质点力学习题&eg

质点力学习题&eg

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习题课 1 章: 质点运动学 2 章: 质点动力学
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解题方法是求导; 求任一时刻的 、已知运动方程 a v t r ), ( ) 1 ( 解题方法是积分 求运动方程, 及初始条件 、已知 0 0 ) 2 ( v r a v dt r d v = 2 2 dt r d dt v d a = = dt v r d = dt a v d = = t t v v dt a v d 0 0 = t t r r dt v r d 0 0 计算在坐标系中进行! 质点运动的问题分成两类
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已知质点运动方程 j t i t r ) 4 ( 2 2 + = (SI) ,求: ①质点的初速度和初加速度; ②质点从 t =1s t =2s 的平均速度; t = 1s 时的切向加速度和法向加速度。 j a j t i t r v 2 , 2 2 d d ) 1 ( = = = ) s m ( 2 ), s m ( 2 -2 0 -1 0 = = j a i v ) s m ( 3 2 , 3 2 4 3 2 ) 2 ( 1 - 1 2 2 1 = = = = = + = j i t r v j i r r r i r j i r 1 解: ) s (m 2 1 2 ) s (m 2 1 2 d d 1 2 ) 3 ( 2 - 2 2 2 2 - 2 2 2 2 = + = = = + = = + = + = t a a a t t t v a t v v v t n t y x 小为 因为任一时刻的速度大
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一质点在 xy 平面内作曲线运动,其加速度是时间的函数。 已知 a x =2 a y =36 t 2 (SI) ,当 t =0 时质点静止在坐标原点,求: ①此质点的运动方程;②此质点的轨迹方程 t v a x x d d ) 1 ( = 由定义有 解: 2 0 0 0 d 2 d d t x t t t v x t t x x = = = 得到 同理有 (2) 质点的轨迹方程 3 12 t v y = 4 0 3 0 0 3 d 12 d d t y t t t v y t t y y = = = 得到 j t i t r 4 2 3 + = 质点的运动方程 0 2 4 6 8 10 0 50 100 150 200 250 300 y = 3 x 2 y x 2 dt a dv x x = t v t t a v x t t v x x 2 d 2 d d 0 0 x 0 = = = 2 3 x y =
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假定某日刮北风,风速大小为 u 。一运动员在风中跑步, 他对地的速度大小为
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质点力学习题&eg

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