Jai Guru DevaSri Ganesaya NamahaJai Hind

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Course Offered by:

Prof. Bhanuprakash Tallapragada

Dept. of Marine Engineering

College of Engineering

Andhra University

Visakhapatnam - 530003

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Engineering Mechanics – II (Dynamics)

Chapter – 12

Lecture 2

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General Curvilinear Motion

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Curvilinear Motion occurs when the particle moves

along a curved path.

Since the path described is in Three-Dimensions,

vector analysis is used to formulate the particle’s

position, velocity, and acceleration.

In this lecture we look into the general aspects of

curvilinear motion.

Introduction

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Position

Consider a particle located at P on a space curve

defined by the path function ‘s’.

The position vector is r = r (t)

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Displacement During a small interval of time

∆t, particle moves by a

distance ∆s along the curve to

a new position P/ defined by

The displacement ∆r

represents the change in the

particle’s position

r r r′ = + ∆

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Velocity

0lim

avg

t

rvt

r drvt dt∆ →

∆=

∆∆

= =∆

1. dr is tangent to the curve at P

2. Direction of v is also tangent to the curve

3. As ∆t → 0, ∆r approaches ∆s, hence speed (v – scalar) is given by dsv

dt= Thus speed can be obtained by

differentiating path function wrt time

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avgvat

∆=

∆

AccelerationParticle has vel v at ‘t’, at t + ∆t, velocity is v/ = v + ∆v

2

2

dv d radt dt

= =

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Hodograph• Hodograph is the locus of the arrowhead of the velocity

vector

• ∆v will approach the tangent to the hodograph as ∆tgoes to zero

• Acceleration will always be tangent to the hodographand never tangential to the path

• Velocity is always tangential to the path

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1. Position:

2. Here x = x(t), y = y(t), z = z(t), so r = r (t)

3. The magnitude of r is always positive and defined by

4. Direction of r is given by unit vector ur = r/r

Curvilinear Motion : Rectangular Components

r xi yj zk= + +

2 2 2r x y z= + +

i, j, k frame of reference is fixed

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Velocity( ) ( ) ( )

( )

x y z

dr d d dv xi yj zkdt dt dt dt

d dx dixi i xdt dt dt

drv v i v j v kdt

= = + +

= +

= = + +

Magnitude is given by

2 2 2x y zv v v v= + +

Direction is given by uv = v/v

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Acceleration

, ,

x y z

x x y y z z

dva a i a j a kdt

where a v a v a v

= = + +

= = =

2 2 2x y za a a a= + + Magnitude

Direction is specified by ua = a/a

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Suppose you are shooting a projectile up the hill, what angle must you shoot so that it will go the maximum distance along the hill?

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Here’s a classic example that every beginning physics student has

traditionally learned.

A hunter sees a monkey in a tree, and decides to shoot it. He

knows that this particular species of monkey always falls

from the tree at the instant the shot is fired. At what angle

must the hunter aim, in order to hit the monkey?

Monkey and Hunter

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Monkey and Hunter

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The first step is to solve for the time at which

impact occurs.

The x-component of the velocity of the

projectile is v0 cos θ,

The distance in the x-direction is D

so the time of impact is

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At the time of impact, the z-coordinates of themonkey and bullet must be the same:

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This says that the hunter should aim directly at

the monkey!

The downward acceleration of the monkey

exactly compensates for the downward

acceleration of the bullet, as long as they start

falling at the same time.

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It may at first seem a bit surprising that this result

doesn’t depend on the speed of the bullet

But it makes sense when you think about it.

The faster the bullet, the less time the bullet has to fall,

but the monkey also has less time to fall, so the speed

has no effect.

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Example:

Estimate the maximum distance a long jumper can

jump. Determine some reasonable values to use in

this problem.

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First assume vI = 30 ft/sec (a world class sprinter). Let θ = 45° to maximize the range.

Finally use g = 32 ft/sec2. The range equation gives

R = 28.125 ft

World Record by Bob Beamon in

Mexico Olympics in 1968 is 29ft 2.5 in

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Normal and Tangential Components When path is known – n-t coordinate system

convenient Origin located at the particle

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Normal and Tangential Components (cont)1. Planar Motion:

2. Particle P is moving in a plane along a fixed curve

3. P is located at distance ‘s’ from a point ‘O’

4. Coordinate system is considered with Origin at P.

5. t-axis is tangent to the curve

6. Positive t-direction represented by ut (unit vector)

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The normal coordinate will be directed towards O/, the centre of

curvature of the plane curve.

This is always on the concave side of the curve

Positive direction is un.

The plane containing the n-t axes is called the osculating plane,

and in this case is fixed in the plane of motion

Normal and Tangential Components (cont)

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Velocity:1. Velocity is always tangential to the path2. Path s = s(t)3. Magnitude of Velocity = v = ds/dt4. V= v (scalar) x ut (Unit vector - direction)

Normal and Tangential Components (cont)

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Normal and Tangential Components (Cont)

Acceleration

Acceleration is the time

rate of change of velocity

v v= = +t ta v u u

v= tV u

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Normal and Tangential components (Cont)

1. When particle moves along arc ds in time dt, magnitude of ut does not change (magnitude = 1)

2. Its direction changes to u/t

3. Thus u/t = ut + dut

4. Magnitude of dut= ut dθ =1x dθ

5. Direction is perpendicular to ut i.e., along un

6. Therefore dut = dθ un

n tv v a a= = + = +t t n ta v u u u u

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Normal and Tangential Components ( cont)

t n t

2

n

2 2t n

sds=ρdθ; θ=ρ

v

dv dv ds dt dv ds dva a where a =v = × × = × =vdt dt dt ds ds dt ds

va =ρ

a= a a

t n n n

t n

su u u u

a u u

= θ = =ρ ρ

= + =

+

Since

ds

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Normal and Tangential Components (cont)Summary

1. Straight line motion ρ = ∞, an = 0, at = dv/dt

2. Curved motion constant speed at = dv/dt = 0, an = v2/ρ –

centripetal acceleration

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Normal and Tangential components (Cont)

Radius of curvature is given by

32 2

2

2

dy1+dx

ρ=d ydx

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Cylindrical Components

1. In Some Engineering Problems, it is often

convenient to use Cylindrical Coordinate system

r, θ, z

2. If the problem is in plane we use Polar Coordinates

r, θ

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Cylindrical Components (contd)

Polar Coordinates

R extends outwards from origin to the particle

θ is measured ccw between fixed reference line and r-axis

Positive directions of r and θ are given by unit vectors ur, uθ

These are perpendicular to one another

ur extends from P along increasing r when θ is held fixed

uθ extends from P in a direction that occurs when r is held fixed and θ is increased

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Position: At any instant the position of the particle is defined by the position vector

rrr u=

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VelocityInstantaneous velocity is obtained by taking time derivative of r.

r rr rv r u u= = +

0

rr t t tx

uu lim lim uθ∆ → →∞

∆ ∆θ = = ∆ ∆

rθv vrv u uθ= +ru uθ= θ

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Velocity Graphically the velocity components are shown

rv r and v rθ= = θ

Magnitude of V is given by 2 2v= (r +(rθ)

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Acceleration

r r r r rr ra v u u u u uθ θ θ= = + + θ + θ + θ

0 0t t

r

r

lim limt t

ua a

r

r

uu u

ua u u

θθ ∆ → ∆ →

θ

θ θ

∆ ∆θ = = − ∆ ∆ = −θ

= +

2 2ra r r a r rθ= − θ = θ + θ

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Cylindrical Coordinates

In a space curve we have three

coordinates (3 DOF) for a

particle and so we use r, θ, z

( ) ( )2

r z

r r z

r-rθ rθ+2rθ z

P r z

r z

r z

r u u

v u u u

a u u uθ

θ

= +

= + θ +

= + +

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Time Derivatives ( Issues)

1. If the coordinates are defined as time parametricequations r = r(t), θ= θ(t), then derivatives are easy

2. In some cases the relationship between r and θ maybe given

Thus of the four time derivatives

namely if we know two,

then we can get the other two from

these equations

r,r,θ,θ