, xA = 0, vAx = 150 (4/5) m/s

Range, R will be R = 120 tAB

2) Now write a vertical motion equation. Use the distance equation.

+ yB = yA + vAy tAB – 0.5 g tAB2

where yB = – 150, yA = 0, and vAy = 150(3/5) m/s

We get the following equation: –150 = 90 tAB + 0.5 (– 9.81) tAB2

APPLICATIONS (continued)

A roller coaster travels down a

hill for which the path can be

approximated by a function

y = f(x).

The roller coaster starts from rest

and increases its speed at a

constant rate.

How can we determine its velocity

and acceleration at the bottom?

Why would we want to know

these values?

12.7 NORMAL AND TANGENTIAL COMPONENTS

When a particle moves along a curved path, it is sometimes convenient

to describe its motion using coordinates other than Cartesian. When the

path of motion is known, normal (n) and tangential (t) coordinates are

often used.

In the n-t coordinate system, the

origin is located on the particle

(the origin moves with the

particle).

The t-axis is tangent to the path (curve) at the instant considered,

positive in the direction of the particle’s motion.

The n-axis is perpendicular to the t-axis with the positive direction

toward the center of curvature of the curve.

NORMAL AND TANGENTIAL COMPONENTS (continued)

The position of the particle at any

instant is defined by the distance, s, along the curve from a

fixed reference point.

The positive n and t directions are

defined by the unit vectors un and ut,

respectively.

The center of curvature, O’, always

lies on the concave side of the curve.

The radius of curvature, r, is defined

as the perpendicular distance from

the curve to the center of curvature at

that point.

VELOCITY IN THE n-t COORDINATE SYSTEM

The velocity vector is always

tangent to the path of motion

(t-direction).

The magnitude is determined by taking the time derivative of

the path function, s(t).

v = v ut where v = s = ds/dt .

Here v defines the magnitude of the velocity (speed) and

ut defines the direction of the velocity vector.

ACCELERATION IN THE n-t COORDINATE SYSTEM

Acceleration is the time rate of change of velocity:

a = dv/dt = d(vut)/dt = vut + vut

. .

Here v represents the change in

the magnitude of velocity and ut

represents the rate of change in

the direction of ut.

.

.

.

a = v ut + (v2/r) un = at ut + an un.

After mathematical manipulation,

the acceleration vector can be

expressed as:

ACCELERATION IN THE n-t COORDINATE SYSTEM

(continued)

So, there are two components to the

acceleration vector:

a = at ut + an un

• The normal or centripetal component is always directed

toward the center of curvature of the curve. an = v2/r

• The tangential component is tangent to the curve and in the

direction of increasing or decreasing velocity.

at = v or at ds = v dv .

• The magnitude of the acceleration vector is

a = [(at)2 + (an)

2]0.5

SPECIAL CASES OF MOTION

There are some special cases of motion to consider.

1) The particle moves along a straight line.

r => an = v2/r = 0 => a = at = v

The tangential component represents the time rate of change in

the magnitude of the velocity.

.

2) The particle moves along a curve at constant speed.

at = v = 0 => a = an = v2/r .

The normal component represents the time rate of change in the

direction of the velocity.

SPECIAL CASES OF MOTION (continued)

3) The tangential component of acceleration is constant, at = (at)c.

In this case,

s = so + vo t + (1/2) (at)c t2

v = vo + (at)c t

v2 = (vo)2 + 2 (at)c (s – so)

As before, so and vo are the initial position and velocity of the

particle at t = 0. How are these equations related to projectile

motion equations? Why?

4) The particle moves along a path expressed as y = f(x).

The radius of curvature, r, at any point on the path can be

calculated from

r = ________________ ]3/2 (dy/dx)2 1 [ +

2 d2y/dx

THREE-DIMENSIONAL MOTION

If a particle moves along a space

curve, the n and t axes are defined as

before. At any point, the t-axis is

tangent to the path and the n-axis

points toward the center of curvature.

The plane containing the n and t axes

is called the osculating plane.

A third axis can be defined, called the binomial axis, b. The

binomial unit vector, ub, is directed perpendicular to the osculating

plane, and its sense is defined by the cross product ub = ut × un.

There is no motion, thus no velocity or acceleration, in the

binomial direction.

EXAMPLE

Given: A boat travels around a

circular path, r = 40 m, at a

speed that increases with

time, v = (0.0625 t2) m/s.

Find: The magnitudes of the boat’s

velocity and acceleration at

the instant t = 10 s.

Plan:

The boat starts from rest (v = 0 when t = 0).

1) Calculate the velocity at t = 10 s using v(t).

2) Calculate the tangential and normal components of

acceleration and then the magnitude of the

acceleration vector.

EXAMPLE (continued)

Solution:

1) The velocity vector is v = v ut , where the magnitude is

given by v = (0.0625t2) m/s. At t = 10s:

v = 0.0625 t2 = 0.0625 (10)2 = 6.25 m/s

2) The acceleration vector is a = atut + anun = vut + (v2/r)un. .

Tangential component: at = v = d(.0625 t2 )/dt = 0.125 t m/s2

At t = 10s: at = 0.125t = 0.125(10) = 1.25 m/s2

.

Normal component: an = v2/r m/s2

At t = 10s: an = (6.25)2 / (40) = 0.9766 m/s2

The magnitude of the acceleration is

a = [(at)2 + (an)

2]0.5 = [(1.25)2 + (0.9766)2]0.5 = 1.59 m/s2

APPLICATIONS

The cable and pulley system shown

can be used to modify the speed of

the mine car, A, relative to the speed

of the motor, M.

It is important to establish the

relationships between the various

motions in order to determine the

power requirements for the motor

and the tension in the cable.

For instance, if the speed of the cable (P) is known because we

know the motor characteristics, how can we determine the

speed of the mine car? Will the slope of the track have any

impact on the answer?

12.10 RELATIVE POSITION

The absolute position of two

particles A and B with respect to

the fixed x, y, z reference frame are

given by rA and rB. The position of

B relative to A is represented by

rB/A = rB – rA

Therefore, if rB = (10 i + 2 j ) m

and rA = (4 i + 5 j ) m,

then rB/A = (6 i – 3 j ) m.

RELATIVE VELOCITY

To determine the relative velocity of B

with respect to A, the time derivative of

the relative position equation is taken.

vB/A = vB – vA

or

vB = vA + vB/A

In these equations, vB and vA are called absolute velocities

and vB/A is the relative velocity of B with respect to A.

Note that vB/A = - vA/B .

RELATIVE ACCELERATION

The time derivative of the relative

velocity equation yields a similar

vector relationship between the

absolute and relative accelerations

of particles A and B.

These derivatives yield: aB/A = aB – aA

or

aB = aA + aB/A

SOLVING PROBLEMS

Since the relative motion equations are vector equations,

problems involving them may be solved in one of two ways.

For instance, the velocity vectors in vB = vA + vB/A could be

written as two dimensional (2-D) Cartesian vectors and the

resulting 2-D scalar component equations solved for up to

two unknowns.

Alternatively, vector problems can be solved “graphically” by

use of trigonometry. This approach usually makes use of the

law of sines or the law of cosines.

Could a CAD system be used to solve these types of problems?

LAWS OF SINES AND COSINES

a b

c

C

B A

Since vector addition or subtraction forms

a triangle, sine and cosine laws can be

applied to solve for relative or absolute

velocities and accelerations. As a review,

their formulations are provided below.

Law of Cosines: A bc c b a cos 2 2 2 2 - + =

B ac c a b cos 2 2 2 2 - + =

C ab b a c cos 2 2 2 2 - + =

Law of Sines: C

c

B

b

A

a

sin sin sin = =

EXAMPLE

Given: vA = 650 km/h

vB = 800 km/h

Find: vB/A

Plan:

a) Vector Method: Write vectors vA and vB in Cartesian

form, then determine vB – vA

b) Graphical Method: Draw vectors vA and vB from a

common point. Apply the laws of sines and cosines to

determine vB/A.

EXAMPLE (continued)

Solution:

vA = (650 i ) km/h

vB = –800 cos 60 i – 800 sin 60 j

= ( –400 i – 692.8 j) km/h

a) Vector Method:

vB/A = vB – vA = (–1050 i – 692.8 j) km/h

v A B /

= (-1050)2 +(-692.8)2 = 1258 km/h

= tan-1( ) = 33.4 1050

692.8

EXAMPLE (continued)

b) Graphical Method:

Note that the vector that measures the tip of B relative to A is vB/A.

vB

vA

120

vB/A

Law of Cosines:

(vB/A)2 = (800) 2 + (650) 2 − (800) (650) cos 120

vB/A = 1258 km/h

vB/A vA

sin(120 ) sin

= or = 33.4

Law of Sines:

Rectilinear Kinematics – Continuous Motion

82

Important Points

• Dynamics is concerned with bodies that have accelerated motion.

• Kinematics is a study of the geometry of the motion.

• Kinetics is a study of the forces that cause the motion.

• Rectilinear kinematics refers to straight-line motion.

• Speed refers to the magnitude of velocity.

• Average speed is the total distance traveled divided by the total

time. This is different from the average velocity, which is the

displacement divided by the time.

• A particle that is slowing down is decelerating.

• A particle can have an acceleration and yet have zero velocity.

• The relationship a ds = v dv is derived from a = dv/dt and v = ds/dt,

by eliminating dt.

83

Procedures 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, sand 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 = 0. 84

Chap. 13: Kinetics of a Particle

105

F = ma or Fx i + Fy j + Fz k = m(ax i + ay j + az k)

or, as scalar equations, Fx = max, Fy = may, and Fz = maz.

Example 13.2

Example 13.2 (continued)

Example 13.2 (continued)

13.5 NORMAL & TANGENTIAL COORDINATES

When a particle moves along a curved

path, it may be more convenient to

write the equation of motion in terms of

normal and tangential coordinates.

The normal direction (n) always points toward the path’s center of curvature.

In a circle, the center of curvature is the center of the circle.

The tangential direction (t) is tangent to the path, usually set as positive in

the direction of motion of the particle.

Ftut + Fnun+ Fbub = mat+man

NORMAL AND TANGENTIAL ACCERLERATIONS

The tangential acceleration, at = dv/dt, represents the time rate

of change in the magnitude of the velocity. Depending on the

direction of Ft, the particle’s speed will either be increasing or

decreasing.

The normal acceleration, an = v2/r, represents the time rate of

change in the direction of the velocity vector. Remember, an

always acts toward the path’s center of curvature. Thus, Fn

will always be directed toward the center of the path.

Recall, if the path of motion is

defined as y = f(x), the radius of

curvature at any point can be

obtained from

Example 13.9

Example 13.9 (continued)

14.1 WORK OF A FORCE

A force does work on a particle when the particle undergoes a displacement

along the line of action of the force.

Work is defined as the product of force and

displacement components acting in the same

direction. So, if the angle between the force and

displacement vector is , the increment of work

dU done by the force is

dU = F ds cos

By using the definition of the dot product and

integrating, the total work can be written as

r2

r1

U1-2 = F • dr

Principle of Work and Energy

114

Impulse and Momentum

115