Rotational Inertia · 转动惯量
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| rotational inertia/rəʊˈteɪʃənl ɪˈnɜːʃə/ | 转动惯量 | zhuǎn dòng guàn liàng |
| parallel-axis theorem/ˈpærəlel ˈæksɪs ˈθɪərəm/ | 平行轴定理 | píng xíng zhóu dìng lǐ |
| distribution of mass/ˌdɪstrɪˈbjuːʃn ɒv mæs/ | 质量分布 | zhì liàng fēn bù |
Why a skater spins faster with arms tucked in
- A spinning figure skater pulls their arms in and suddenly whirls much faster.
- They added no push -- they only changed where their mass is.
- Spinning has its own kind of "inertia," and it depends on shape, not just amount.
- This is the quantity that makes rotation feel heavy or light.
滑冰者收臂后为什么转得更快
- 旋转的花样滑冰者把手臂收进来,忽然就转得快多了。
- 他们没有额外用力——只是改变了质量在哪里。
- 旋转有它自己的一种"惯性",而且它取决于形状,不只是多少。
- 这就是使旋转显得沉重或轻盈的那个量。
Rotational inertia
- Rotational inertia 转动惯量 $I$ is the resistance to a change in spinning -- the rotational twin of mass.
- A large $I$ means a torque produces only a small angular acceleration.
- Unlike ordinary mass, it depends on how the mass is arranged.
转动惯量
- 转动惯量 $I$ 是对旋转变化的抵抗——质量的旋转孪生。
- 大的 $I$ 意味着一个力矩只产生很小的角加速度。
- 与普通质量不同,它取决于质量如何排布。
Select all · 所有 true statements about rotational inertia. · 选择关于转动惯量的所有正确陈述。
Rotational inertia resists angular acceleration and depends on mass distribution -- shape matters, so the third is false. · 转动惯量抵抗角加速度并取决于质量分布——形状很重要，因此第三项是错误的。
Distance squared
- For point masses, $I = \sum m_i r_i^2$ -- each bit's distance from the axis is squared.
- Mass far from the axis counts far more than mass near it.
- Doubling a mass's distance quadruples its contribution to $I$.
距离的平方
- 对于质点,$I = \sum m_i r_i^2$——每一小块到轴的距离都被平方。
- 离轴远的质量比离轴近的质量重要得多。
- 把一个质量的距离加倍,它对 $I$ 的贡献就变成四倍。
Two $3\ \text{kg}$ masses sit $2\ \text{m}$ from an axis on a light rod. Total rotational inertia (in kg·m^2)? · 两个$3\ \text{kg}$质量的物体位于轻杆上距轴$2\ \text{m}$处。总转动惯量（单位 kg·m^2）是多少？
$I = \sum m r^2 = 3(2)^2 + 3(2)^2 = 12 + 12 = 24\ \text{kg}\cdot\text{m}^2$.
Moving a mass twice as far from the axis changes its contribution to $I$ by a factor of... · 将质量移至距轴两倍远处，其对$I$的贡献变为原来的...倍。
$I \propto r^2$, so doubling $r$ multiplies the contribution by $2^2 = 4$. · $I \propto r^2$，因此使$r$加倍会使贡献变为$2^2 = 4$倍。
Continuous bodies
- For a solid object, integrate: $I = \int r^2\,dm$.
- The parallel-axis theorem 平行轴定理, $I = I_{cm} + Md^2$, shifts a known $I$ to a parallel axis a distance $d$ away.
- Standard shapes (rod, disk, sphere) have tabulated formulas.
连续物体
- 对于实心物体,做积分:$I = \int r^2\,dm$。
- 平行轴定理 $I = I_{cm} + Md^2$,把已知的 $I$ 移到相距 $d$ 的平行轴上。
- 标准形状(杆、盘、球)有现成的公式。
More or less rotational inertia? · 转动惯量更大还是更小？
Rotational inertia depends on how far the mass sits from the axis. Sort each case. · 转动惯量取决于质量距离轴的远近。对每种情况进行排序。
To shift a known rotational inertia to a parallel axis, use the ____-axis theorem. · 要将已知转动惯量平移到平行轴，请使用____轴定理。
$I = I_{cm} + Md^2$ is the parallel-axis theorem. · $I = I_{cm} + Md^2$是平行轴定理。
Why distribution matters
- The distribution of mass 质量分布 is everything: same mass, different shape, different $I$.
- Pulling the skater's arms in shrinks $r$, so $I$ drops and the spin speeds up.
- A flywheel puts its mass at the rim to make $I$ -- and its stored spin -- as large as possible.
为什么分布很重要
- 质量分布就是一切:相同的质量,不同的形状,不同的 $I$。
- 收起滑冰者的手臂缩小了 $r$,所以 $I$ 减小,旋转加快。
- 飞轮把质量放在边缘,使 $I$——以及它储存的旋转——尽可能大。
A skater pulling their arms in decreases their rotational inertia. · 滑冰者收拢手臂会减小其转动惯量。
Arms in means mass is closer to the axis (smaller $r$), so $I$ drops -- and the spin speeds up. · 手臂内收意味着质量更靠近轴（较小的$r$），因此$I$下降——转速加快。
Two objects have equal mass. Which has the larger rotational inertia about its center? · 两个物体质量相等。关于其质心，哪个具有更大的转动惯量？
The hoop's mass sits at large $r$, and $r^2$ weights it heavily, so its $I$ is larger. · 圆环的质量位于大$r$处，且$r^2$对其权重较大，因此其$I$更大。
Two $2\ \text{kg}$ masses sit on a light rod, each $0.5\ \text{m}$ from the axis.
- $I = \sum m r^2 = 2(0.5)^2 + 2(0.5)^2 = 1\ \text{kg}\cdot\text{m}^2$.
- Slide them out to $1\ \text{m}$ and $I$ jumps to $2(1)^2 + 2(1)^2 = 4\ \text{kg}\cdot\text{m}^2$ -- four times larger.
两个 $2\ \text{kg}$ 的质量在一根轻杆上,各离轴 $0.5\ \text{m}$。
- $I = \sum m r^2 = 2(0.5)^2 + 2(0.5)^2 = 1\ \text{kg}\cdot\text{m}^2$。
- 把它们滑到 $1\ \text{m}$,$I$ 跳到 $2(1)^2 + 2(1)^2 = 4\ \text{kg}\cdot\text{m}^2$——大了四倍。
Rotational inertia is not just "how much mass." Two objects of equal mass can have very different $I$: a hoop (mass at the rim) resists spinning far more than a solid disk of the same mass, because the $r^2$ weights the outer mass heavily.
转动惯量不只是"有多少质量"。质量相等的两个物体可以有非常不同的 $I$:一个圆环(质量在边缘)比同质量的实心圆盘更难旋转,因为 $r^2$ 重重地加权了外侧的质量。
Rotational inertia $I$ is the resistance to angular acceleration -- $I = \sum m r^2$ (or $\int r^2\,dm$), so mass far from the axis counts most. The distribution of mass decides it, and the parallel-axis theorem shifts the axis. It is why a skater spins faster with arms pulled in.
转动惯量 $I$ 是对角加速度的抵抗——$I = \sum m r^2$(或 $\int r^2\,dm$),所以离轴远的质量最重要。质量分布决定它,而平行轴定理用来移轴。这就是滑冰者收臂后转得更快的原因。