Chapte 1. Kinetics of Paticles: Newton s Second Law Intoduction Newton s Second Law of Motion Linea Momentum of a Paticle Systems of Units Equations of Motion Dynamic Equilibium Angula Momentum of a Paticle Equations of Motion in Radial & Tansvese Components Consevation of Angula Momentum Newton s Law of Gavitation Tajectoy of a Paticle Unde a Cental Foce Application to Space Mechanics Keple s Laws of Planetay Motion
Kinetics of Paticles We must analyze all of the foces acting on the aceca in ode to design a good tack As a centifuge eaches high velocities, the am will expeience vey lage foces that must be consideed in design.
Intoduction Σ Fma 1.1 Newton s Second Law of Motion If the esultant foce acting on a paticle is not zeo, the paticle will have an acceleation popotional to the magnitude of esultant and in the diection of the esultant. Must be expessed with espect to a Newtonian (o inetial) fame of efeence, i.e., one that is not acceleating o otating. This fom of the equation is fo a constant mass system
1.1 B Linea Momentum of a Paticle Replacing the acceleation by the deivative of the velocity yields dv F m dt d dl ( mv) dt dt L linea momentum of the paticle Linea Momentum Consevation Pinciple: If the esultant foce on a paticle is zeo, the linea momentum of the paticle emains constant in both magnitude and diection.
1.1C Systems of Units Of the units fo the fou pimay dimensions (foce, mass, length, and time), thee may be chosen abitaily. The fouth must be compatible with Newton s nd Law. Intenational System of Units (SI Units): base units ae the units of length (m), mass (kg), and time (second). The unit of foce is deived, 1N ( 1kg) m 1 s kg m 1 s U.S. Customay Units: base units ae the units of foce (lb), length (m), and time (second). The unit of mass is deived, 1lbm 1lb 3. ft s 1lb 1slug 1ft s lb s 1 ft
1.1 D Equations of Motion Newton s second law F ma Fee-body diagam ~ Kinetic diagam Can use scala component equations, e.g., fo ectangula components, Rectangula components ( Fi ) ( ) x + Fy j+ Fk z m ai x + ay j+ ak z F ma F ma F ma x x y y z z F mx F my F mz x y z Tangential and nomal components, Radial and tansvese components
Fee Body Diagams and Kinetic Diagams The fee body diagam is the same as you have done in statics; we will add the kinetic diagam in ou dynamic analysis. 1. Isolate the body of inteest (fee body). Daw you axis system (e.g., Catesian, pola, path) 3. Add in applied foces (e.g., weight) 4. Replace suppots with foces (e.g., eactions :nomal foce) 5. Daw appopiate dimensions (usually angles fo paticles)
5 N x y 5 o N mg F f Put the inetial tems fo the body of inteest on the kinetic diagam. 1. Isolate the body of inteest (fee body). Daw in the mass times acceleation of the paticle; if unknown, do this in the positive diection accoding to you chosen axes
5 N x y ma y 5 o ma x N mg F f Σ F ma Daw the FBD and KD fo block A (note that the massless, fictionless pulleys ae attached to block A and should be included in the system).
Daw the FBD and KD fo the colla B. Assume thee is fiction acting between the od and colla, motion is in the vetical plane, and q is inceasing
1. Isolate body. Axes 3. Applied foces 4. Replace suppots with foces 5. Dimensions 6. Kinetic diagam e θ e ma θ ma F f θ θ mg N
Sample Poblem 1.1 A 80-kg block ests on a hoizontal plane. Find the magnitude of the foce P equied to give the block an acceleation of.5 m/s to the ight. The coefficient of kinetic fiction between the block and plane is m k 0.5. STRATEGY: Resolve the equation of motion fo the block into two ectangula component equations. Unknowns consist of the applied foce P and the nomal eaction N fom the plane. The two equations may be solved fo these unknowns.
MODELING and ANALYSIS: Resolve the equation of motion fo the block into two ectangula component equations. F x ma : P cos30 0.5N 80 kg.5m s F y 0: 00 N N Psin30 785N 0 Unknowns consist of the applied foce P and the nomal eaction N fom the plane. The two equations may be solved fo these unknowns.
N Psin30 785N P Pcos30 0.5 sin 30 785 N 00 N P 535N REFLECT and THINK When you begin pushing on an object, you fist have to ovecome the static fiction foce (F μ s N) befoe the object will move. Also note that the downwad component of foce P inceases the nomal foce N, which in tun inceases the fiction foce F that you must ovecome.
Sample Poblem 1.3 The two blocks shown stat fom est. The hoizontal plane and the pulley ae fictionless, and the pulley is assumed to be of negligible mass. Detemine the acceleation of each block and the tension in the cod. STRATEGY: Wite the kinematic elationships fo the dependent motions and acceleations of the blocks. Wite the equations of motion fo the blocks and pulley. Combine the kinematic elationships with the equations of motion to solve fo the acceleations and cod tension.
m B MODELING and ANALYSIS: Wite the kinematic elationships fo the dependent motions and acceleations of the blocks. yb 1 xa a 1 B a A Wite equations of motion fo blocks and pulley. ( ) A B B ( )( ) 300 kg 9.81m s T ( 300 kg) T T F x maaa : T1 100 kg a F y mbab : g T m a 940N - F m a y T 1 C 0 C a B ( 300kg) a B 0 :
Combine kinematic elationships with equations of motion to solve fo acceleations and cod tension. yb 1 xa a 1 B a A T 940 N T T 1 ( 100 kg) a A 940N - ( 300kg) 940N - a B ( 300kg)( 1 a ) T 1 0 ( 150kg) ( 100 kg) 0 a A a A a a T T 1 A B 1 A 8.40m s a 4.0m s ( 100kg) a 840 N T 1680 N 1 A A
REFLECT and THINK Note that the value obtained fo T is not equal to the weight of block B. Rathe than choosing B and the pulley as sepaate systems, you could have chosen the system to be B and the pulley. In this case, T would have been an intenal foce.
Sample Poblem 1.5 The 6-kg block B stats fom est and slides on the 15-kg wedge A, which is suppoted by a hoizontal suface. Neglecting fiction, detemine (a) the acceleation of the wedge, and (b) the acceleation of the block elative to the wedge. STRATEGY: The block is constained to slide down the wedge. Theefoe, thei motions ae dependent. Expess the acceleation of block as the acceleation of wedge plus the acceleation of the block elative to the wedge. Wite the equations of motion fo the wedge and block. Solve fo the acceleations.
y MODELING and ANALYSIS: The block is constained to slide down the wedge. Theefoe, thei motions ae dependent. ab aa + ab A Wite equations of motion fo wedge A and block B. A: - N 1 F ma x A A sin30 ma 1 : ( ) 0.5 N m a (1) A A A A x
B F ma x B x m g sin 30 m a cos30 a B B A BA a a cos30 g sin 30 BA A () Solve fo the acceleations. F ma y B y ( ) N1 m gcos30 m a sin30 (3) B B A a. Acceleation of Wedge A Substitute fo N 1 fom (1) into (3)
( ) ma mgcos30 m a sin30 A A Solve fo a B B A A and substitute the numeical data a A 1.545 m/ s b. Acceleation of Block B Relative to A Substitute aa into ()-> a a cos30 + g sin 30 a BA BA A ( ) ( ) 1.545m s cos30 + 9.81m s sin 30 a BA 6.4m s
REFLECT and THINK Many students ae tempted to daw the acceleation of block B down the incline in the kinetic diagam. It is impotant to ecognize that this is the diection of the elative acceleation. Rathe than the kinetic diagam you used fo block B, you could have simply put unknown acceleations in the x and y diections and then used you elative motion equation to obtain moe scala equations.
Fo tangential and nomal components, F ma F F t t ma t dv m dt F F n n ma n v m ρ
Sample Poblem 1.6 The bob of a -m pendulum descibes an ac of a cicle in a vetical plane. If the tension in the cod is.5 times the weight of the bob fo the position shown, find the velocity and acceleation of the bob in that position. STRATEGY: Resolve the equation of motion fo the bob into tangential and nomal components. Solve the component equations fo the nomal and tangential acceleations. Solve fo the velocity in tems of the nomal acceleation.
MODELING and ANALYSIS: Resolve the equation of motion fo the bob into tangential and nomal components. Solve the component equations fo the nomal and tangential acceleations. F t ma t : mg sin 30 ma a t g sin 30 t a t 4.9 m s F n ma Solve fo velocity in tems of nomal acceleation. v ( )( ) an v ρan m 16.03m s ρ n :.5mg mg cos 30 ma a n (.5 cos 30 ) g v ±5.66 m s n a 16.03m s n
REFLECT and THINK: If you look at these equations fo an angle of zeo instead of 30 o, you will see that when the bob is staight below point O, the tangential acceleation is zeo, and the velocity is a maximum. The nomal acceleation is not zeo because the bob has a velocity at this point.
Sample Poblem 1.7 Detemine the ated speed of a highway cuve of adius 10 m banked though an angle q 18 o. The ated speed of a banked highway cuve is the speed at which a ca should tavel if no lateal fiction foce is to be exeted at its wheels. STRATEGY: The ca tavels in a hoizontal cicula path with a nomal component of acceleation diected towad the cente of the path. The foces acting on the ca ae its weight and a nomal eaction fom the oad suface. Resolve the equation of motion fo the ca into vetical and nomal components. Solve fo the vehicle speed.
SOLUTION: MODELING and ANALYSIS: The ca tavels in a hoizontal cicula path with a nomal component of acceleation diected towad the cente of the path. The foces acting on the ca ae its weight and a nomal eaction fom the oad suface.
Resolve the equation of motion fo the ca into vetical and nomal components. F 0 : y F n ma n R cosθ W 0 W R cosθ : W Rsinθ a g W Wv sinθ cosθ g ρ Solve fo the vehicle speed. n v 19.56m s 70.4km h
REFLECT and THINK: Fo a highway cuve, this seems like a easonable speed fo avoiding a spinout. If the oadway wee banked at a lage angle, would the ated speed be lage o smalle than this calculated value? Fo this poblem, the tangential diection is into the page; since you wee not asked about foces o acceleations in this diection, you did not need to analyze motion in the tangential diection.
Kinetics: Radial and Tansvese Coodinates Hydaulic actuatos, extending obotic ams, and centifuges as shown below ae often analyzed using adial and tansvese coodinates.
Eqs of Motion in Radial & Tansvese Components Conside paticle in pola coodinates, F F θ ma ma θ ( ) θ m ( θ + θ ) m
Sample Poblem 1.10 STRATEGY: A block B of mass m can slide feely on a fictionless am OA which otates in a hoizontal plane at a constant ate Knowing that B is eleased at a distance 0 fom O, expess as a function of a) the component v of the velocity of B along OA, and b) the magnitude of the hoizontal foce exeted on B by the am OA. Wite the adial and tansvese equations of motion fo the block. Integate the adial equation to find an expession fo the adial velocity. Substitute known infomation into the tansvese equation to find an expession fo the foce on the block.
MODELING and ANALYSIS: Wite the adial and tansvese equations of motion fo the block. Integate the adial equation to find an expession fo the adial velocity. dv dv d dv v v dt d dt d dv dv d dv v v dt d dt d v dv θ d θ d v vdv θ0 0 0 d Substitute known infomation into the tansvese equation to find an expession fo the foce on the block. 0 v θ 0 ( ) m θ F F m a m θ a θ ( ) 0 : 0 : F ( θ + θ ) m 0 ( ) 1 F mθ 0
1. Angula Momentum and Obital Motion Satellite obits ae analyzed using consevation of angula momentum.
Eqs of Motion in Radial & Tansvese Components Conside paticle in pola coodinates, ( ) F ma m θ F θ ma θ ( θ + θ ) m This esult may also be deived fom consevation of angula momentum, H O F F θ θ m d dt θ ( ) m θ ( ) θ + θ m ( θ + θ ) m
A. Angula Momentum of a Paticle H O mv moment of momentum o the angula momentum of the paticle about O. H O is pependicula to plane containing H O mv sinφ mv m θ θ H O i x y z mv and mv Deivative of angula momentum with espect to time, H O mv + mv V mv + ma F M O x It follows fom Newton s second law that the sum of the moments about O of the foces acting on the paticle is equal to the ate of change of the angula momentum of the paticle about O. j mv y k mv z
B. Consevation of Angula Momentum When only foce acting on paticle is diected towad o away fom a fixed point O, the paticle is said to be moving unde a cental foce. Since the line of action of the cental foce passes though O, (1.) Position vecto and motion of paticle ae in a plane pependicula to M O H O 0 and mv H constant O H O. * Magnitude of angula momentum, H O mv sinφ constant o mv 0 0 sin φ 0 (1.3)
H O H m O m θ θ h constant angula momentum unit mass (1.4) Radius vecto OP sweeps infinitesimal aea da Define aeal velocity da dt 1 dθ 1 dθ 1 θ dt Recall, fo a body moving unde a cental foce, h θ constant When a paticle moves unde a cental foce, its aeal velocity is constant.
C Newton s Law of Gavitation *Gavitational foce exeted by the sun on a planet o by the eath on a satellite is an impotant example of gavitational foce. *Newton s law of univesal gavitation - two paticles of mass M and m attact each othe with equal and opposite foce diected along the line connecting the paticles, Mm F G G constant of gavitation 66.73 10 Fo paticle of mass m on the eath s suface, 1 m 3 kg s MG W m mg g R 34.4 10 9 ft 4 lb s m ft 9.81 3. s s 4
Sample Poblem 1.1 A satellite is launched in a diection paallel to the suface of the eath with a velocity of 30,000 km/h fom an altitude of 400 km. Detemine the velocity of the satellite as it eaches it maximum altitude of 4000 km. The adius of the eath is 6370 km. STRATEGY: Since the satellite is moving unde a cental foce, its angula momentum is constant. Equate the angula momentum at A and B and solve fo the velocity at B.
MODELING and ANALYSIS: Since the satellite is moving unde a cental foce, its angula momentum is constant. Equate the angula momentum at A and B and solve fo the velocity at B. mvsinφ H constant mv O A A B B A vb va B mv ( ) ( + ) ( + ) 6370 400 km 30,000km h 6370 4000 km v 19,590km h B
REFLECT and THINK: Note that in ode to incease velocity, a spacecaft often applies thustes to push it close to the eath. This cental foce means the spacecaft s angula momentum emains constant, its adial distance deceases, and its velocity v inceases.
*1.3 APPLICATIONS OF CENTRAL FORCE MOTION Tajectoy of a Paticle Unde a Cental Foce m Fo paticle moving unde cental foce diected towads foce cente, ( ) θ F F m( θ + θ ) Fθ 0 Second expession is equivalent to fom which, h h d 1 θ and dθ Afte substituting into the adial equation of motion and simplifying, d u F 1 + u whee u dθ mh u If F is a known function of o u, then paticle tajectoy may be found by integating fo u f(θ ), with constants of integation detemined fom initial conditions.
Application to Space Mechanics *Conside eath satellites subjected to only gavitational pull of the eath, d u F 1 GMm + u whee u F dθ mh u d u GM + u constant dθ h Solution is equation of conic section,(1.37) 1 GM u h GMmu ( 1+ ε cosθ ) ε eccenticity Oigin, located at eath s cente, is a focus of the conic section. Tajectoy may be ellipse, paabola, o hypebola depending on value of eccenticity. Ch GM
Tajectoy of eath satellite is defined by 1 GM h Ch GM ( 1+ ε cosθ ) ε eccenticity (1.37) *Hypebola, e > 1 o C > GM/h. The adius vecto becomes infinite fo + ± 1 1 ± 1 ε cosθ1 0 θ1 cos cos ε GM C h *Paabola, e 1 o C GM/h. The adius vecto becomes infinite fo 1+ cosθ 0 θ 180 1 *Ellipse, e < 1 o C < GM/h. The adius vecto is finite fo θ and is constant, i.e., a cicle, fo e 0.
Integation constant C is detemined by conditions at beginning of fee flight, θ 0, 0, 1 GM Ch 1 cos 0 + 0 h GM 1 GM 1 GM C h v ( ) 0 0 0 0 Satellite escapes eath obit fo ε 1 o C GM h GM ( 0v0 ) GM vesc v0 0 Tajectoy is elliptic fo v 0 < v esc and becomes cicula fo e 0 o C 0,
v cic GM 0 Recall that fo a paticle moving unde a cental foce, the aeal velocity is constant, i.e., Peiodic time o time equied fo a satellite to complete an obit is equal to aea within the obit divided by aeal velocity, da dt h constant 1 1 θ whee a b 1 π ab π ab τ h h + ( ) 0 0 1 1
Sample Poblem 1.14 A satellite is launched in a diection paallel to the suface of the eath with a velocity of 36,900 km/h at an altitude of 500 km. Detemine: a) the maximum altitude eached by the satellite, and b) the peiodic time of the satellite. STRATEGY: Tajectoy of the satellite is descibed by 1 GM + C cosθ h Evaluate C using the initial conditions at θ 0. a)detemine the maximum altitude by finding at θ 180 o.
With the altitudes at the peigee and apogee known, the peiodic time can be evaluated. MODELING and ANALYSIS: Tajectoy of the satellite is descibed by 1 GM + C cos θ h v 0 0 ( ) 6370 + 500 km 6 6.87 10 m km 1000 m/km 36,900 h 3600s/h Evaluate C using the initial conditions at θ 0. h GM 0 0 3 10.5 10 m s 6 3 ( 6.87 10 m)( 10.5 10 m s) v 9 70.4 10 m s ( 9.81m s )( 6.37 10 m) 6 gr 1 3 398 10 m s 1 GM C h 0 1 398 10 m s 1 3 ( ) 6 6.87 10 m 70.4 m s 65.3 10 m 9-1
Detemine the maximum altitude by finding 1 at θ 180 o. 1 3 1 GM 398 10 m s 9 1 C 65.3 10 1 h 70.4 m s m b) With the altitudes at the peigee and apogee known, the peiodic time can be evaluated. a b 1 ( + ) 1 ( 6.87 + 66.7) 0 0 1 π ab τ h 1 π 6 6.87 66.7 10 6 6 ( 36.8 10 m)( 1.4 10 m) 70.4 10 ( ) 1 66.7 10 m 66,700 km max altitude 66,700-6370 km 60,300 km ( ) 9 6 10 m 1.4 10 m 6 s m 36.8 10 6 m 6 m τ 70.3 10 3 s 19h 31min
REFLECT and THINK: The satellite takes less than one day to tavel ove 60,000 km fom the eath and back. In this poblem, you stated with Eq. 1.37, but it is impotant to emembe that this fomula was the solution to a diffeential equation that was deived using Newton s second law.
Keple s Laws of Planetay Motion Results obtained fo tajectoies of satellites aound eath may also be applied to tajectoies of planets aound the sun. Popeties of planetay obits aound the sun wee detemined astonomical obsevations by Johann Keple (1571-1630) befoe Newton had developed his fundamental theoy. Each planet descibes an ellipse, with the sun located at one of its foci. The adius vecto dawn fom the sun to a planet sweeps equal aeas in equal times. The squaes of the peiodic times of the planets ae popotional to the cubes of the semimajo axes of thei obits.