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原题:如图(a),一个质量为m,半径为R的均匀球,由桌子边缘A无初速滚下,试问:当球脱离桌边缘时速度为多少? 有文对此题作了如下的解答:小球脱离前是一个圆周运动,设脱离时球半径与竖直方向成a角[如图(b)],由机械能守恒定律得: mgR-mgRcosa=1/2mv~2, 由小珠圆周运动受力分析得: mgcosa-N=mv~2/R 脱离时N=0,将以上两式联立,解方程得:cosa=2/3,所以球脱离桌边缘时速度应为v=(2/3gR)~(1/2)。并指出,学生易误答为v=(2gh)~(1/2)。笔者认为,v=(2/3gR)~(1/2)的解答仍是错误的。因为
Original question: As shown in figure (a), a uniform sphere with a mass of m and a radius of R. Rolled off from the edge A of the table without an initial velocity. What is the speed when the ball leaves the edge of the table? The answer is: Before the ball breaks off, it is a circular motion. It is assumed that the radius of the ball when leaving the ball is a angle with the vertical direction [Figure (b)], which is obtained by the conservation of mechanical energy: mgR-mgRcosa=1/2mv~2, from small The force of the bead’s circular motion analysis is: When mgcosa-N=mv~2/R is disengaged, N=0, the above two equations are combined and the solution equation is: cosa=2/3, so the speed of the ball when leaving the edge of the table should be v =(2/3gR)~(1/2). And pointed out that students easily mistaken for v = (2gh) ~ (1/2). The author believes that the answer to v=(2/3gR)~(1/2) is still wrong. because