Lecture Notes 12.4-12.5

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CURVILINEAR MOTION:GENERAL & RECTANGULAR COMPONENTS

Todays Objectives:Students will be able to:1. Describe the motion of a

particle traveling along acurved path.

2. Relate kinematic quantitiesin terms of the rectangularcomponents of the vectors.

In-Class Activities: Reading Quiz

Applications General Curvilinear Motion Rectangular Components of

Kinematic Vectors Concept Quiz Group Problem Solving In-Class Quiz

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READING QUIZ

1. In curvilinear motion, the direction of the instantaneous

velocity is alwaysA) tangent to the hodograph.B) perpendicular to the hodograph.C) tangent to the path.

D) perpendicular to the path.

2. In curvilinear motion, the direction of the instantaneousacceleration is always

A) tangent to the hodograph.B) perpendicular to the hodograph.C) tangent to the path.D) perpendicular to the path.

What is a hodograph?

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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 afunction of time.

How can we determine the velocityor acceleration of the plane at anyinstant?

Differentiate its position vector once for v and twice for a.

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APPLICATIONS (continued)

A roller coaster car travels downa fixed, helical path at a constantspeed.

How can we determine its position or acceleration at anyinstant? Position is a function of the trig functions ofangles (sine and cosine), the angular velocity which we knowis constant, time, and the helical radius of motion.Acceleration is the second derivative of the position.

If you are designing the track, why is it important to beable to predict the acceleration of the car? The mass of the car andcontents times the acceleration (towards the center of curvature and the z axis) must be reacted by the track

structure to withstand the load of the car/contents to prevent structural failure.

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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, vectors are used todescribe 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 thecurve during time interval t, thedisplacement is determined by vector subtraction : 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 particleduring the time increment t isv avg = r /t .

The instantaneous velocity is thetime-derivative of positionv = d r /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 sapproaches the magnitude of r as t0, the speed can beobtained 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 thevelocity of a particle.

If a particles velocity changes from v to v over atime increment t, the average acceleration duringthat increment is:

a avg = v/t = ( v - v)/tThe instantaneous acceleration is the time-derivative of velocity:

a = d v/dt = d 2 r /dt 2

A plot of the locus of points defined by the arrowheadof the velocity vector is called a hodograph . Theacceleration vector is tangent to the hodograph, butnot, 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 interms of its x, y, z or rectangular components , relative to a fixedframe of reference .

The magnitude of the position vector is: r = (x 2 + y 2 + z 2)0.5

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

The position of the particle can bedefined at any instant by the

position vector r = x i + y j + z k .

The x, y, z components may all befunctions of time, i.e.,

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

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RECTANGULAR COMPONENTS: VELOCITY

The magnitude of the velocityvector is

v = [(v x)2 + (v y)2 + (v z)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 = d r /dt = d(x i)/dt + d(y j)/dt + d(z k)/dt

Since the unit vectors i, j, k are constant in magnitude anddirection , this equation reduces to v = v x i + v y j + v z k

where v x = = dx/dt, v y = = dy/dt, v z = = dz/dtx y z

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RECTANGULAR COMPONENTS: ACCELERATION

The direction of a is usuallynot tangent to the path of the

particle.

The acceleration vector is the time derivative of thevelocity vector (second derivative of the position vector):

a = d v/dt = d 2 r /dt 2 = a x i + a y j + a z k

where a x = = = dv x /dt, a y = = = dv y /dt,

az = = = dv z /dt

vx x vy y

vz z

The magnitude of the acceleration vector is

a = [(a x)2 + (a y)2 + (a z)2 ]0.5

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EXAMPLE 1

Given: The motion of two particles (A and B) is described bythe position vectors r A = [3t i + 9t(2 t) j] m and r B = [3(t 2 2t +2) i + 3(t 2) j] m.

Find: The point at which the particles collide and theirspeeds just before the collision.

Plan: 1) The particles will collide when their positionvectors are equal, or r A = r B .

2) Their speeds can be determined by differentiatingthe position vectors.

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EXAMPLE 1 (continued)

1) The point of collision requires that r A = r B,so x A = x B and y A = y B .

Solution:

Set the x-components equal: 3t = 3(t 2 2t + 2)Simplifying: t 2 3t + 2 = 0Solving: 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: 3t 2 5t 2 = 0Solving: 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 commontime). Substituting this value into r A or r B yields

xA = x B = 6 m and y A = y B = 0

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2) Differentiate r A and r B to get the velocity vectors.

Speed is the magnitude of the velocity vector.

vA = (3 2 + 18 2) 0.5 = 18.2 m/svB = (6 2 + 3 2) 0.5 = 6.71 m/s

v B = d r B/dt = x B i + y B j = [ (6t 6) i + 3 j ] m/s

At t = 2 s: v B = [ 6 i + 3 j ] m/s

v A = d r A/dt = = [ 3 i + (18 18t) j ] m/s

At t = 2 s: v A = [ 3 i 18 j ] m/s

j yA i xA . +

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CONCEPT QUIZ

1. If the position of a particle is defined byr = [(1.5t 2 + 1) i + (4t 1) j ] (m), its speed at t = 1 s is

A) 2 m/s B) 3 m/s

C) 5 m/s D) 7 m/s

2. The path of a particle is defined by y = 0.5x 2. If thecomponent of its velocity along the x-axis at x = 2 m isvx = 1 m/s, its velocity component along the y-axis at this

position is

A) 0.25 m/s B) 0.5 m/s

C) 1 m/s D) 2 m/s

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EXAMPLE 2

Find: Determine the particles distance from the origin O and themagnitude of its velocity and acceleration when t = 1 s.

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Plan:


1) Integrate the x-velocity relation to determine the

particles x-position.

2) Differentiate the path relation to determine thevelocity and acceleration.

Solution:

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= 5 2 + 12.5 2 = 13.5 ft/s

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GROUP PROBLEM SOLVING

Given: The velocity of the particle isv = [ 16 t 2 i + 4 t 3 j + (5 t + 2) k] m/s.When t = 0, x = y = z = 0.

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

Plan: Note that velocity vector is given as a function of time.1) Determine the position and acceleration by

integrating and differentiating v, respectively, usingthe initial conditions.2) Determine the magnitude of the acceleration vector

using t = 2 s.

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GROUP PROBLEM SOLVING (continued)

Solution:1) x-components:

2) y-components:

Velocity known as: v x = x = dx/dt = (16 t 2 ) m/s

Position: = (16 t 2) dt x = (16/3)t 3 = 42.7 m at t = 2 s

Acceleration: a x = x = v x = d/dt (16 t 2) = 32 t = 64 m/s 2

x

dx0

t

0

Velocity known as: v y = y = dy/dt = (4 t 3 ) m/s

Position: = (4 t 3) dt y = t 4 = (16) m at t = 2 s

Acceleration: a y = y = v y = d/dt (4 t 3) = 12 t 2 = 48 m/s 2

y

dy0

t

0

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GROUP PROBLEM SOLVING (continued)

3) z-components:

Position vector: r = [ 42.7 i + 16 j + 14 k] m.

Acceleration vector: a = [ 64 i + 48 j + 5 k] m/s 2

Magnitude: a = (64 2 + 48 2 +5 2)0.5 = 80.2 m/s 2

4) The position vector and magnitude of the acceleration vectorare written using the component information found above.

Velocity is known as: v z = z = dz/dt = (5 t + 2) m/s

Position: = (5 t + 2) dt z = (5/2) t 2 + 2t = 14 m at t=2s

Acceleration: a z = z = v z = d/dt (5 t + 2) = 5 m/s 2

z

dz0

t

0

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1. If a particle has moved from A to B along the circular path in

4s, what is the average velocity of the particle ?A) 2.5 i m/s (r = 0i+0j, r=10i+0j; r/t=v avg=(10/4)i=2.5i)

B) 2.5 i +1.25 j m/s

C) 1.25 i m/sD) 1.25 j m/s

2. The position of a particle is given as r = (4t 2 i - 2x j) m.Determine the particles acceleration.

A) (4 i +8 j ) m/s 2 B) (8 i -16 j ) m/s 2

C) (8 i) m/s 2 D) (8 j ) m/s 2 x = 4t 2, y = 2x = 8t 2, r = 4t 2i 8t 2 j, dr/dt = v = 8ti 16tj; dv/dt = a = 8i-16j

A B

R=5mx

y

IN-CLASS QUIZ

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Lecture Notes 12.4-12.5