Drew Youngren dcy2@columbia.edu
where the bounds of integration for $\mathcal R$ or $\mathcal E$ are translated appropriately.
Find the volume of the "stadium" defined as the region above the $xy$-plane bound by a sphere of radius 2, a cylinder of radius 4, and the cone $x^2 + y^2 = z^2$, cut as shown.
\[ \int_{\pi/2}^{2\pi} \int_{\pi/4}^{\pi/2}\int_2^{4\csc \phi} \rho^2 \sin\phi \,d\rho\,d\phi\,d\theta = \]
\[ = (32 - 2\sqrt{2})\pi \]
The whole is the sum of its parts.
Density, a rate of $\frac{\text{mass}}{\text{volume}}$ can vary continuously as a function of location $\mu(x,y,z)$. \[ \text{mass}_{\text{total}} = \iiint\limits_{\mathcal E} dm = \iiint\limits_{\mathcal E} \mu \,dV \]
but density can be more general: $\frac{\text{stuff}}{\text{space}}$.
Suppose point masses $m_1$ and $m_2$ are located at positions $x_1$ and $x_2$, resp., along a line. The center of mass $\bar x$ is the position that balances the "moments".
\[m_1 (x_1 - \bar x) + m_2(x_2 - \bar x) = 0\]
Suppose point masses $m_1$ and $m_2$ are located at positions $x_1$ and $x_2$, resp., along a line. The center of mass $\bar x$ is the position that balances the "moments".
\[\bar x = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\]
Suppose point masses $m_1$ and $m_2$ are located at positions $x_1$ and $x_2$, resp., along a line. The center of mass $\bar x$ is the position that balances the "moments".
\[\bar x = \frac{m_1}{m_1 + m_2} x_1 + \frac{m_2}{m_1 + m_2} x_2\]
aka, a weighted average of the positions.
Suppose mass is distributed continuously along the interval $a \leq x \leq b$ with density $\mu(x)$. The center of mass $\bar x$ is the position that balances the "moments".
\[\int_a^b (x - \bar x) \mu(x)\,dx = 0\]
Suppose mass is distributed continuously along the interval $a \leq x \leq b$ with density $\mu(x)$. The center of mass $\bar x$ is the position that balances the "moments".
\[\bar x = \frac{\int_a^b x \mu(x)\,dx}{\int_a^b \mu(x)\,dx} \]
aka, a weighted average of the positions.
A solid region $\mathcal E$ with continuously varying density (mass per unit volume) $\mu(x,y,z)$ has total mass \[M = \iiint_\mathcal E \mu(x,y,z)\,dV.\]
The center of mass $(\bar{x},\bar y,\bar z)$ is the (weighted) average position of the mass in the object. More concisely, \[\bv{\bar{x} \\ \bar{y} \\ \bar{z}} = \frac1M \bv{\iiint_\mathcal E x \mu(x,y, z)\,dV \\ \iiint_\mathcal E y \mu(x,y, z)\,dV \\ \iiint_\mathcal E z \mu(x,y, z)\,dV}\]
The nose of a race car is modeled by half of a right cone with height $h$ and radius $R$. Assuming uniform density, how high is the center of mass from the bottom.
CoM measures where mass is located on average. It can't measure how far it is spread out.
The moment of inertia does this. For a mass distribution $\mu(x,y,z)$ and a particular axis of rotation, \[ I = \iiint\limits_{\mathcal E} (\text{distance to axis})^2 \mu(x,y,z)\,dV \]
Find the center of mass and the moments of inertia about the standard axes for a mass distribution $\mu$ on the unit square in $xy$.
\[\mu(x,y) = x + 2 y^2 \] in units of mass/area.
\[\bar{x} = \frac{\int_0^1\int_0^1 x (x + 2y^2)\,dy\,dx}{\int_0^1\int_0^1 (x + 2y^2)\,dy\,dx} = \frac47\]
\[\bar{y} = \frac{\int_0^1\int_0^1 y (x + 2y^2)\,dy\,dx}{\int_0^1\int_0^1 (x + 2y^2)\,dy\,dx} = \frac{9}{14}\]
\[I_{x} = \int_0^1\int_0^1 y^2 (x + 2y^2)\,dy\,dx = \frac{17}{30}\]
\[I_{y} = \int_0^1\int_0^1 x^2 (x + 2y^2)\,dy\,dx = \frac{17}{36}\]
Find an expression for the moment of inertia $I$ (about a central axis) in terms of the total mass $M$ (with uniform density) for the following shapes: