t///Example 3: The slider A slides in the slide with1Avelocity a, hinged on a homogeneous rod AB with77mass m and length I , and the rod rotates around A withangular velocity O. As shown in the figure, calculate theQkinetic energy of the rod when the Angle between rodAB and the plumb line is P.Answer: The bar AB moves in a plane, the velocity ofits center of mass C isVc =VA +VCA1The velocity composition vector diagram is shown0Bv2 = v2 +vcA - 2vAVca cos(180° -)=v? +(lo) +2vAlo cosp =v? ++I'? +lov cospSo the kinetic energy of the bar is T = mve +Jco?=m(v+'? +lovA cos )+(ml)=m(v +'? +lovA cos)
Example 3: The slider A slides in the slide with velocity , hinged on a homogeneous rod AB with mass and length , and the rod rotates around A with angular velocity . As shown in the figure, calculate the kinetic energy of the rod when the Angle between rod AB and the plumb line is . A v m l A B A v l A B A v A v C C v CA v Answer : The bar AB moves in a plane, the velocity of its center of mass C is C A CA v v v = + The velocity composition vector diagram is shown. ( ) 2 cos cos 2 cos(180 ) 2 2 4 2 1 2 2 1 2 2 1 2 2 2 A A A A C A CA A CA v l v l v l l v v v v v v = + + = + + = + − − So the kinetic energy of the bar is 2 2 2 1 2 T = 1 mvC + JC ( cos ) ( cos ) ( ) 2 2 3 2 1 2 1 2 2 1 2 1 2 2 2 1 4 2 1 2 1 A A A A m v l l v m v l l v ml = + + = + + +
I . The kinetic energy theorem for particlesTake the vector form of the differential equation for particle motiond=FmdtDot dr on both sides of this equation, and we can getdvdr = F.drmdtDue to dr = vdt, So the above equation can be written asmv.dv = F.drd(mv2) = SW2That is, the differential kinetic energy of the particle is equal tothe element work of the force acting on the particle, which is thedifferential form of the kinetic energy theorem of the particle
Ⅰ. The kinetic energy theorem for particles Take the vector form of the differential equation for particle motion F dt dv m = Dot on both sides of this equation, and we can get dr dr F dr dt dv m = Due to dr vdt , So the above equation can be written as = mv dv F dr = d mv ) = W 2 1 ( 2 That is, the differential kinetic energy of the particle is equal to the element work of the force acting on the particle, which is the differential form of the kinetic energy theorem of the particle
I . The kinetic energy theorem for particlesd(-mv2)=WIn the integral, we get1=W12mymv22That is, in a certain process of particle motion, thechange in kinetic energy of the particle is equal to thework done by the force acting on the particle, whichis the integral form of the kinetic energy theorem oftheparticle
In the integral, we get 12 2 ) 2 1 ( 2 1 d mv W v v = 12 2 1 2 2 2 1 2 1 mv − mv = W That is, in a certain process of particle motion, the change in kinetic energy of the particle is equal to the work done by the force acting on the particle, which is the integral form of the kinetic energy theorem of the particle Ⅰ. The kinetic energy theorem for particles