Kinematics

• Newton's laws of motion
• Trajectory, speed, and acceleration
• Constant versus non constant acceleration
• Gravitational motion and ballistics
• Angular speed and acceleration
• Satellites, orbits
• Slope and movement decomposition

Newton's 3 Laws of Motion

• I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.
• II. The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors; in this law the direction of the force vector is the same as the direction of the acceleration vector.
• III. For every action there is an equal and opposite reaction.

Trajectory, Speed,

• Problem. Describe the movement of a point-wise object through space.
• Trajectory - a curve in space s(x(t), y(t), z(t)) where the parameter t is the time.
• Speed (velocity) - the distance covered in a unit of time v = distance/time (m/s)
• Momentary speed - limit of the speed when the lapse of time goes to 0.
• v(t) = s'(t)

... and Acceleration

• Acceleration - the variation of the speed in a unit of time.
• a(t) = v'(t) = s''(t)
• Our goal: compute the trajectory of an object based on the following data:
• origin of the movement
• initial speed as a vector
• some description of the acceleration (gravitation, input from the user for car race or flight simulation, etc.)
• other physical constraints (the form of the road, degrees of freedom for articulated objects, etc.)

Types of Motion Calculation

• Newtonian motion where Dt is the lapse of time between the two points on the trajectory:
• s(t + Dt) = s(t) + v(t) Dt + 0.5 a(t) (Dt)²
  v(t + Dt) = v(t) + a(t) Dt
• Lagrangian motion:
• Lagrangian: L = T - U, where T is the kinetic energy and U is the potential energy.
• T = 0.5 m v².
• U = m * g * height
• Equation of motion:
  

Constant Acceleration

• We can simply plot the position and speed as known functions of time.
• Considering that the time starts from 0, that the object's initial position is s₀, and that its initial speed is v₀, then we have
• s(t) = s₀ + v₀ t + a t²/2
• v(t) = v₀ + a t

Non Constant Acceleration

• We can describe the acceleration as a function.
• This gives us a system of partial differential equations - PDEs with given initial values for t = 0 (initial value problem - IVP).
• Usually it is solved by approximating s and v on small intervals of time.
• s'(t) = v(t), s(0) = s₀
• v'(t) = a(v, s, t, ...), v(0) = v₀
• If a is just a function of t, then we can compute
• v(t) = Integral₀^t a(u) du,
  s(t) = Integral₀^t v(u) du

Example - Drag Force

• The drag force (and acceleration) is proportional to the square of the speed.
• a = -kv² = dv/dt
• -k dt = dv/v²
• By integrating between 0 and t on the left, and between v₀ = v(0) and v₁ = v(t) on the right, we obtain
• -k t = -1/v₁ + 1/v₀
• v(t) = v₀/(1 +v ₀ k t)
• s(t) = s₀ + v₀ t /(1 + v₀ k t) - kv²t²/2

Newton's Law of Gravitation

• Every object in the universe attracts every other object with a force directed along the line of centers of the two objects that is proportional to the product of their masses and inversely proportional to square of the distance between them.
• If the two objects have masses m₁ and m₂ and are at a distance of r, then the attraction force between them is
  F = G m₁ m₂ / r²
• where G = 6.67 x 10^-11 N m² /kg² is the universal gravitational constant

Earth's Gravitation

• To define the gravitational acceleration at the surface of the Earth, we need
• M = 5.98 x 1024 kg, the mass of the Earth, and
  R = 6.38 x 106 m, the radius of the Earth
• we can define the gravitational acceleration g on the surface of the Earth as (approximate sea level)
  g = G M / R = 9.8 m / s²
• Then for every object of mass m, the gravitational force is F = g m.

Ballistic Trajectories

• Problem. The object is projected from the ground at a given angle a and with given initial speed v₀.
  
• We decompose v₀0 in
  v_0x = |v₀| sin a v_0y = |v₀| cos a
• If we consider again that the time starts from 0, we must compute the position and the speed at time t.
• We will consider the initial position to be 0.
• The horizontal acceleration is equal to 0, so the horizontal speed is constant.
• The vertical acceleration is equal to -g.
• x(t) = v_x t,
  v_x(t) = v_0x
  y(t) = v_0y t - 0.5 g t²,
  v_y(t) = v_0y - g t
• The trajectory is parabolic.

Total Distance and Time

• How long will it take for the object to fall down again? Where will it land? How high will it go?
• If we impose v_y(t₁) = 0, we have
• v_0y = g t₁, t₁ = v_0y/g
• The total time will be twice that much T = 2 v_0y/g
• The horizontal distance covered in this time is
• X = v_0x 2 v_0y/g
• The vertical distance covered in t₁ is
• H = v_0y v_0y/g - 0.5 v_0y²/g = 0.5 v_0y²/g

Angular Speed

• When an object moves on a circular trajectory, the angular speed is the rate at which the angle changes.
• w = d a / dt
• The rate at which the angular velocity changes is the angular acceleration.
• W = d w / dt
• If r is the radius of the circle and v is the speed that is tangent to the circle, then we can compute
• w = v/r, W = a/r.

Satellite

• When an object arrives in the neighborhood of a second massive object with a speed perpendicular to the line between their centers, the first object may become a satellite.
• When an object rotates, the centrifugal force is pushing it towards the exterior of the circle.
• The gravitation must equal this force for the object to be a satellite. It will play the role of the centripetal force.
• Fc = m w² r = m v² /r
• m v² / r = G m M / r ²
• v = sqrt(G M / r)

Gravitational Orbits

• The orbits of the planets generally are ellipses.
• The orbits of some of the planets (e.g., Venus) are ellipses of such small eccentricity that they are essentially circles, and we can put artificial satellites into orbit around the Earth with circular orbits if we choose.
• Some comets have parabolic orbits; this means that they pass the Sun once and then leave the Solar System, never to return. Other comets have elliptical orbits and thus orbit the Sun with specific periods.
• The gravitational interaction between two passing stars generally results in hyperbolic trajectories for the two stars.
  

Slope

• Consider an object moving down a path with a given slope.
• Decompose the gravitational acceleration into a component that is tangent to the surface, and another that is normal to the surface.
• The tangent component will influence the speed in the tangent direction.
• The normal component may slow it down because of the friction (otherwise it has no effect on the movement).
• g_t = g sin a
• g_p = g cos a
  

Friction Force

• A force occurring when two solid objects are in contact, applied against their relative movement.
• The frictional resistance is proportional to the force which presses the surfaces together as well as the roughness of the surfaces. If we denote by N this force (like the gravitation), then the frictional resistance force may then be written:
• Ff = mu N
• where mu is a constant depending on the two surfaces.

Friction and Slope

• For a terrain that is not flat, the normal force to the surface based on gravitation is
• G_n = m g_p = m g cos alpha
• where m is the mass of the object.
• The friction force will then be
• FF = mu G_n = m mu g cos(alpha)
• The resulting acceleration
• a_f = g mu cos(alpha) a
• This will have a sign that is opposite to the direction of movement.

Change of Slope

• Passing from a surface of a given slope to a surface of different slope.
• If v is the speed at the passage point, we decompose it in a speed that is tangent to the new surface (v_t), and a speed that is normal to the new surface (v_p).
• The normal component is lost for the movement and it is transformed in impact momentum.
• The computations on the new surface start with the tangent speed.
  

Application - Slide

• By a smooth change in the slope, transform the almost vertical speed in an almost horizontal speed at the end of the trajectory. This way we avoid damaging the object by collision with the ground (slide, roller coaster).
  

Terrain-Dependant Slope

Solving the equation: the projection of the direction vector on the terrain plane must be orthogonal to the plane normal. The scalar product of the 2 vectors must be 0.