Chapter 8KINEMATICS of a PARTICLE

BAA1113 Engineering

Mechanics

Topics COIn Your Text Book

(R.C. Hibbeler)

8.1 Rectilinear Kinematics:

Continuous/Erratic motion*CO3

12.1 – 12.3

8.2 Curvilinear Motion: Normal

and Tangential Component12.4 – 12.7

Chapter 8 KINEMATICS of a PARTICLE

* Analyze the kinematics of motion that involves force & acceleration and work & energy principle.

Lesson Outcomes• Describe the concept of position,

displacement, velocity and acceleration.

• Analyze particle motion along a straight line and represent the motion graphically.

• Analyze particle motion along a curved path using different coordinate systems.

APPLICATIONS

A sports car travels along a straight road.

Can we treat the car as a particle?

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

determine its position and velocity at some instant?

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.

RECTILINEAR KINEMATICS: CONTINIOUS MOTION

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:

òò =t

o

v

vo

dtadv òò =s

s

v

v oo

dsadvvor òò =t

o

s

so

dtvds

Position:

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:

ta v v co+=yields=òò

t

oc

v

v

dtadvo

2coo

s

t(1/2) a t v s s ++=yields=òòt

os

dtvdso

)s - (s2a )(v v oc

2

o2 +=yields=òò

s

sc

v

v oo

dsadvv

The car moves in a straight line such that for a short time its velocity is defined by v = (0.9t2 + 0.6t) m/s where t is in sec. Determine it position and acceleration when t = 3s. When t = 0, s = 0.

Example 12.1

Coordinate SystemPositive is to the right.

PositionWhen s = 0 when t = 0, we have

When t = 3s, s = 10.8m

ttdt

dsv 6.09.0 2

23

0

23

00

2

03.03.03.03.06.09.0 ttttsdtttds

tsts

Solution

Acceleration

Knowing v = f(t), the acceleration is determined from a = dv/dt, since this equation relates a, v and t.

When t = 3s, a = 6m/s2

6.08.16.09.0 2 tttdt

d

dt

dva

Solution

A rocket travel upward at 75m/s. When it is 40m from the ground, the engine fails. Determine max height sB reached by the rocket and its speed just before it hits the ground.

Example 12.3

Coordinate SystemOrigin O with positive upward.

Maximum HeightWe have:t = 0 vA = +75m/s, s = sB vB = 0, aC = -9.81m/s2

m 327)(222 BABCAB sssavv

Solution

VelocityThe negative root was chosen since the rocket is moving downward.

smsmv

ssavv

C

BCCBC

/1.80/1.80

)(222

Solution

Rectilinear Kinematics: Erratic Motion

APPLICATION

In many experiments, a velocity versus position (v-s) profile is obtained.

If we have a v-s graph for the tank truck, how can we determine its acceleration at position s = 1500 feet?

ERRATIC MOTION

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.

A-S GRAPH

a-s graph

½ (v1² – vo²) = = area under theòs2

s1

a ds

A more complex case is presented by the acceleration versus position or a-s graph. The area under the a-s curve represents the change in velocity

(recall ò a ds = ò v dv ).

This equation can be solved for v1, allowing you to solve for the velocity at a point. By doing this repeatedly, you can create a plot of velocity versus distance.

V-S GRAPH

Another complex case is presented by the velocity vs. distance or v-s graph. By reading the velocity v at a point on the curve and multiplying it by the slope of the curve (dv/ds) at this same point, we can obtain the acceleration at that point. Recall the formula

a = v (dv/ds).

Thus, we can obtain an a-s plot from the v-s curve.

A bicycle moves along a straight road such that it position is described by the graph as shown. Construct the v-t and a-t graphs for 0 ≤ t ≤ 30s.

Example 12.6

v-t GraphBy differentiating the eqns that defining the s-t graph, we have

By measuring the slope of the s-t graph at a given time instant,

6306;3010

6.03.0;100 2

dt

dsvtssts

tdt

dsvtsst

smt

svt /6

1030

30150 s;20

Solution

a-t GraphBy differentiating the eqns defining the lines of the v-t graph,

06;3010

6.06.0;100

dt

dvavst

dt

dvatvst

Solution

A test car starts from rest and travels along a straight track such that it accelerates at a constant rate for 10 s and then decelerates at a constant rate. Draw the v-t and s-t graphs and determine the time t’ needed to stop the car. How far has the car traveled?

Example 12.7

v-t GraphUsing initial condition v = 0 when t = 0,

When t = 10s, v = 100m/s,

When t = t’, v = 0 t’ = 60 s

tvdtdvasttv

10,10;1010000

1202,2;2;1010100

tvdtdvattstv

Solution

s-t Graph.Using initial conditions s = 0 when t = 0,

When t = 10s, s = 500m,

When t’ = 60s, s = 3000m

2

005,10;10;100 tsdttdstvst

ts

600120

1202;1202;6010

2

10500

tts

dttdstvststs

Solution

The v-s graph describing the motion of a motorcycle as shown. Construct the a-s graph of the motion and determine the time needed for the motorcycle to reach the position s = 120 m.

Example 12.8

a-s Grapha-s graph can be determined using a ds = v dv,

0

;15;12060

6.004.0

32.0;600

ds

dvva

vmsm

sds

dvva

svms

Solution

TimeWhen s = 0 at t = 0,

For second segment of motion,

s 8.05 t m, 60 sAt

3ln5)32.0ln(532.0

32.0;32.0;600

0

sts

dsdt

ds

v

dsdtsvms

st

o

s 12.05 t m, 120 sAt

05.41515

15;15;12060

6005.8

s

tds

dt

ds

v

dsdtvms

st

Solution

General Curvilinear Motion

APPLICATIONS

The path of motion of a plane can be tracked with radar and its x, y, and z coordinates (relative to a point on earth) recorded as a function of time.

How can we determine the velocity or acceleration of the plane at any instant?

APPLICATIONS

A roller coaster car travels down a fixed, helical path at a constant speed.

How can we determine its position or acceleration at any instant?

If you are designing the track, why is it important to be able to predict the acceleration of the car?

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 vectorr = 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 Ds along the curve during time interval Dt, the displacement is determined by vector subtraction: D 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 Dt isvavg = Dr/Dt .

The instantaneous velocity is the time-derivative of positionv = 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 Ds approaches the magnitude of Dr 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 Dt, the average acceleration during that increment is:

aavg = Dv/Dt = (v - v’)/Dt

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.

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/dtx 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

At any instant the horizontal position of the weather balloon is defined by x = (9t) m, where t is in second. If the equation of the path is y = x2/30, determine the distance of the balloon from the station at A, the magnitude and direction of the both the velocity and acceleration when t = 2 s.

Example 12.9

PositionWhen t = 2 x = 9(2)= 18 m, y = (18)2/30 = 10.8m The straight-line distance from A to B is

VelocityWe have

Solution

m 218.1018 22 r

smxdt

dyv

smtdt

dxv

y

x

/8.1030/

/99

2

VelocityWhen t = 2 s, the magnitude of velocity is

The x is tangent to the path is

AccelerationWe have

thus,

smv /1.148.109 22

2.50tan 1

x

yv v

v

2/4.5 and 0 smvava yyxx

222 /4.54.50 sma

900

4.5tan 1

a

Solution

For a short time, the path of the plane in the figure is described by y = (0.001x2)m. If the plane is rising with a constant velocity of 10m/s , determine the magnitudes of the velocity and acceleration of the plane when it is at y= 100 m

Example 12.10

When y = 100m x = 316.2mWhen vy = 10m/s t = 10s

VelocityUsing the chain rule,

The magnitude is

m/s 81.152.316002.010

002.0001.0 2

xx

xy

vv

xvxdt

dyv

m/s 7.1822 yx vvv

Solution

AccelerationUsing the chain rule,

When x = 316.2m, vx = 15.81m/s,

Magnitude of the plane’s acceleration is

xxxxy xavvxvxva 2002.0002.0002.0

22 m/s 791.02.31681.15002.00

0

xx

yy

aa

av

222 m/s 791.0 yx aaa

Solution

Motion of a Projectile

APPLICATIONS

A good kicker instinctively knows at what angle, q, and initial velocity, vA, he must kick the ball to make a field goal.

For a given kick “strength”, at what angle should the ball be kicked to get the maximum distance?

APPLICATIONS

A basketball is shot at a certain angle. What parameters should the shooter consider in order for the basketball to pass through the basket?

Distance, speed, the basket location, … anything else ?

APPLICATIONS

A firefighter needs to know the maximum height on the wall she can project water from the hose. What parameters would you program into a wrist computer to find the angle, q, that she should use to hold the hose?

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) tWhy 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)

For any given problem, only two of these three equations can be used. Why?

A sack slides off the ramp with a horizontal velocity of 12 m/s. If the height of the ramp is 6 m from the floor, determine the time needed for the sack to strike the floor and the range R where the sacks begin to pile up.

Example 12.11

Coordinate System

Origin at point A. Initial velocity is (vA)x = 12 m/s and (vA)y = 0m/sAcceleration between A and B is ay = -9.81 m/s2. Since (vB)x = (vA)x = 12 m/s, the 3 unknown are (vB)y, R and tAB

Solution

Vertical MotionVertical distance from A to B is known,

Horizontal Motion

sttatvyy ABABcAByAAB 11.12

1)( 2

m3.13

1.11120

)(

R

R

tvxx ABxAA

Solution

The chipping machine is designed to eject wood at chips vO = 7.5 m/s. If the tube is oriented at 30° from the horizontal, determine how high, h, the chips strike the pile if they land on the pile 6 m from the tube.

Example 12.12

Coordinate System

3 unknown h, tOA and (vB)y. Taking origin at O, for initial velocity of a chip,

Also,(vA)x = (vO)x = 6.5 m/s and ay = -9.81 m/s2

smv

smv

yO

xO

/75.3)30sin5.7()(

/5.6)30cos5.7()(

Solution

Horizontal Motion

Vertical MotionRelating tOA to initial and final elevation of the chips,

st

tvxx

OA

OAxA

9231.0

)( 00

mh

tatvyhy OAcOAyOA

38.12

1)(1.2 2

0

Solution

Curvilinear Motion:Normal and Tangential

Components

APPLICATIONS

Cars traveling along a clover-leaf interchange experience an acceleration due to a change in velocity as well as due to a change in direction of the velocity.

If the car’s speed is increasing at a known rate as it travels along a curve, how can we determine the magnitude and direction of its total acceleration?

Why would you care about the total acceleration of the car?

APPLICATIONS

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?

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

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) andut 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

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 MOTIONThere are some special cases of motion to consider.

1) The particle moves along a straight line.r => an = v2/ = 0 =>r 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

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)21[ +

2d2y/dx

When the skier reaches the point A along the parabolic path, he has a speed of 6m/s which is increasing at 2m/s2. Determine the direction of his velocity and the direction and magnitude of this acceleration at this instant. Neglect the size of the skier in the calculation.

Example 12.14

Coordinate SystemEstablish n, t axes at the fixed point A.Determine the components of v and a.

VelocityThe velocity is directed tangent to the path where

V make an angle of θ = tan-1 = 45° with the x axis

2

10

1, 1

20 x

dyy x

dx

smvA /6

Solution

Acceleration

Since ,the radius of curvature is

The acceleration becomes

mdxyd

dxdy28.28

/

])/(1[22

2/32

2

2

/}273.12{ smuu

uv

uva

nt

ntA

2

2

1

10

d y

dx

Solution

AccelerationThe magnitude and angle is

Thus, 57.5° – 45 ° = 12.5 °a = 2.37 m/s2

5.57

237.1

2tan

/37.2237.12

1

222

sma

Solution

Race car C travels round the horizontal circular track that has a radius of 90 m. If the car increases its speed at a constant rate of 2.1 m/s2, starting from rest, determine the time needed for it to reach an acceleration of 2.4 m/s2. What is its speed at this instant?

Example 12.15

Coordinate SystemThe origin of the n and t axes is coincident with the car at the instant.

AccelerationThe magnitude of acceleration can be related to its components using , thus

The velocity is

We have

22nt aaa 2.1 m/sta

0 ( ) 2.1t cv v a t t

222

/049.0 smtv

an

Solution

AccelerationThe time needed for the acceleration to reach 2.4m/s2 is

The speed at time t = 4.87s is

2 2

22 22.4 2.4 0.049 4.87 s

t na a a

t t

smtv /2.101.2

Solution

END