Dynamics Lecture3 Curvilinear Motion - Normal and Tangent Components

Curvilinear Motion: Normal and Tangential Components

• When the path of motion of a particle is known, describe the path using n and t coordinates which act normal and tangent to the path• Consider your origin for the n and t axes to be located at the particle

Planar Motion• Consider particle P which is moving in a plane along a fixed curve, such that at a given instant it is at position s, measured from point O

• Consider a coordinate system that has origin at a fixed point on the curve on the curve, and at the instant, considered this origin happen to coincide with the location of the particle

• t axis is tangent to the curve at P and is positive in the direction of increasing s

• Designate this positive position direction with unit vector ut

• For normal axis, note that geometrically, the curve is constructed from series differential arc segments• Each segment ds is formed from the arc of an associated circle having a radius of curvature ρ (rho) and center of curvature O’

• Normal axis n is perpendicular to the t axis and is directed from P towards the center of curvature O’• Positive direction is always on the concave side of the curve, designed by un

• Plane containing both the n and t axes is known as the oscillating plane and is fixed on the plane of motion

Velocity.• Since the particle is moving, s is a function of time• Particle’s velocity v has direction that is always tangent to the path and has a magnitude that is determined by taking the time derivative of the path function s = s(t)

sv

uvv t

where

Acceleration• Acceleration of the particle is the time rate of change of velocity

tt uvuvva


• As the particle moves along the arc ds in time dt, ut preserves its magnitude of unity• When particle changes direction, it becomes ut’

ut’ = ut + dut

• dut stretches between the arrowhead of ut and ut’, which lie on an infinitesimal arc of radius ut = 1

nnnt uv

us

uu


2va

vdvdsava

uauaa

n

tt

nntt

• Magnitude of acceleration is the positive value of

22nt aaa

where

and


Consider two special cases of motion• If the particle moves along a straight line, then ρ → ∞ and an = 0. Thus , we can conclude that the tangential component of acceleration represents the time rate of change in the magnitude of velocity.

• If the particle moves along the curve with a constant speed, then and

vaa t

0vat /2vaa n


• Normal component of acceleration represents the time rate of change in the direction of the velocity. Since an always acts towards the center of curvature, this component is sometimes referred to as the centripetal acceleration• As a result, a particle moving along the curved path will have accelerations directed as shown


Three Dimensional Motion• If the particle is moving along a space curve, at a given instant, t axis is completely unique• An infinite number of straight lines can be constructed normal to tangent axis at P


• For planar motion, - choose positive n axis directed from P towards path’s center of curvature O’- The above axis also referred as principle normal to curve at P-ut and un are always perpendicular to one another and lies in the osculating plane


• For spatial motion, a third unit vector ub, defines a binormal axis b which is perpendicular to ut and un• Three unit vectors are related by vector cross

productub = ut X un• un is always on the concave side


PROCEDURE FOR ANALYSISCoordinate System• When path of the particle is known, establish a set of n and t coordinates having a fixed origin which is coincident with the particle at the instant• Positive tangent axis acts in the direction of the motion and the positive normal axis is directed toward the path’s center of curvature• n and t axes are advantageous for studying the velocity and acceleration of the particle


Velocity• Particle’s velocity is always tangent to the path• Magnitude of the velocity is found from the derivative of the path function

Tangential Acceleration • Tangential component of acceleration is the result of the time rate of change in the magnitude of velocity

sv


• Tangential component acts in the positive s direction if the particle’s speed is increasing and in the opposite direction if the seed is decreasing• For rectilinear motion,

vdvdsava tt • If at is constant,

)()(2

)(

)(21

020

2

0

200

ssavv

tavv

tatvss

cc

cc

cc


Normal Acceleration• Normal component of acceleration is the result of the time rate of change in the direction of the particle’s velocity• Normal component is always directed towards the center of curvature of the path along the positive n axis• For magnitude of the normal component,

2van


• If the path is expressed as y = f(x), the radius of the curvature ρ at any point on the path is determined from

22

2/32

/

])/(1[

dxyd

dxdy


EXAMPLE 12.14

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.

Coordinate System. Establish the origin of the n, t axes at the fixed point A on the path and determine the components of v and a along these axes.

Velocity. The velocity is directed tangent to the path.

1,201

10

2 xdx

dyxy

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

smvA /6

EXAMPLE 12.14

Acceleration. Determined from nt uvuva)/( 2

mdxyd

dxdy28.28

/

])/(1[22

2/32

The acceleration becomes

2

2

/}273.12{ smuu

uv

uva

nt

ntA

EXAMPLE 12.14

5.57

327.12

tan

/37.2237.12

1

222

sma

Thus, 57.5° – 45 ° = 12.5 °

a = 2.37 m/s2

EXAMPLE 12.14

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 System. The origin of the n and t axes is coincident with the car at the instant. The t axis is in the direction of the motion, and the positive n axis is directed toward the center of the circle.

Acceleration. The magnitude of acceleration can be related to its components using 22

nt aaa

t

tavv ct

1.2

)(0

EXAMPLE 12.15

222

/049.0 smtv

an

Thus,

The time needed for the acceleration to reach 2.4m/s2 is 22

nt aaa

Solving for t = 4.87 s

Velocity. The speed at time t = 4.87 s is

smtv /2.101.2

EXAMPLE 12.15

The box travels along the industrial conveyor. If a box starts from rest at A and increases its speed such that at = (0.2t) m/s2, determine the magnitude of its acceleration when it arrives at point B.

EXAMPLE 12.16

Coordinate System. The position of the box at any instant is defined by s, from the fixed point A. The acceleration is to be determined at B, so the origin of the n, t axes is at this point.

Acceleration. Since vA when t = 0

2

00

1.0

2.0

2.0

tv

dttdv

tvatv

t

(1)

(2)

EXAMPLE 12.16

The time needed for the box to reach point B can be determined by realizing that the position of B is sB = 3 + 2π(2)/4 = 6.142 m, since sA = 0 when t = 0

st

dttds

tdtds

v

B

tB

690.5

1.0

1.0

02142.6

0

2

EXAMPLE 12.16

Substituting into eqn (1) and (2),

smv

smva

B

BtB

/238.3)69.5(1.0

/138.1)690.5(2.0)(2

2

At B, ρB = 2 m, 2

2/242.5)( sm

va

B

BnB

2

22

/36.5

)242.5()138.1(

sm

aB

EXAMPLE 12.16

Dynamics Lecture3 Curvilinear Motion - Normal and Tangent Components

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