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The light bar can have both tension and support for the object, and the light rope is replaced by a light rod. The pendulum ball can make a complete circular motion in the vertical plane. At the bottom point, the velocity v0 of the ball should be at least Analyze, the ball can be subjected to the pulling force of the light bar at the highest point, and can also be supported by the light bar, so the small ball at the highest point can be zero. Therefore, the ball has the lowest velocity at the highest point and does not deviate from the track, and this speed can be zero. The velocity value v=gL of the small ball at the highest point is the critical value at which the small ball is subjected to the change of the elastic direction of the light bar at its highest point. Ie v Lightly on the object can have both tension and support, but the length changes with the change of force. Example 3 has a light spring with the original length L0, the stiffness coefficient is k, one end is an object of mass m, the other end Fixed at point O on the turntable. The block rotates at an angular velocity along with the turntable, and the maximum static friction between the block and the turntable is fm, and the position range of the block on the turntable is determined. According to the meaning of the problem, there is a maximum static friction force fm between the block and the turntable. When the radius of rotation of the block is the smallest, it is set to r1. At this time, the amount of spring compression is L0-r1, and for the block, it is pointed The maximum static friction force fm of the center of the circle and the spring force F of the spring, and F=k(L0-r1), then fm-k(L0-r1)=mr12, and r1=fm-kL0m2-k is obtained. When the radius of rotation of the block is the largest, it is set to r2, and the elongation of the spring is (r2-L0). For the block, the spring force F and the maximum static friction force fm of the spring pointing to the center are received, and F= k(r2-L0), then k(r2-L0)-fm=mr2, and r2=fm+kL0k-m2 is obtained. Therefore, the position of the block is fm-kL0m2-k≤r≤fm+kL0k-m2.
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