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§ 3 、

§ 3 、. 动量 矩 守恒定律. 一、力矩 M. 已知 : 力 F ,作用点位矢 r ,力臂为 d. M. =. F. d. =. F. r. sin. j. M. F. F. r. M. =. F. ×. j. r. 0. r. 四指由 r 转向 F. d. 拇指为 M 正方向. 单位: kgm 2 s -2. 则:力对原点力矩. 方向:用右手定则判断:. 二、动量矩 L. 已知 : 质点质量为 m, 位矢为 r , 速度为 v. mv. P. L. =. =. m. P.

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§ 3 、

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  1. §3、 动量矩守恒定律 一、力矩 M 已知:力F,作用点位矢r,力臂为d M = F d = F r sin j M F F r M = F × j r 0 r 四指由 r转向 F d 拇指为 M正方向 单位:kgm2s-2 则:力对原点力矩 方向:用右手定则判断:

  2. 二、动量矩 L 已知: 质点质量为m,位矢为r,速度为v mv P L = = m P v d j r 0 m d L P r L = P = r × mv × r 比较 r M = F × 单位:kgm2s-1 则:质点的动量 则:对原点的动量矩 = mv r sinj 动量矩 力矩 方向:用右手定则判断。

  3. 三、 动量矩守恒定律 r ( r×P ) d d L L mv d d d = = mv + r × × dt dt dt dt dt = v × mv + r ×F = M M 0 = c L = 则: 即: = 0 若: 若系统的给定点的所受外力矩为零, 系统对该定点的动量矩守恒

  4. L d dt d P M = = F d t 即:系统的合外力为零,总动量守恒。 即:若系统的给定点所受外力矩为零, 则系统对该定点的动量矩守恒

  5. 质量为m的质点以速度沿一直线运动, 则它对直线上任一点的角动量为何值? m m A r r r L L = = P = r × mv P = r × mv × × v A d = mv r sinj L v j [例1] = 0 [例2]质量为m的质点以速度沿一直线运动, 则它对与直线垂直距离为d的一点的角动量 大小是多少? = mvd

  6. 已知地球的质量为m,太阳的质量为M, 地心与日心的距离为 R,引力常数为 G, 求:地球绕太阳作圆周运动的轨道角动量。 解: r L = P = r × mv × F = G = m v2 vR = GMR Mm R R2 = m GMR [例3] L = R mv

  7. F 30 103 由图得: F = 30- ( t-3 )2 t 0 3 6 t = 3 - 3 = 1.27 s 质量的木箱在水平拉力F的作用下,由静止开始运动,若F随时间的变化关系如图中抛物线所表示那样,若木箱与地面之间的静摩擦系数,滑动摩擦系数,如重力加速度以计算,求: (1)写出力F随时间的变化函数;(2)木箱何时开始运动? (3)6s时木箱的运动速度。 [2-36] 解:设: F = a + b( t+c )2 (2) F = m1mg = 20 N 解得: 木箱开时开始运动

  8. 6 ò I =DP = F合dt 1.27 103 109 = 20t- ( t-3 )3 = 20- ( t-3 )2 = 30 +50 3 /3 F合= F – m2mg (3) = 59 Ns v = I/m = 5.9 m/s

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