学 §13-2动量矩定理 一.质点的动量矩定理 d(mv) =F 2 两边叉乘矢径F,有F d(my h×F 7mm)=rxm左边可写成 mny ×一 =(F×mv dt dt t X 而xm=Xmv=0,F×F=m0(F) dt 故: 下Xm)F×F,a[m(m)=m(F 质点对任一固定点的动量矩对时间的导数,等于作用在质 点上的力对同一点之矩。这就是质点对固定点的动量矩定理。 16
16 F dt d mv = ( ) §13-2 动量矩定理 一．质点的动量矩定理 两边叉乘矢径 , 有 r F dt d mv r = ( ) r 左边可写成 mv dt dr r mv dt d dt d mv r = ( ) − ( ) mv v mv 0 , r F m (F ), dt dr 而 = = = O ( ) , [m (mv )] m (F ) dt d r mv r F dt d = O = O 质点对任一固定点的动量矩对时间的导数，等于作用在质 点上的力对同一点之矩。这就是质点对固定点的动量矩定理。 故：
Dynarnics Projecting the equation given above on the axes of a Cartesian coordinate system through a fixed center we obtain m,(mv)=m,(F)d m1(m)=m,(F)m2(m)=m2(F These equations are the moment of momentum theorem of a particle with respect to a fixed axis They are also called the projection forms of the principle of moments with respect to a fixed axis. The derivative of the moment of momentum of a particle with respect to any fixed axis is equal to the moment of the force acting on the particle with respect to the same axis If m,(F)=0(m (F)=O)then mo(mv)=const. (m(mv)=const) These results express the conservation law of the moment of momentum of a particle 7
17 Projecting the equation given above on the axes of a Cartesian coordinate system through a fixed center O we obtain ( ) ( ), ( ) ( ), m (mv ) m (F) dt d m mv m F dt d m mv m F dt d x = x y = y z = z These equations are the moment of momentum theorem of a particle with respect to a fixed axis. They are also called the projection forms of the principle of moments with respect to a fixed axis. The derivative of the moment of momentum of a particle with respect to any fixed axis is equal to the moment of the force acting on the particle with respect to the same axis. If These results express the conservation law of the moment of momentum of a particle. m (F )=0 (m (F )=0) O z then mO (mv )= const. (m (mv) = const) z
力单 将上式在通过固定点O的三个直角坐标轴上投影,得 m,(mv)=m,(F)d m1(m)=m,(F),m2(m)=m2(F 上式称质点对固定轴的动量矩定理,也称为质点动量矩定 理的投影形式。即质点对任一固定轴的动量矩对时间的导数, 等于作用在质点上的力对同一轴之矩。 若m(F)=0(m(F)=0)则m(m)=常矢量(m(m)=常量) 称为质点的动量矩守恒。 18
18 将上式在通过固定点O的三个直角坐标轴上投影，得 ( ) ( ), ( ) ( ), m (mv ) m (F) dt d m mv m F dt d m mv m F dt d x = x y = y z = z 上式称质点对固定轴的动量矩定理，也称为质点动量矩定 理的投影形式。即质点对任一固定轴的动量矩对时间的导数， 等于作用在质点上的力对同一轴之矩。 称为质点的动量矩守恒。 若 m (F )=0 (m (F )=0) O z 则 mO (mv )= 常矢量 (m (mv)=常量) z
Dynarnics EXample 2 a pendulum of mass m and length l is released from =o at t=0. Deduce its equation of motion Solution Choose the small ball m as a particle The force analysis is shown in figure mo(F)=mo()+mo(mg )=-mglsi mg Ino by the analysis of motion we obtain v=lp, OM mo(mv)=mol= ml applying the moment of momentum theorem we hay mo(mv)=m(f), hence (m/)=-mglsin (+sin(=0 When the pendulum swing with small amplitude we can usesin x Leto 2g then (+09=0.Solving the differential equation and taking att=0,=0o, o=0 into account, we obtain the equation of motion of the simple pendulum for small amplitude as =o cos gt The period of the pendulum is 7=2z18 19
19 The force analysis is shown in figure: By the analysis of motion we obtain 2 m (mv) ml l ml O = = v = l , ⊥OM Applying the moment of momentum theorem we have , hence . When the pendulum swing with small amplitude we can use . Let , then .Solving the differential equation and taking at into account, we obtain the equation of motion of the simple pendulum for small amplitude as The period of the pendulum is m (mv ) m (F ) dt d O = O ( ) sin , sin 0 2 = − + = l g ml mgl dt d sin l g n = 2 0 2 +n = t l g cos =0 l g T = 2 mO (F )=mO (T )+mO (mg)=−mglsin Solution Choose the small ball m as a particle. [Example 2] A pendulum of mass m and length l is released from = 0 at t =0. Deduce its equation of motion. t = 0, =0 , 0 = 0
学 例2单摆已知m,l,t=0时∝=q,从静止 开始释放。求单摆的运动规律。 解:将小球视为质点。 受力分析;受力图如图示。 T mo(F)=mo(T)+mo(mg)=-mglsinp 运动分析:v=1,⊥OMm0(m)=mly=m2 tmg 由动量矩定理m2(m)=mn(F) ap a(ml0)=-mglsin, i+ sin=0 微幅摆动时,sinφ≈φ,并令n 2=1 则+O2g=0 解微分方程,并代入初始条件(t=0,q=卯o,0=0)则运动方程 0=90V7,摆动周期7=2( 20
20 运动分析： 。 2 m (mv) ml l ml v = l , ⊥OM O = = 由动量矩定理 即 m (mv ) m (F ) dt d O = O ( ) sin , sin 0 2 = − + = l g ml mgl dt d 微幅摆动时， sin , 并令 l ，则 g n = 2 0 2 +n = 解微分方程,并代入初始条件 (t = 0, =0 , 0 = 0) 则运动方程 t l g cos =0 ，摆动周期 l g T = 2 mO (F )=mO (T )+mO (mg)=−mglsin 解：将小球视为质点。 受力分析；受力图如图示。 [例2] 单摆 已知m，l，t =0时= 0，从静止 开始释放。 求单摆的运动规律