华南农业大学物理力学（英文版）：《 MECHANICS》

)革款大串 修德博学 创新 South China Agr icultura l Uni versity MECHANICS Kinematics of particles ° Kinetics of particles Geostatics and Hydromechanics Special relativity Dep physics

MECHANICS • Kinematics of particles • Kinetics of particles • Geostatics and Hydromechanics • Special relativity Dep. physics

苹南癢掌大 但博学求回新 South chi na Agricul tural Uni versity Chap. 1 Kinematics of particles describing motion of particles 1. Base conception 2. Important conception 3. Discussion of curve motion 4 Relative motion

D e p. p h y sic s D e p. p h y sic s D e p. p h y sic s 2 Chap.1 Kinematics of particles ——describing motion of particles 1. Base conception 2. Important conception 3. Discussion of curve motion 4. Relative motion

革擦大寧 但博学一求宣倒新 South chi na Agricul tural Uni versity 1. Base conception Particles to describe an object whose parts all move in exactly the same way. Even a complex object can be treated as a particle if there are no internal motions Reference frame and coordinate frame Vectors a physical quantity that requires the specification of both direction and magnitude

D e p. p h y sic s D e p. p h y sic s D e p. p h y sic s 3 1. Base conception • Particles to describe an object whose parts all move in exactly the same way. Even a complex object can be treated as a particle if there are no internal motions. • Reference frame and Coordinate frame • Vectors a physical quantity that requires the specification of both direction and magnitude

③革由接掌大孝 但博学求回新 South chi na Agricul tural Uni versity 2. Important conception Position vector r(t r=x(1)+y(t)j+(1)k Displacement vector △r=r(t+△t)-r(t)

Dep. physics Dep. physics Dep. physics 4 2. Important conception • Position vector • Displacement vector Z Y X O rZ Y X O 1r 2r r   r  r(t  t)  r(t)      r  r ( t )   r  x(t)i  y(t) j  z(t)k    

苹南癢掌大 但博学求回新 South chi na Agricul tural Uni versity ∫=x2+y1+k a=x,i +yaj+zak △ △F=乃h (b -xai +(b-ya)j +(可b-a)k =△x+△y+△xk A=(△)2+(4y2+(△x)2=X=△s cosa=△x △ , CoSy △ F/,cOS △F 5

D e p. p h y sic s D e p. p h y sic s D e p. p h y sic s 5 r   x x i b a   (  ) xi yj zk          z z k b a   (  ) br  ar   y y j b a  (  ) rb xb i yb j zbk        ra xa i ya j zak        2 2 2 r  (x)  (y)  (z)  r z ,cos r y cos  ,cos               r x  r  s o br r  ar  s a b

革擦大寧 信德博学 求创案 South chi na Agricul tural Uni versity Velocity vector ① average velocity △ 2 instantaneous velocity △ v=lim △1→>0△tdt ·Acce| eration vector ① average acceleration △ 2 instantaneous acceleration △v =im △口→0△tdt 6

D e p. p h y sic s D e p. p h y sic s D e p. p h y sic s 6 • Velocity vector ① average velocity ② instantaneous velocity • Acceleration vector ① average acceleration ② instantaneous acceleration r v t      0 lim t r d r v  t dt         v a t      0 lim t v d v a  t dt        

革擦大寧 信德博学 求创案 South chi na Agricul tural Uni versity d i+j+k=v i+v,j+v,k 2,2,dh = ,+卩 )2+( ds (≠ COSY=

D e p. p h y sic s D e p. p h y sic s D e p. p h y sic s 7 v i v j v k x y z       2 2 2 2 2 2 ( ) ( ) ( ) dt dz dt dy dt dx v v v v v   x  y  z     k dt dz j dt dy i dt dx v        dt ds  v v v v v v x y z cos  , cos  , cos   ( ) dt dr 

革擦大寧 但博学求回新 South chi na Agricul tural Uni versity dp d = dy +",j+2k d l+ 十 k dt dt dt x 氵+a,j+ 2 +a,+a g cosa=x C0s阝= CoSy= 8

D e p. p h y sic s D e p. p h y sic s D e p. p h y sic s 8 dt dv a    k dt dv j dt dv i dt dv a x y z        k dt d z j dt d y i dt d x    2 2 2 2 2 2    a x i a y j a z k       2 2 2 a  a  ax  ay  az cos a   ax cos a ay   a az cos   2 2 dt d r  

③举南演掌大 但博学求回新 South chi na Agricul tural Uni versity (例)以下为质点的位置矢量: F=( Rcos a)i+扩(R、o为正常数) 求质点的速度、加速度,并分别求出速度、加速度的大小和方向。 解: ax v=vI+ vy y 已知:x= RcOs O,y=t v=vi+v,j=-(Rosin ot)i+j i√- Rosin at+1=√Ro2sin2o+1 -Rosin ot COSC cOS B r202 Sin ot +1 Ro Sin ot +1

D e p. p h y sic s D e p. p h y sic s D e p. p h y sic s 9 (例) 以下为质点的位置矢量： r  (Rcost)i  tj    （R、 为正常数） 求质点的速度、加速度，并分别求出速度、加速度的大小和方向。 解： x y d r d x d y v v i v j i j d t d t d t             已知：x  R cost，y  t 2 2 2 2 | v | (R sint) 1  R  sin t 1  ( sin ) x y v  v i  v j   R t i  j      2 2 2 2 2 2 sin 1 cos cos sin 1 sin 1 R t R t R t             

③举南演掌大 但博学求回新 South chi na Agricul tural Uni versity di di a=aia y dt dt 据上知:v=-( Rosin ot),vn=1 a=-(ro cos at )i a=rocos at 方向为x轴方向 10

Dep. physics Dep. physics Dep. physics 10 2 2 ( sin ) 1 ( cos ) | | | cos | x y x y x y dv dv dv a a i a j i j dt dt dt v R t v a R t i a R t                            据上知： ， 方向为x轴方向