作业帮（我们竞赛老师出的,挺有挑战性）在直角坐标系中重力加速度竖直向下为g,有一光滑长杆绕原点匀速逆时针转动,角速度为w,上面套有一个可以自由滑动的圆环（一开始与长杆相对静止）,问圆
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/04 06:04:49
（我们竞赛老师出的,挺有挑战性）在直角坐标系中重力加速度竖直向下为g,有一光滑长杆绕原点匀速逆时针转动,角速度为w,上面套有一个可以自由滑动的圆环（一开始与长杆相对静止）,问圆
（我们竞赛老师出的,挺有挑战性）
在直角坐标系中重力加速度竖直向下为g,有一光滑长杆绕原点匀速逆时针转动,角速度为w,上面套有一个可以自由滑动的圆环（一开始与长杆相对静止）,问圆环的运动轨迹（可以直接求解析式,也可以分解运动再描述）.
你那个加速度的方程我也会列,问题是加速度决定位移,位移决定旋转半径,旋转半径又决定离心力……加速度（参数包含含位移）二重积分出来又是位移……我不知道该怎么解这种方程……
（我们竞赛老师出的,挺有挑战性）在直角坐标系中重力加速度竖直向下为g,有一光滑长杆绕原点匀速逆时针转动,角速度为w,上面套有一个可以自由滑动的圆环（一开始与长杆相对静止）,问圆
Ok, so you have already know how to derive the formula, the big problem now is how to solve the second order differential equation.
Actually, you can find some books and read yourself which is really benefical.
But here I will tell you the answer.
The hoop will undergo a uniform circular motion on the radical.
But on the normal,
the answer should be
y= normal part C1*e^w+C2*e^(-w)+ particular part (g/1+w^2)*cos(x)
where c1 and c2 are arbitrary constant
because you want to find the particular answer but not the normal answer. the third speaker is right if we do not consider the particular answer.
Hope helps
Cheers