Motion in a plane in space

An ordered pair of coordinates, e.g. (7.3m, 2.5m), determines a point in a plane and an arrow from the origin to that point  is described as a position vector and can be used to describe the location of any point. The length of the position vector (the magnitude) records the distance from the origin to the point. An orientation records the angle between the position vector and some specified line (usually the x-axis).

The position vector of any point (x,y) can be specified as r = (x, y)

The origin is known as the zero vector, 0

The longest side of any right angled triangle, known as the hypotenuse, is given by Pythagoras’s Theorem,

c2 =  a2 + b2 or c = (a2 +b2)1/2

In addition, the ratios of the lengths a, b and c define the trigonometric ratios provided that θ < 90°

sin θ = b/c

cos θ = a/c

tan θ = b/a

A useful result that follows from Pythagoras’s Theorem is that (sin θ)2 + (cos θ)2 = 1

This is called an identity since it is true for all values of θ, and is usually written in the form,

sin2 θ + cos2 θ = 1

Some useful values (worth remembering) are;

sin 0° = 0 cos 0° = 1 tan 0° = 1
sin 30° = ½ cos 30° = √3/2 tan 30° = 1/√3
sin 45° = 1/√2 cos 45° = 1/√2 tan 45° = 1
sin 60° =  √3/2 cos 60° = ½ tan 60° = √3
sin 90° = 1 cos 90° = 0 tan 90° is undefined

In the case of a position vector it follows from Pythagoras’s theorem that

r = (x2 + y2)1/2

and

x = r cos θ

y = r sin θ

θ = arctan (y/x)

The displacement, sx = Δx = x2-x1 and sy = Δy = y2-y1

So the displacement vector s = (sx, sy) = Δr = r2-r1 = (x2-x1, y2-y1)

The magnitude of the position vector is,

s = (sx2 – sy2)1/2

And when the angle between the x-axis and the displacement is θ

sx = s cos θ

sy = s sin θ

If a and b are displacement vectors, if the arrow representing b is drawn from the head of the arrow representing a, then an arrow drawn from the tail of a to the head of b represents their resultant, the displacement c = a + b. This is sometimes known as the triangle rule for adding displacement vectors.

c = a + b means that cx = ax + bx and cy = ay + by

If a displacement vector represents the change of position in a time interval, the average velocity vector in that time interval is

<v> = Δ r / Δ t = (Δx/Δt, Δy/Δt)

i.e.

v = dr/dt = (dx/dt,dy/dt)

The speed of a particle is given by the magnitude of the vector,

v =  (vx2 + vy2)

The instantaneous acceleration vector is the derivative of the velocity vector,

a = dv/dt = (dvx/dt, dvy/dt)

and the acceleration has a magnitude,

a = (ax2 + ay2)1/2

and the acceleration components are,

ax – a cos θ and ay = a sin θ

The essential points that Galileo had grasped was that, the horizontal and vertical movements are independent, apart from the fact that they must both have the same direction 🙂

The vertical component of a projectile’s motion is an example of uniformly accelerated motion, in which the constant acceleration is due to the Earth’s gravity, directed downward, and has a magnitude of g (≈ 9.8ms-1). The horizontal component is an example of uniform motion, in which the constant horizontal velocity is determined when the projectile is launched.

The time of a projectile’s flight is,

T = (2u sinθ)/g

The range is

R = | (2u2 sinθ cosθ )/g|

And for a fixed launch speed u, the maximum range Rmax is

Rmax = u2/g

And is achieved when the launch angle is 45°

The path, or trajectory, followed by a projectile is

sy = asx2 + bsx

where a and b are constants given by,

a = -g/(2ux2) and  b = uy/ux

Hopefully, you can see this form of equation is a quadratic function and therefore; A projectile travelling close to the Earth (so that g remains constant), and in the absence of air resistance, follows a trajectory that has the form of a parabola. A Quadratic Equation has the form

ax2 + bx + c = 0

its two solutions are always given by

x = (-b± (b2-4ac)1/2)/2a

Note:

if b2 >  4ac then the formula gives two solutions

if b2 = 4ac the two solutions are identical

if b2 < 4ac then the root are complex  (i.e. as in complex number, not as in ‘complicated’ 🙂 )

Quadratics can also be solved by factorization, i.e. written in the form

a(x – α)(x – β) = 0

where α + β = -b/a and αβ = c/a

For bodies launched from a height,

T = (uy + (uy2 + 2gh)1/2)/g

And

R = |(uxuy + ux(uy2+2gh)1/2)/g|

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~ by jamesdow2013 on March 22, 2013.

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