Impulse and Momentum Relationship

Impulse and Momentum Relation

• The change in a system's momentum equals the integral of force with respect to time.
• Equation 4.7 contains the same physical information as equation 4.6
• It presents a new perspective on force, as the change in momentum equals the time integral of force.
• Producing a specific change in momentum over a time interval t only requires that ∫F dt has the appropriate value.
  • This means that a small force acting for a long time or a large force acting briefly can achieve the same change.
• Impulse refers to the integral ∫ F dt, which is the change in momentum.

Rubber Ball Rebound Example

• A rubber ball with a mass of 0.2 kg strikes the floor at 8 m/s and rebounds at about the same speed.
• High speed photographs show the ball is contact with the floor for Δt = 10⁻³ s.
• The momentum of the ball just before hitting the floor is Pa = −1.6k kg m/s, and its momentum 10⁻³ later is Pb = +1.6k kg m/s.
• Using ∫F dt = Pb - Pa gives ∫F dt = 1.6k - (−1.6k) = 3.2k kg m/s.
• The collision time being Δt = tb - ta, the average force Fav during the collision is expressed as FavΔt = ∫ta+Δtta F dt.
• The average force is Fav = (3.2k kg m/s) / (10⁻³ s) = 3200k N, directed upward.
• 3200 N is about 720 lb, and the instantaneous force on the ball peaks even higher.
• A softer surface would result in a longer collision time, decreasing peak force.
• Total force F is the sum of Ffloor and Fgrav, where F = Ffloor - Mgk.
• The impulse equation is ∫ floor Ffloor dt - ∫ floor Mg k dt = 3.2 k kg m/s.
• The gravitational force contributes an impulse of − Mg∫floor k dt =− Mgk∫floor dt = −(0.2)(9.8)(10⁻³) k = −1.96 × 10⁻³ k kg m/s.
• The impulse from gravity is less than one-thousandth of the total impulse.
• Contact forces during short collisions are large enough that the impulse from moderate forces like gravity or friction is negligible.

Spacecraft Momentum Problem

• A spacecraft of mass M(t) moves through space at a constant velocity v.
• Encounters dust particles at a rate of dm/dt, with the dust having a velocity u.
• The external force F needed to keep the spacecraft moving uniformly can be determined by focusing on a short time interval between t and t + Δt.
• System consists of M(t) and the mass increment Δm.
• Initial momentum is P(t) = M(t)v + (Δm)u.
• Final momentum is P(t + Δt) = M(t)v + (Δm)v.
• The rate of change of momentum is approximately ΔP/Δt = (v - u)Δm/Δt.
• The exact rate of change of momentum, in the limit as Δt approaches 0, is dP/dt = (v - u)dm/dt.
• The required external force is F = (v - u)dm/dt.
• F can be either positive or negative according to the stream of mass direction.
  • If u = v, the momentum is constant, and F = 0.

Vector Nature of Angular Velocity and Angular Momentum

• Describing the most general rotational motion requires suitable coordinates and a position vector r = xî + yĵ + zk.
• Velocity and acceleration are found by successively differentiating r with respect to time.
• An attempt to use an analogous procedure for rotational motion using angular coordinates is incorrect.
• Angular position cannot be represented by a vector, but angular velocity is a good vector.
• Angular velocity is defined as ω = dθx/dt î + dθy/dt ĵ + dθz/dt k = ωx î + ωy ĵ + ωz k.
• Rotations through finite angles do not commute, while infinitesimal rotations like Δθx, Δθy, and Δθz do.
• Angular velocity is a vector because ω = lim (Δθ/Δt) as Δt→0 represents a true vector.
• Translational velocity of a particle in a rotating rigid body is related to the angular velocity of the body.
• A vector r from the origin to any particle sweeps out a cone as the body rotates about an axis.
• The angle φ between în and r is constant, and the tip of r moves on a circle of radius r sin φ.
• The magnitude of the displacement |Δr| is 2r sin φ sin (Δθ/2).
• If Δθ is very small, |Δr| ≈ r sin φ Δθ.
• If rotation through angle Δθ occurs in time Δt, |Δr|/Δt ≈ r sin φ (Δθ/Δt).
• In the limit as Δt approaches 0, |dr/dt| = r sin φ dθ/dt.
• The magnitude |dr/dt| = r sin φ dθ/dt, and the direction of dr/dt and în are given correctly by dr/dt = ôn × r dθ/dt.
• Therefore, dr/dt = v = ω × r.
• Average velocity ω = θ/t

Angular speed questions

• Angular speed of a rotating top is π*3 = 3π if it travels π radians in a third of a second
• The units of omega, or angular velocity, are radians per second.
• Angular speed is similar to linear speed, but uses degrees or radians instead of distance per unit time, with the formula: Angular speed= θ/t
• A car wheel of radius 30 inches rotates at 10 revolutions per second, so the angular speed of the tyre would be 20πrad/s, since a full rotation (360°) equals 2π.

Integrating Equations of Motion in One Dimension

• For problems with a single variable, the equation of motion reduces to m(d²x/dt²) = F(x) or m(dv/dt) = F(x).
• The equation can be solved for v by integrating m dv/dt = F(x) with respect to x: m ∫dv/dt dx = ∫F(x) dx.
• The integral on the right can be evaluated if F(x) is known.
• The integral on the left, ∫dv/dt dxcan be integrated by changing the variable from x to t using differentials, dx = (dx/dt) dt = v dt.
• This makes m ∫dv/dt dx = m ∫dv v dt = m ∫ d(½v²) dt = ½mv².
• Putting these results into the original equation gives ½mvb² - ½mva² = ∫F(x) dx.
• Alternatively, we can use indefinite upper limits:½ mv² - ½mva² = ∫F(x) dx.
• v is the speed of the particle at position x.
• Equation 5.5 gives v as a function of x, enough for finding x as a function of t.

Mass Thrown Upward:

• For a mass m thrown vertically upward with initial speed vo, the maximum height can be found with constant gravity and by negating air friction.
• Taking the z axis vertically upward, F = -mg and using the work energy equation gives: ½ mv₁² - ½ mvo² = ∫z₀ z₁ F dz = -mg∫z₀ z₁ dz = -mg(z₁ - z₀).
• At the peak, v₁ = 0, so mg(z₁ - z₀) = ½ mvo², giving z₁ = z₀ + vo²/2g.

Integrating Equations of Motion in Several Dimensions

• The equation of motion of a particle acted on by a force depending on position is F(r) = m(dv/dt).
• Consider the particle moving a small distance Δr.
• Assuming that Δr is small, F is effectively constant over this displacement and can be obtained with the scalar product of F• Δr = m(dv/dt) • Δr.
• For a short length of path, v is approximately constant, where Δr = v Δt, the time the particle requires to travel Δr.
• This gives m(dv/dt) • Δr = m(dv/dt) • vΔt.
• The equation can transform the vector identity 2A•dA/dt = dA²/dt like this: v • dv/dt = ½d/dt(v²).
• Equation 5.9 becomes F • Δr = m/2 d/dt (v²)Δt.
• To find the relationship for an entire trajectory, divide the trajectory from initial position ra to final position rb into N short segments of length Δrj, where j is an index. For each segment, F(rj) • Δrj = m/2 d/dt(vj²)Δtj.
• Adding the equations of all segments gives the sum of force times displacement= sum of the change in kinetic energy =ΣF(rj) • Δrj = Σ m/2 d/dt (vj²)Δtj.
• In the limit (as segment length approaches zero and number of segments approaches infinity), the sums becomes integrals. ∫ F(r) • dr = ∫ m/2 d/dt (v²) dt, where ta and tb are the times corresponding to ra and rb.
• This integral can be evaluated, for this equation: ½mvb² - ½mva².
• For one dimension v² = vx².
• This generates a line integral where the integration is carried out on a path.
• This represents a simple generalization of the result we found for one dimension

The Work-Energy Theorem

• The integral F · dr = ½mvb² - ½mva² is known as the work-energy theorem.
• Kinetic energy (K) describes the quantity ½mv².
• The integral ∫F · dr is called the work (Wba).
• Wba = Kb - Ka.

The Work and Total Force

• The work done by a force F in a small displacement is given by Δ𝑊 = 𝐅 ⋅ Δr = 𝐹 cos 𝜃Δr
• If 𝐅 is the sum of several forces 𝐅 = ΣFi, total work done is the sum of work done by each force from point a to b.
• 𝑊𝑏𝑎=Σ(𝑊i)𝑏𝑎 where (𝑊i)𝑏𝑎 =∫Fᵢ ⋅ dr Integrated from rₐ to rb.
• Center of mass of an extended system moves according to the equation of motion: F = MR̈ = M dV/dt
• Integrating with respect to position gives: ∫RbRaF · dR = ½MVb² - ½MVₐ²
• dR = Vdt is the displacement of the center of mass in time dt.

Spacecraft escape velocity problems

• The force on a mass is 𝐹 = -GMₑm/r²
• The problem is one dimensional, so the kinetic energy at distance r by the work-energy theorem can be determined simply: 𝐾(𝑟) − 𝐾(rₑ) =∫F(r) dr Integrated from 𝑅ₑ to 𝑟 = −GMem ∫dr/r² Integrated from 𝑅ₑ to r
• Given ½mv(r)² − ½mVₒ² = 𝐺𝑀ₑm(1/r-1/R𝑒) The maximum height can be found when 𝑣(𝑟) = 0
• Given 𝑣ₒ² = 2𝐺𝑀𝑒(1/𝑅𝑒-1/rmax) Due to familiarity of constants, gravity can be written as 𝑔 = 𝐺𝑀𝑒/𝑅ₑ² so 𝑣ₒ² = 2𝑔𝑅𝑒² (1/𝑅𝑒-1/rmax) = 2𝑔𝑅𝑒 (1- 𝑅e/rmax) and 𝑟 max = 𝑅𝑒/(1-𝑣0²/2𝑔𝑅e)

Kepler's Laws of Planetary motion

• Johannas Kepler, a 16th century astronomer established 3 laws which govern planetary motion around the sun.
• Law of orbits : Planets move in elliptical orbits with the sun at one foci
• Sun at one foci of the elliptical orbit : an elliptical path has two foci where the sun is located at one of them and allows us to determine if other stars have planets
• Law of areas: line connecting sun and planet equal areas in equal intervals of time
• Because planets revolve around sun, a planet moves faster when closer to the sun and slower when its further
• Law of periods : the square of the period of each planet is proportional to size of semi major axis of its orbit
• Kepler discovered planetary motion, but Newton discovered planets move due to a gravitational force which leads to the law of universal gravitation

Kepler's First Law: Eccentricity

• The shape of an ellipse is defined by its eccentricity (e), which ranges from 0 to nearly 1: e = √(1 - (b²/a²)).
• Given the distance of the perihelion and aphelion, you could calculate the semi-major axis, and the eccentricity.
• If a comet's orbit has a semi-major axis of 10 AU and a semi-minor axis of 2 AU, then its eccentricity is √(1- (2²/10²)) = √(1- 4/100) = √0.96 ≈ 0.98
• Given the semi-major axis and eccentricity, one can find the perihelion (closest distance to the Sun) and aphelion (farthest distance):
  • Perihelion = a(1 - e)
  • Aphelion = a(1 + e)

Kepler's Second Law

• Kepler's second law is about the conservation of angular momentum.
• Planets move faster when closer to the Sun (perihelion) and slower when farther away (aphelion).
• The angular momentum of an orbiting body is constant: 𝐿 = mvr, where m is the mass, v is the velocity and r is the radius.
• A problem gives you the radius and velocity at two different points of the orbit if the equation m(v1)(r1) = m(v2)(r2) to solve for an unknown variable.

Kepler's Third Law

• The precise form of Kepler's 3rd law: 𝑇² = (4π²/G(M + m))a³, where G is the gravitational constant, M is the mass of the central body, and m is the mass of the orbiting body.
• In planetary problems, m is often negligible compared to M, so commonly T² ∝ a³.
• Gravitational Constant (G) = 6.674 x 10-11 N•m²/kg²

Kepler's Law Practice Problems and Solutions

• For a comet calculate a) the semi-major axis of its orbit and b) The eccentricity of its orbit
• Solution a) Semi major axis = (perihelion + aphelion) / 2 a = (0.5 + 17.5) / 2, a = 9 au
• Solution b) Eccentricity = (aphelion - perihelion) / (aphelion + perihelion e = (17au - 0.5au) / (17.5au +0.5au), e = 0.94
• If an asteroid has a semi major axis of 3.2 and minor axis of 3.0. Calculate the eccentricity
• Eccentricity = square root of ( 1- 3squared/3.2squared, Eccentricity = 0.35
• If a planet's speed at perhelion is 35km/s, and its distance at aphelion is three times the distance at perhelion. Find speed at Aphelion
• Formula angular conservation: v1xr1 = v2xr2 Because v1, 35 = r2, 3r1: v2, 35/3 V2 = 11.67 km/s
• If a planet orbits a sun at a velocity of 8,000 km when it is 7000km from the planets centre. Whats the velocity of the satellite when it is 14,000 km from planets centre ?
• 1vxr1 = v2xr2, v1 = 8,000 km, r1 = 7000, r2 = 14000
• 2 = (8kx7)/14k V2 = 4000 m/s Planet x = Sa that is 4 times that of earth, how many earth years = one earth orbit + SolUTion Keplers third law: (T1^2/ T2^2) = ( Sa1^3/ Sa 2^3 Let earth = 1 Sa = 4 ( 1^2/ T2^2) = SA earth / Sa x) ^3 T = SA RT (164 = √64

· A satellite orbits with radius of period of 8 hours, if another satellite orbits the same radius in 8 hours what is the SA · Solution use equation t1^2/t2^ 2 = SA1^3/sat2^3 T1. 8 hours. A1, 1000km, T1. /729 = 86413 Sa = a1 ^3 is 11/3 from 11,393 to 10 9. Is to 20,500 km

Conservation of Angular Momentum

• If there is no net external torque around a point in a fixed inertial reference frame, the angular momentum of a system of particles around that point is conserved. dÎ/dt = 0
• L = Î₁ + Î₂ + · · · + ÎN = constant

Angular Momentum of a Rotating Rigid Body

• Each particle of an object rotates in the xy plane about the z-axis with an angular velocity ω.
• The angular velocity vector points along the rotation axis, according to the right hand rule.
• To find the angular momentum of the entire object a d the momenta of all the individual particles by breaking up a hoop of mass M into N small pieces each of mass mi.
• The hoop rotates with angular speed ω, and 𝑣 = ωR.
• To find the total angular momentum of the object, we sum up all N pieces.
• Lz = ΣRm₁vᵢsin(𝜃ᵢ).
• Because it's a circle, all sin(θᵢ) == 1 and all position vectors have magnitude R.
• Lz = ΣRm₁vᵢ = R ΣmᵢωR = ωR² Σmᵢ = ωMR² = Iω
• So although we showed this for a specific shape, the angular momentum of a rigid object rotating about CM) is the momentum of inertiaof the object about the rotation axis times the angular speed.
• In vector form: L = Iω
• Because the angular velocity points along the rotation axis according to the right hand rule, it is parallel to the angular momentum.

Satellite Orbits

• An orbit is the curved path that an object in space takes around another object due to gravity.
• Objects of similar mass orbit each other with neither object at the center, whilst small objects orbit around larger objects.

Types of Orbit:

• Geostationary Orbits (GEO) circle the Earth above the equator following the Earth's rotation, taking 23 hours 56 minutes and 4 seconds.
  • Satellite speed should be around 3 km per second at an altitude of 35 786 km.
• Low Earth Orbits (LEO) are close to the Earth's surface, normally at an altitude of less than 1000 km, but could be as low as 160 km.
• Medium Earth Orbits (MEO) comprise orbits anywhere between LEO and GEO.
• Polar Orbits and Sun-synchronous Orbits (SSO) travel past the Earth from north to south rather than from west to east, passing roughly over the Earth's poles.
  • Can deviate by 20 to 30 degrees is still classed as a polar orbit.
  • Polar orbits are a type of LEO, as they are at low altitudes between 200 to 1000 km.
• Transfer Orbits and Geostationary Transfer Orbits (GTO) get satellites from one orbit to another, are carried to space with launch vehicles, and are placed on a transfer orbit so they are able to can move from one orbit to another.