Motion in 2D o Vectors o Projectile Motion Physics -101 Piri Reis University 2010

o Aim of the lecture  Cover material relevant to 2-D and 3-D kinematics: speed, acceleration vector representation projectile motion o Main learning outcomes  familiarity with  vectors  vector algebra and calculus  equations of motion using vectors  projectile motion  ability to  solve kinetics problems in multiple dimensions  compute projectile trajectories Lecture 3

Review of some Mathematics Vectors written as a a a a a & others They all mean the same thing – a direction and magnitude

X (east) y (north) The vector (5,4) means

X (east) y (north) The vector (5,4) means km in the x direction (east)

X (east) y (north) The vector (5,4) means km in the y direction (north) (5,4) total

X (east) y (north) Can also make (5,4) by going north first: (5,4) is the sum of (5,0) and (0,4) And the order does not matter (5,4) = (5,0) + (0,4) = (0,4) + (5,0)

X (east) y (north) In fact (5,4) can be made by adding lots of vectors: (5,4) The sum of lots of ‘short walks’ is the shortest distance between The starting and ending place.

X (east) y (north) What does it mean to subtract vectors? a b -b-b a+ba+b a-ba-b Subtracting means to go in the opposite direction but for the same distance.

Two vectors are added by adding the components a = (3,4) b = (6,9) then a + b = (3+6,4+9) Similarly to subtract them a – b = (3-6,4-9) X (east) y (north) a b a+ba+b a-ba-b

Review of some Mathematics

Its usual to use a ˆ above unit vectors, but not always, eg i,j

Review of some Mathematics

Vectors in 3-D Vectors in 3-D are an extension of vectors in 2-D Three components r = (x,y,z) o The formalism is identical  There is no way to tell from a vector equation if it is 2-D or 3-D  It doesn’t really matter because all the algebra is the same  The same relationships exist between vectors  The same rules, but with one more component  This is one of the advantages of working with vectors effectively you are manipulating all the dimensions simultaneously z x y z x y r

Review of some Mathematics

Product of vectors Multiplying vectors is more complicated than multiplying simple numbers o Multiplying by a scalar simply keeps the direction the same, but makes it longer:  if a = (3,5) then  a = (3 ,5  ) o The scalar product, written a.b can be defined as  a.b = IaI IbI cos(  ) where  is the angle between the two vector directions o The scalar product is also called the ‘dot’ product o The scalar product is a scalar, it is an ordinary number o The scalar product can also be evaluated from components:  If a = (7,2) and b = (4,9), then a.b = 7x4 + 2x9  Generally a.b = a 1 b 1 + a 2 b 2 where a = (a 1,a 2 ) and b = (b 1,b 2 )

Motion in 2-D - Projectiles X (east) y (north) r1r1 r2r2  r = r 2 -r 1 If a particle starts from position r 1 at time t = t 1 and moves to position r 2 at time t = t 2 then its displacement vector is  r = r 2 -r 1  which it moved in time  t = t2 – t1

Motion in 2-D - Projectiles X (east) y (north) r1r1 r2r2  r = r 2 -r 1 Its average velocity is then defined as v =  r /  t Velocity is a vector quantity, its magnitude IvI is the speed of motion its direction is the direction of motion.

Motion in 2-D - Projectiles X (east) y (north) r1r1 rdrd  r = r 2 -r 1 As the change becomes small, part of a path then v =  r /  t becomes the instantaneous velocity v = dr/dt which is the derivative of displacement with respect to time.  r = r d -r 1

Motion in 2-D - Projectiles X (east) y (north) The tangent to the curve is the direction of the velocity the magnitude CANNOT be determined from this plot as the time is not given Consider a particle following the dotted trajectory Here the particle stopped going south and started to go north

Projectiles Motion if an object is in motion, its position can be described by the equation x = vt + x 0 where x is its location at time t, v is its velocity, x 0 its starting displacement if the velocity is not constant then x = x 0 + ∫ vdt (this was discussed last week) A special case is when the acceleration, a, is constant, in which case x = x 0 + ∫ vdt = x 0 + v 0 t + ½at 2 where v 0 is the starting velocity or  x = v 0 t + ½at 2 this is often written as s = ut + ½at 2 for the distance traveled in time t in 1-D

Projectiles Motion These equations all have 1-D equivalents – vectors are just a ‘shorthand’ x = vt + x 0 is the same as x = v x t = x 0 and y = v y t + y 0 where x, y are locations at time t, x 0 starting x position etc.. if the velocity is not constant then x = x 0 + ∫ v x dt and y = y 0 + ∫ v y dt A special case is when the acceleration is constant, in which case x = x 0 + ∫ v x dt = x 0 + v x0 t + ½a x t 2 & y = y 0 + ∫ v y dt = y 0 + v y0 t + ½a y t2 or  x = v x0 t + ½a x t 2 &  y = v y0 t + ½a y t 2 Each dimension has its own set of equations BUT TIME IS SAME FOR BOTH

This is NOT what falling apples do, They fall directly downwards and the reason is gravity. They do however get faster and faster as they fall, they move with constant acceleration

Falling apples obey the equations y = y 0 + ½at 2 v = at but a is caused by gravity and is m/s 2 we use the symbol g for the acceleration caused by gravity at the surface of the earth. The minus sign just means that the acceleration is downwards

y = y 0 + ½at 2 v = gt 2 so y = 8 – 9.81t 2 m 2 When the apple hits the ground 0 = 8 – 9.81t 2 2 so t = √(16 / 9.81) = 1.28 s 8m y = y 0 + ½at 2 v = gt more generally t = √(2y 0 /g) so the speed of impact on the ground is v ground = gt = g√(2y 0 /g) = √(2gy 0 ) Check the dimensions v - in [m/s] √(2gy 0 ) g in [ms -2 ] y 0 in [m] so √([ms -2 ][m])

2-D projectile In the x direction the equation of motion is x = v x0 t v x = v x0 (constant speed in x direction) In the y direction it is the same as for the falling apple, except that the initial velocity is not zero and the apple starts from ground level For falling apple: y = y 0 + ½at 2 v y = at But now y 0 is 0 as the apple starts from ground And the initial velocity, vy0 is no longer zero. So: y = v y0 t + ½at 2 v y = v y0 + at

2-D projectile y = v y0 t + ½at 2 v y = v y0 + at x = v x0 t v x = v x0 t = x/v x0 y = v y0 {x/v x0 } + ½a{x/v x0 } 2 but a = g = m/s so: y = {v y0 /v x0 }x - {4.9/v x0 2 }x 2 Which describes a parabola

2-D projectile y = v y0 t + ½at 2 v y = v y0 + at x = v x0 t v x = v x0 y = {v y0 /v x0 }x - {4.9/v x0 2 }x 2 If ax 2 +bx + c = 0 then x = -b±√(b 2 -4ac) 2a So the value for x for a given y can be evaluated by regarding y as a constant and solving as a quadratic v y = v y0 + gt = v y0 + g{x/v x0 } = v y0 + {g/v x0 }x = v y0 – {9.81/v x0 }x

2-D projectile y = {v y0 /v x0 }x - {4.9/v x0 2 }x 2 v y = v y0 - {9.81/v x0 }x v x = v x0 Note that because of the negative sign (which came from g) the vertical speed starts out +ve but becomes negative The horizontal speed never changes – this assumes no air resistance

Projectile motion can be described using a vector equation instead r = r 0 + v 0 t + ½at 2 x y r More on gravity in another lecture

x y r o There are several closely related projectile problems  the differences are essentially the ‘boundary conditions’,  which means the starting values.  the equations are the same, but the values at the start are different x y x y