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Theoretical Mechanics KINEMATICS * Navigation: Right (Down) arrow – next slide Left (Up) arrow – previous slide Esc – Exit Notes and Recommendations: ruschev@tu- plovdiv.bg Technical University of Sofia Branch Plovdiv

Theoretical Mechanics KINEMATICS * Navigation: Right (Down) arrow next slide Left (Up) arrow previous slide Esc Exit Notes and Recommendations:

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Lecture 2 Determination of the Motion of a Particle Coordinate methods Rectangular coordinates If the particle is at point ( x, y, z ) on the curved path s shown in figure, then its location is defined by the position vector: When the particle moves, the x, y, z components of r will be functions of time; i.e., x = x(t),y = y(t), z = z ( t ), so that r = r(t). At any instant the magnitude of r is defined as And the direction of r is specified by the unit vector

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Page 1: Theoretical Mechanics KINEMATICS * Navigation: Right (Down) arrow  next slide Left (Up) arrow  previous slide Esc  Exit Notes and Recommendations:

Theoretical MechanicsKINEMATICS

* Navigation: Right (Down) arrow – next slide Left (Up) arrow – previous slide Esc – Exit Notes and Recommendations: [email protected]

Technical University of SofiaBranch Plovdiv

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Lecture 2

Vector methodConsider a particle located at a point on a space curve. The position of the particle, measured from a fixed point O, will be designated by the position vector r. Notice that both the magnitude and direction of this vector will change as the particle moves along the curve. This vector is a function of time:

The relation, which defines the position of the particle in an arbitrary time t, is called a law of the motion in a vector form or position vector equation.

r r t

Curvilinear Motion

Determination of the Motion of a ParticleParticle moving along a curve other than a straight line is in curvilinear motion.

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Lecture 2

Determination of the Motion of a ParticleCoordinate methodsRectangular coordinatesIf the particle is at point (x, y, z) on the curved path s shown in figure, then its location is defined by the position vector:

When the particle moves, the x, y, z components of r will be functions of time; i.e., x = x(t) ,y = y(t), z = z ( t ), so that r = r(t). At any instant the magnitude of r is defined as

And the direction of r is specified by the unit vector

r x i y j z k

2 2 2r x y z

/ru r r

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Lecture 2

Determination of the Motion of a ParticleCoordinate methods

s s t

Natural coordinatesWhen the path along which a particle travels is known, then it is often convenient to describe the motion using n and t coordinate axes which act normal and tangent to the path, respectively, and at the instant considered have their origin located at the particle.Law of motion in natural form is

where s is the curvilinear coordinate of the particle and t is the time. At a given instant of time s is located along a fixed line measured from a fixed point O to the particle.

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Lecture 2

Velocity and AccelerationVelocityVector method

• Consider a particle which occupies position P defined by at time t and P’ defined by at t + t, if , then

r

r

dtds

tsv

dtrd

trv

t

t

0

0

lim

lim

instantaneous velocity (vector)

instantaneous speed (scalar)

rrr

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Lecture 2

Velocity and AccelerationAccelerationVector method

dt

vdtva

t

0lim

instantaneous acceleration (vector)

• Consider velocity of particle at time t and velocity at t + t, where

v v

• In general, the acceleration vector is not tangent to the particle path and velocity vector.

vvv

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Lecture 2

Velocity and AccelerationRectangular Components of Velocity & Acceleration

• When position vector of particle P is given by its rectangular components, kzjyixr

• Velocity vector,

x y zdx dy dzv i j k xi y j zk v i v j v kdt dt dt

//

/

x

y

z

V dx dt xV dy dt y

V dz dt z

• Acceleration vector,

2 2 2

2 2 2

x y z

d x d y d za i j kdt dt dt

xi y j zk a i a j a k

2 2

2 2

2 2

/ /

/ /

/ /

x x

y y

z z

a dV dt d x dt x

a dV dt d y dt y

a dV dt d z dt z

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Lecture 2

Velocity and AccelerationTangential and Normal Components

• The velocity vector of a particle is tangent to the particle path. In general, the acceleration vector is not. Wish to express the acceleration vector in terms of the tangential and normal components.

• are tangential unit vectors for the particle path at P and P’. When drawn with respect to the same origin, and is the angle between them.

tt ee and

ttt eee

0 0

2sin 2

sin 2lim lim

2

1 02

t

tn n

tn

t ttn t t

t n

e

e e e

deed

d e edee e e

d de e

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Lecture 2

Velocity and AccelerationTangential and Normal Components

• With the velocity vector expressed asthe particle acceleration may be written as

t tdv dv de dv de d dsa e v e vdt dt dt dt d ds dt

butt

nde dse d ds vd dt

• Tangential component of acceleration reflects change of speed and normal component reflects change of direction.

• Tangential component may be positive or negative. Normal component always points toward center of path curvature.

After substituting,2 2

t n t ndv v dv v

a e e a adt dt

tevv

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Lecture 2

Velocity and AccelerationTangential and Normal Components

2 2

t n t ndv v dv va e e a adt dt

• Relations for tangential and normal acceleration also apply for particle moving along space curve.

• Plane containing tangential and normal unit vectors is called the osculating plane.

b t ne e e

• The normal to the osculating plane is found from

n

b

e principal normale binormal

• Acceleration has no component along the binormal.

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Lecture 2

Velocity and AccelerationSample Problem 1

A automobile is traveling on curved section of highway at 90 km/h. The driver applies brakes causing a constant deceleration rate.

Knowing that after 8 s the speed has been reduced to 72 km/h, determine the acceleration of the automobile immediately after the brakes are applied.

SOLUTION:

• Calculate tangential component and maximal normal component of acceleration.

• Determine acceleration magnitude and direction with respect to tangent to curve.

VA=95 km/h

830 m

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Lecture 2

Velocity and AccelerationSample Problem 1

90 km/h 25m/s72 km/h 20 m/s

SOLUTION:• Calculate tangential and normal components of

acceleration.

2

22max

max 2

20 25 ft s m0.6258 s s25m s m0.753830 m s

t

n

vat

va

• Determine acceleration magnitude and direction with respect to tangent to the curve.

22 2 20.625 0.753t na a a 2

m0.979s

a

1 1 0.753tan tan0.625

n

t

aa

50.3

at=0.625 m/s2

an=0.753 m/s2

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Lecture 2

Velocity and AccelerationSample Problem 2

The boxes travel along the industrial conveyor. If a box starts from rest at A and increases its speed such that

at = (0.2t) m/s2 , where t is in seconds, determine the magnitude of its acceleration when it arrives at point B.

SOLUTION:The position of the box at any

instant is defined from the fixed point A using the position or path coordinate s.

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Lecture 2

Velocity and AccelerationSample Problem 2

2 2

6.1422 3

0 0

0.1 0.1

0.1 6.142 0.0333

5.69

Bt

B

B

dsv t ds t dtdt

ds t dt t

t s

2 22

2 2 2 2 2

3.238 5.242 /2

1.138 5.242 5.36 /

BB n

B B Bt n

va m s

a a a m s

2

0 0

0.2 0.2

0.2 0.1

t

v t

dva v t dv t dtdt

dv t dt v t

To determine the acceleration components andit is first necessary to formulate v and so that they may be evaluated at B.

ta v 2 /na v v

The position of B is 2 2

3 6.1424Bs m

2

2

0.2 5.69 1.138 /

0.1 5.69 3.238 / , 2

B t

B

a m s

v m s m