Moment of Inertia: Cylinder转动惯量:圆柱体
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Moment of Inertia: Cylinder转动惯量:圆柱体
The expression for the moment of inertia of a solid cylinder can be built up from the moment of inertia of thin cylindrical shells.
Show form of integral展示积分形式
实心圆柱体的转动惯量可以从薄圆柱壳的转动惯量构建起来。 |
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Moment of Inertia: Hollow Cylinder转动惯量:空心圆柱
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Moment of Inertia: Cylinder About Perpendicular Axis转动惯量:关于垂直轴的圆柱体
The development of the expression for the moment of inertia of a cylinder about a diameter at its end (the x-axis in the diagram) makes use of both the parallel axis theorem and the perpendicular axis theorem. The approach involves finding an expression for a thin disk at distance z from the axis and summing over all such disks. 推导圆柱体关于其端面直径(图中x轴)的转动惯量的表达式时,同时运用了平行轴定理和垂直轴定理。该方法涉及找到距离轴心距离为z的薄盘的表达式,并对所有这样的薄盘求和。
![]() Obtaining the moment of inertia of the full cylinder about a diameter at its end involves summing over an infinite number of thin disks at different distances from that axis. This involves an integral from z=0 to z=L. For any given disk at distance z from the x axis, using the parallel axis theorem gives the moment of inertia about the x axis. 计算完整圆柱体关于其端面直径的转动惯量需要对无限多个位于不同距离处的薄盘进行求和。这涉及从z=0到z=L的积分。对于任意一个位于距离x轴z处的薄盘,使用平行轴定理可以求得其关于x轴的转动惯量。
![]() Now expressing the mass element dm in terms of z, we can integrate over the length of the cylinder. 将质量元素 dm 表达为 z 的函数,我们可以对圆柱体的长度进行积分。
![]() This form can be seen to be plausible it you note that it is the sum of the expressions for a thin disk about a diameter plus the expression for a thin rod about its end. If you take the limiting case of R=0 you get the thin rod expression, and if you take the case where L=0 you get the thin disk expression. 这种形式可以被看出是合理的,如果你注意到它是薄盘绕其直径的表达式与薄棒绕其一端的表达式的和。如果你取半径R为零的极限情况,就会得到薄棒的表达式;而当长度L为零时,就会得到薄盘的表达式。
The last steps make use of the polynomial forms of integrals. 最后几步利用了积分的多项式形式。
Thin disc moments of inertia薄盘的惯性矩
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