半圆转动惯量计算

Câu hỏi

一个半圆形薄板质量为M，半径为R。当它以直径为轴转动时，转动惯量为多大

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Noticing properties

I'm examining the moment of inertia for a semicircular lamina. By contrasting it with a full disk, I aim to derive a straightforward expression by utilizing known moments of inertia for disks and semicircular laminae.

$I_{semi\, circle} = \frac{1}{2} \cdot I_{full\, disk}$

Considering the cylinder effect

I'm curious about a full disk's moment of inertia: $\frac{1}{4} M R^2$ about any diameter. This matches earlier findings, aligning with circular symmetry.

Understanding the moment of inertia

I'm mapping out the half-disk's moment of inertia, integrating for symmetry, and carefully considering the axis through the center to ensure accurate mass distribution.

Plotting the axis

I’m setting up the semicircle on a coordinate system with the x-axis as the axis of rotation. I’m integrating to find the moment of inertia using uniform mass density and polar coordinates.

Piecing together polar coordinates

I’m simplifying the integral calculations using polar coordinates for the semicircle aligned with the x-axis. This ensures we correctly account for the moment of inertia.

Staying symmetrical

I’m thinking through the half-disk's symmetry about the x-axis. Given the equal mass distribution above and below this axis, the final moment of inertia calculation should stay balanced and accurate.

Determining the half-disk's inertia

I’m mapping out that the half-disk's moment of inertia is half that of a full disk with double the mass and radius, considering symmetry and distribution of mass above the x-axis.

Ensuring precision in mass

I'm recalculating the moment of inertia for the half-disk by correctly adjusting the mass to $M$ , aligning it with the original specifications for accurate integration.

Taking a closer look

I'm recalculating the moment of inertia for the half-disk, ensuring consistency with mass labeling and integrating over the top half, leading to an accurate value for $I_x$ .

Confirming the outcome

I’ve confirmed that the moment of inertia for the half-disk aligns with earlier findings. This consistency reaffirms that integrating over the top half and applying symmetry yields accurate results.

Thought about half-circle moment of inertia，持续 44 秒

Trả lời

解答过程详述：

已知一个半圆形薄板，其质量为 $M$ ，半径为 $R$ 。我们需要求其绕直径为轴旋转时的转动惯量。

1. 基本思路

半圆形薄板是从一个圆形薄板(质量均匀分布)的一半而来。如果我们先考虑一个完整的圆形薄板，然后利用对称性和比例关系，就可以较为简化地得出半圆的转动惯量。

对于一个半径为 $R$ 、质量为 $M_{\text{full}}$ 的完整圆形薄板，已知其绕任一过圆心的直径轴的转动惯量为（这是标准结果）：

I_{\text{full disk}} = \frac{1}{4}M_{\text{full}}R^{2}.

2. 全圆与半圆的关系

设完整圆形薄板的质量为 $M_{\text{full}}$ ，则半圆形薄板的质量为 $M = \frac{1}{2}M_{\text{full}}$ 。也就是说，如果原本的完整圆盘质量为 $M_{\text{full}} = 2M$ 。

将该完整圆盘关于一条直径轴分割为上下两个完全对称的半圆形薄板，每一个半圆的质量为 $M$ 。

由于圆盘密度均匀，对称分割后，上半圆和下半圆关于此直径轴的质量分布与形状完全相同。转动惯量的计算中，距离轴的垂直距离平方项 $y^2$ 在对称分布中关于中轴线具有对称性。

3. 利用对称性

完整圆盘绕该直径轴的转动惯量为：

I_{\text{full disk}} = \frac{1}{4}(2M)R^{2} = \frac{1}{2}MR^{2}.

这里我们已经用到了 $M_{\text{full}} = 2M$ 。

现在，这个完整圆盘被上下对称地分为两半，每半的面积、质量分布（关于该轴）是对称的，因此每个半圆所贡献的转动惯量应是完整圆盘转动惯量的一半。故半圆绕同一轴的转动惯量为：

I_{\text{half disk}} = \frac{1}{2}I_{\text{full disk}} = \frac{1}{2}\left(\frac{1}{2}MR^{2}\right) = \frac{1}{4}MR^{2}.

4. 最终结果

半圆形薄板（质量 $M$ ，半径 $R$ ）绕其直径为轴旋转的转动惯量为：

\boxed{I = \frac{1}{4}MR^{2}}.

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