ME 2226: DYNAMICS

CHAPTER 12.1 12.2CHAPTER 12.1 12.2INTRODUCTION& RECTILINEAR KINEMATICS: CONTINUOUSMOTION

Ryan Taylor

Mechanical and Industrial Engineering Dept.

rdtaylor@d.umn.edu

1

COURSE INTRODUCTION

Go over syllabus

Homework Format

Draw any necessary diagrams and tabulate the problem data

List knowns and unknowns from the problem List knowns and unknowns from the problem

Identify the formulas used and show all steps

Consistent significant figures

Circle or box your answers

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OBJECTIVES

Find the kinematic quantities (position, displacement, velocity, and acceleration) of a particle traveling along a straight path.

Relations between s(t), v(t), and a(t) for general rectilinear motion.rectilinear motion.

Relations between s(t), v(t), and a(t) when acceleration is constant.

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An Overview of Mechanics

Mechanics: The study of how bodies react to forces acting on them.

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

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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

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.

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

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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.

The average velocity of a particle during a time interval t is

vavg = r / tvavg = r / t

The instantaneous velocity is the time-derivative of position.

v = dr / dt

v = ds / dt

Average speed is the total distance traveled divided by elapsed time: (vsp)avg = sT / t

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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.

The instantaneous acceleration is the time derivative of velocity.

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

As the book indicates, the derivative equations for velocity and acceleration can be manipulated to get a ds = v dv

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

Acceleration can be positive (speed increasing) or negative (speed decreasing).

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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. 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:

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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:

tavv +== tv

dtadv a tavv co +=yields= o

c

v

dtadvo

2coo

s

t(1/2)a t vss ++=yields= t

os

dtvds

o

)s-(s2a)(vv oc2

o

2 +=yields= s

s

c

v

v oo

dsadvv9

a

EXAMPLE PROBLEM

Given: A particle travels along a straight line to the rightwith 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 PROBLEM 2

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 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/s2.11

ME 2226: DYNAMICS

CHAPTER 12.3RECTILINEAR KINEMATICS: ERRATICMOTION

Ryan Taylor

Mechanical and Industrial Engineering Dept.

rdtaylor@d.umn.edu

12

OBJECTIVES

Determine position, velocity, and acceleration of a particle using graphs.

s-t, v-t, a-t, v-s, and a-s diagrams

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ERRATIC MOTION(Section 12.3)

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

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.

concepts of differentiation and integration as used in dynamics.

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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.

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V-T GRAPH

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)

Also, the distance moved (displacement) of the particle is the area under the v-t graph

during time t.

Therefore, the acceleration vs. time (or a-t) graph can be constructed by finding the slope at various points along the v-t graph.

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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.

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A-S GRAPH

(v v ) = = area under the

s1

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 a ds = v dv ).

a-s graph

(v1 vo) = = area under thes0

a ds

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.

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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.

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EXAMPLE PROBLEM 1

Given: The s-t graph for a sports car moving along a straight road.

Find: The v-t graph and a-t graph over the time interval shown.

What is your plan of attack for the problem? 20

EXAMPLE PROBLEM 2

Given: The v-t graph shown.

Find: The a-t graph, average speed, and distance traveled for the 0 - 90 s interval.

Find slopes of the v-t curve and draw the a-t graph. Find the area under the curve. It is the distance traveled. Finally, calculate average speed (using basic definitions!).

Plan:

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ME 2226: DYNAMICS

CHAPTER 12.4 12.5CHAPTER 12.4 12.5CURVILINEARMOTION: GENERAL & RECTANGULAR COMPONENTS

Ryan Taylor

Mechanical and Industrial Engineering Dept.

rdtaylor@d.umn.edu

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OBJECTIVES

Describe the motion of a particle traveling along a curved path

Relate kinematic quantities in terms of the rectangular components of the vectors

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GENERAL CURVILINEAR MOTION(Section 12.4)

A particle moving along a curved path undergoes curvilinear motion. Since the motion is often three-dimensional, vectorsare used to describe the motion.

A particle moves along a curve defined by the path function, s.

The position of the particle at any instant is designated by the vectorr = r(t). Both the magnitude and direction of rmay vary with time.

If the particle moves a distance s along the curve during time interval t, the displacement is determined by vectorsubtraction: r = r - r

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VELOCITY

Velocity represents the rate of change in the position of a particle.

The average velocity of the particle during the time increment t isvavg = r/t .The instantaneous velocity is the time-The instantaneous velocity is the time-derivative of positionv = dr/dt .

The velocity vector, v, is alwaystangent to the path of motion.

The magnitude of v is called the speed. Since the arc length s approaches the magnitude of r as t0, the speed can be obtained by differentiating the path function (v = ds/dt). Note that this is not a vector!

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ACCELERATION

Acceleration represents the rate of change in the velocity of a particle.

If a particles velocity changes from v to v over a time increment t, the average accelerationduring that increment is:

aavg = v/t = (v - v)/taavg = v/t = (v - v)/tThe 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.

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CURVILINEAR MOTION: RECTANGULAR COMPONENTS(Section 12.5)

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 position of the particle can be defined at any instant by the position vector

r = x i + y j + z k .

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

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) .

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RECTANGULAR COMPONENTS: VELOCITY

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 + vzk

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

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.

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RECTANGULAR COMPONENTS: ACCELERATION

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 + azk

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

az = = = dvz /dt

vx x vy y

vz z

The direction of a is usually not tangent to the path of the particle.

vz z

The magnitude of the acceleration vector is

a = [(ax)2 + (ay)

2 + (az)2 ]0.5

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EXAMPLE PROBLEM 1

Given: The motion of two particles (A and B) is described by the position vectorsrA = [3t i + 9t(2 t) j] m and rB = [3(t

2 2t +2) i + 3(t 2) j] m.

Find: The point at which the particles collide and 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. 30

EXAMPLE PROBLEM 2

Given: The velocity of the particle is v = [ 16 t2 i + 4 t3 j + (5 t + 2) k] m/s. When t = 0, x = y = z = 0.

Find: The particles coordinate position and the magnitude of its acceleration when t = 2 s.

Plan:Plan:

Note that velocity vector is given as a function of time.

1) Determine the position and acceleration by integrating and differentiating v, respectively, using the initial conditions.

2) Determine the magnitude of the acceleration vector using t = 2 s.

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