Dynarnics 2)Rigid body in fixed-axis rotation: L, =m (m,,=2m r 0=12. a The moment of momentum of a rigid body in fixed-axis rotation with respect to the axis of rotation is equal to the product of the moment of inertia of the rigid body with respect to that axis and its angular velocit 3)Rigid body in plane motion: L-=m(mvc)+Ic.a The moment of momentum of a rigid body in plane motion with respect to an axis perpendicular to the symmetrical plane of mass is equal to the sum of the moment of momentum of the center of mass of the rigid body in translational motion together with the center of mass about that axis and the moment of momentum of the rigid body when rotating around the center of mass with respect to a parallel axis through the center of mass of the body
11 3) Rigid body in plane motion: The moment of momentum of a rigid body in plane motion with respect to an axis perpendicular to the symmetrical plane of mass is equal to the sum of the moment of momentum of the center of mass of the rigid body in translational motion together with the center of mass about that axis and the moment of momentum of the rigid body when rotating around the center of mass with respect to a parallel axis through the center of mass of the body. Lz =mz mi vi =mi ri =Iz 2 ( ) Lz =mz (mvC )+I C 2) Rigid body in fixed-axis rotation: The moment of momentum of a rigid body in fixed-axis rotation with respect to the axis of rotation is equal to the product of the moment of inertia of the rigid body with respect to that axis and its angular velocity
学 2.定轴转动刚体L=m(m)=∑m12O=l0 定轴转动刚体对转轴的动量矩等于刚体对该轴转动惯量与角速 度的乘积。 3.平面运动刚体=m2(mvc)+lcO 平面运动刚体对垂直于质量对称平面的固定轴的动量矩,等于 刚体随同质心作平动时质心的动量对该轴的动量矩与绕质心轴 作转动时的动量矩之和。 12
12 3．平面运动刚体 平面运动刚体对垂直于质量对称平面的固定轴的动量矩，等于 刚体随同质心作平动时质心的动量对该轴的动量矩与绕质心轴 作转动时的动量矩之和。 Lz =mz mi vi =mi ri =Iz 2 ( ) Lz =mz (mvC )+I C 2．定轴转动刚体 定轴转动刚体对转轴的动量矩等于刚体对该轴转动惯量与角速 度的乘积
Dynamics EXample 1 Pulley A: mi, R,R=2R2, 11 Pulley B: m2, R2, 12; M YO A Material body C: m3 Determine the moment of momentum of the system with respect to the axis O 13 Solution o=LoA +oB+loc =1o1+(202+m2V2R)+mn2 1=y2=+m2+m)Rv R 22 Rio 1× R
13 3 2 2 2 1 1 2 1 v = v = R = R ( ) . 2 2 3 2 3 2 2 2 2 1 m m R v R I R I LO = + + + LO =LOA +LOB +LOC 1 1 2 2 2 2 2 3 3 2 = I + (I + m v R ) + m v R Solution： [Example 1] Pulley A: m1，R1，R1=2R2，I1 Pulley B: m2，R2，I2 ； Material body C: m3 Determine the moment of momentum of the system with respect to the axis O
学 例1滑轮A:m,R1,R12R2, 滑轮B:m2,R2,l2;物体C:m3 M YO A 求系统对O轴的动量矩。 解:Lo=LO4+LOg+LoC B =1m1+(202+m22R)+my2R 13 v3=v2=R2O2221 =(n2+2+m2+m3)R23 RR
14 3 2 2 2 1 1 2 1 v = v = R = R 2 2 3 2 3 2 2 2 2 1 ( m m )R v R I R I LO = + + + LO =LOA +LOB +LOC 1 1 2 2 2 2 2 3 3 2 = I + (I + m v R ) + m v R 解： [例1] 滑轮A：m1，R1，R1=2R2，I1 滑轮B：m2，R2，I2 ；物体C：m3 求系统对O轴的动量矩
Dynarnics 813-2 Moment of momentum theorem 1. Moment of momentum theorem of a particle: d(mv)=F Now multiplying both sides of dt this equation by the radius vector r according +(mv) F moomy-rxm to vector multiplication, we get rx F×F dt y The left-hand side can be expressed in the form (mP)=4(×m)-xm But1×m=节×m=0andF×F=mo(F) erefore (F×m)=F×F,[ do(mv l-m(F) The time-derivative of the moment of momentum of a particle with respect to any fixed center with respect to time is equal to the moment of the force acting on the particle with respect to the same 15 center. This is the moment of momentum theorem of a particle
15 §13-2 Moment of momentum theorem 1. Moment of momentum theorem of a particle: mv dt dr r mv dt d dt d mv r = ( ) − ( ) But, mv v mv 0 and r F m (F), dt dr = = = O ( ) , [m (mv )] m (F ) dt d r mv r F dt d = O = O The time-derivative of the moment of momentum of a particle with respect to any fixed center with respect to time is equal to the moment of the force acting on the particle with respect to the same center. This is the moment of momentum theorem of a particle. Therefore: F dt d mv = ( ) r F dt d mv r = ( ) r . Now multiplying both sides of this equation by the radius vector according to vector multiplication, we get . The left-hand side can be expressed in the form