* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Rotational Motion and the Law of Gravity
Faster-than-light wikipedia , lookup
Laplace–Runge–Lenz vector wikipedia , lookup
Inertial frame of reference wikipedia , lookup
Photon polarization wikipedia , lookup
Classical mechanics wikipedia , lookup
Angular momentum operator wikipedia , lookup
Hunting oscillation wikipedia , lookup
Coriolis force wikipedia , lookup
Centrifugal force wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Relativistic angular momentum wikipedia , lookup
Equations of motion wikipedia , lookup
Fictitious force wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Work (physics) wikipedia , lookup
Jerk (physics) wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Classical central-force problem wikipedia , lookup
Rotational Motion and the Law of Gravity -Study of linear motion: Δxdisplacement, v-velocity, a-acceleration -Study of rotational motion: Δθ-angular displacement, ω-angular velocity, αangular acceleration - Angular quantities in Physics must be express in radians • Radian θ=s/r –tha arc length s along a circle divided by the radius r • Ex: 360o= 2π radians 180o= π radians 45o=2πrad/360o=(π/4)rad • If OP-reference line, t=0; P-on reference line, t=Δt; P-new position In this time, the line OP has moved the angle θ with respect of the reference line. θ (SI rad)- angular position (linear motion was –x) An object angular displacement: Δθ= θf- θi SI: radians The average angular speed ωav of a rotating rigid object during the time Δt is defined as the angular displacement Δω divided by Δt: ωav= (θf-θi) /(tf -ti)= Δθ/Δt SI: rad/s The instantaneous speed ω of a rotating rigid object is defined as the limit of average speed: ω =lim(Δt→0)Δθ/Δt An object average angular acceleration αav, during an interval Δt is: αav= (ωf- ωi)/ (tf-ti) =Δω/Δt SI: rad/s2 The instntaneous angular acceleration α α = lim(Δt→0) Δω/Δ t SI: rad/s2 • When a rigid objects rotates about a fixed axis, as does the bicycle wheel, every portion of the object has a same angular speed and the same angular acceleration Rotational motion under constant angular acceleration • Relations between angular and linear quantities: Δθ=Δs/r Δθ/Δt=1/r(Δs/Δt) ω= v/r The tangential speed of a point on a rotating object= the distance oft hat point from the axis of rotation multiply by an angular speed vt= ωr Δvt=rΔω Δvt /Δt=r Δω/Δt at= r α Tha tangential acceleration of a point on a rotating object= distance from the axis of rotation multiplied with α • Centripetal acceleration Fig shows a car moving in a circular path with constant linear speed Even through the car moves at a constant speed, it still has an acceleration aav= (vf-vi) /(tf-ti)= Δv /Δt Δv = (vf-vi) –change of velocity When Δt is very small, Δs and Δθ are very small, vf and viare almost parallel, Δv is perpendicular, and points toward the center of the circle • In fig, the triangle with sides Δs and r are similar to the one formed by the vectors: Δv/v = Δs/r Δv =v/r Δs aav = Δv/ Δt aav = v/r Δv /Δt If Δt becomes very small, Δs /Δt approches instantaneous value of tangential speed v For circular motion at a constant speed, the acceleration vector always points toward the center of the circle, the acceleration is called a centripetal (center-seeking) acceleration aav approaches ac, the instantaneous centripetal acceleration: ac =v2/r but vt=rω ac =r2ω2/r =rω2 at =rω The tangential and centripetal components of acceleration are perpendicular to each other, total acceleration a=√(at2+ac2) • Angular quantities are vectors • The direction of angular velocity ω can be found with right-hand rule (wrap the axis of rotation with your right hand so that fingers wrap in the direction of rotation, your tumbpoints the diretion of ω • Forces causing Centripetal Acceleration An object can have a centripetal acceleration only if some external force acts on it. -for a ball whirling in a circle at the end of a sting, that force is tension in string -for a car moving on a flat circular track, the force is friction between the car and the track - A satellite in circular orbit around the Earth has a centripetal acceleration due to the gravitational force between the satellite and Earth. - “ centripetal force” –mean the force in action acts toward the center Fc= m ac= mv2/r A net force causing a centripetal acceleration, acts toward the center of a circular path and effects a change in the direction of the velocity vector, if the force vanish, the object will leave the circular path, and move a straight line tangent to the circle. • Pb solving strategy: 1.Draw a free-body diagram, labeling the forces act on it 2. Choose a coordinate system (one axis perpendicular to the circular path followed by the object (radial direction) one axis tangent to the circular path (the tangential, or angular direction) The normal direction, perpendicular to the plane of motion, also needed 3. Find the net force Fc toward the center Fc =ΣFr, where ΣFr is the sum of the radial components of the forces (this cause the centripetal acceleration) 4. Use Newton’s 2nd Law, for the radial, tangential, and normal directions: ΣFr=m ac, ΣFt= m at, ΣFr=m an Remember, ac=vt2/r 5. Solve for the unknown quantities • Newtons Law of Universal Gravitation: If two particles with masses m1 and m2 are separated by a distance r, then a gravitational force acts along a line joining them with the magnitude : F=G( m1m2/ r2) G=6.673x10-11kg-1m3s-2 is constant of universal gravitation F- always an attractive force • Gauss’s Law: the gravitational force exerted by a uniform sphere on a particle outside the sphere is the same as the force exerted in the entire mass of the sphere was concentrated at its center. • Near the surface of Earth, F= mg, g varies considerably with altitude (article page 207) • Gravitational PE: PE= mgh –if object near Earth’s surface PE= -G MEm/r; SI: Joules(J) PE2-PE1 = -GMEm/(RE+r)-(-GMEm/RE) =- GMEm[ 1/(RE+r)-1/RE] = GMEmh/[RE(RE+h)] (1/[RE(RE+h)≈1/RE2) ≈GMEmh/RE2 (g = GME/RE2) ≈mgh • Escape speed- if an object is projected upward from Earth’s surface with a large enough speed, it can soar off into space and never return • Escape speed- can be found apllying conservation of energy: KEi +PEi= 1/2mv2- GMEm/RE =0 (vf=0) 1/2mvesc2- GMEm/RE =0 vesc=√ 2GME/RE • Keplers’s Law (article: history/214) 1.All planets move in elliptical orbits with the Sun at one of the focal points. 2.A line drawn from the Sun to any planet sweeps out equal areas in equal intervals 3. The square of the orbital period of any planet is proportional with the cube of the average distance from the planet to the Sun • 1.All planets move in elliptical orbits with the Sun at one of the focal points. Newton’s Law of gravitaion: any object bound by a force that varies as 1/r2, will move in elliptical orbit Sun is in one of the foci. The distance from the sun to the planet continuosly changes • 2. In an interval Δt. The planet moves from A to B, more slowly because it is far away from the Sun In the same interval Δt from C to D the planet moves faster because it is closer to the sun ( conservation of angular momentum) • 3. If a planet of mass MP moving around Sun (MS), because the orbit is circular, the planet moves with a ct. speed v Newton’s Law: MP ac= MPv2/r =(MP/r)(GMS/r)= MPGMS/r2 The speed of the planet = circumference of the orbit/ time required for one revolutions v= 2πr / T (T- period of the planet) GMS /r2= v2/r = (2πr / T)2/r T2 =(4π2/ GMS )r3 = KS r3 T2 = KS r3 Ks = ct. = 4π2/ GMS = 2.97x10-19 s2/m2 The mass of the Sun : MS =4πr / G KS (see table page 216)