Solution

We will use conservation of energy. We must take into account the fact that it was rolling, meaning both linear \(KE\) and rotational \(KE\) are present:

Position	\(PE\)	\(KE_{\text{linear}}\)	\(KE_{\text{rotational}}\)	\(E_{\text{total}}\)
Top	\(mgh_1\)	\( 0 \)	\( 0 \)	\(mgh_1\)
Bottom	\(mgh_2\)	\( \frac{1}{2} m v_2^2 \)	\( \frac{1}{2} I \omega_2^2\)	\(mgh_2 + \frac{1}{2} m v_2^2 + \frac{1}{2} I \omega_2^2\)

Before equating the energy, however, we need to simplify the final \(KE_{\text{rotational}}\) in order to express everything in terms of \(v_2\): $$ \begin{eqnarray} KE_{\text{rotational}} &=& \frac{1}{2} I \omega_2^2 \\ &=& \frac{1}{2} (\frac{2}{5} m r^2) (\frac{v_2}{r})^2\\ &=& \frac{1}{5} m v_2^2 \end{eqnarray} $$ This gives:

$$ \begin{eqnarray} mgh_1 &=& mgh_2 + \frac{1}{2} m v_2^2 + \frac{1}{5} m v_2^2 \\ \Rightarrow mgh_1 - mgh_2 &=& \frac{7}{10} m v_2^2\\ \Rightarrow g(h_1 - h_2) &=& \frac{7}{10} v_2^2\\ v_2 &=& \sqrt{\frac{10}{7}g(h_1 - h_2) } \end{eqnarray} $$ Note that \(m\) and \(r\) both canceled out.

Solution

$I$ through the origin

We use $I = \int r^2 dm$, where $r$ is the distance of each mass element from the axis of rotation ($z$-axis in this case). We have $dm = \frac{M}{L^2} dA$ (note that $\frac{M}{L^2}$ is the mass per unit area) and $dA = dx dy$ is the area element. Therefore, we have: $$ \begin{eqnarray} I &=& \int r^2 dm \\ &=& \int (x^2 + y^2) \frac{M}{L^2} dA \\ &=& \frac{M}{L^2} \int_0^L \int_0^L (x^2 + y^2) dx dy \\ &=& \frac{M}{L^2} \left[ \int_0^L x^2 dx \int_0^L dy + \int_0^L y^2 dy \int_0^L dx \right] \\ &=& \frac{M}{L^2} \left[ \frac{L^3}{3} \cdot L + \frac{L^3}{3} \cdot L \right] \\ &=& \frac{2}{3} M L^2 \\ \end{eqnarray} $$

$I$ through the center of the square

Now, we shift the axis of rotation to the center of the square. The limits of integration now change to $-\frac{L}{2}$ to $\frac{L}{2}$ for both $x$ and $y$. Therefore, we have: $$ \begin{eqnarray} I_{CM} &=& \int r^2 dm \\ &=& \int (x^2 + y^2) \frac{M}{L^2} dA \\ &=& \frac{M}{L^2} \int_{-L/2}^{L/2} \int_{-L/2}^{L/2} (x^2 + y^2) dx dy \\ &=& \frac{M}{L^2} \left[ \int_{-L/2}^{L/2} x^2 dx \int_{-L/2}^{L/2} dy + \int_{-L/2}^{L/2} y^2 dy \int_{-L/2}^{L/2} dx \right] \\ &=& \frac{M}{L^2} \left[ \frac{(L/2)^3 - (-L/2)^3}{3} \cdot L + \frac{(L/2)^3 - (-L/2)^3}{3} \cdot L \right] \\ &=& \frac{M}{L^2} \left[ \frac{L^4}{12} + \frac{L^4}{12} \right] \\ &=& \frac{1}{6} M L^2 \\ \end{eqnarray} $$

Using the parallel axis theorem

The parallel axis theorem states that if the moment of inertia about an axis through the center of mass is $I_{CM}$, then the moment of inertia about a parallel axis a distance $d$ away is given by: $$ I = I_{CM} + M d^2 $$ In our case, we have $I = \frac{2}{3} M L^2$ from part (a), and $d = L \sin 45^\circ = \frac{L}{\sqrt{2}}$ (or you can show using the Pythagorean theorem $d^2 + d^2 = L^2$). Therefore, we have: $$ \begin{eqnarray} I_{CM} &=& I - M d^2 \\ &=& \frac{2}{3} M L^2 - M \left(\frac{L}{\sqrt{2}}\right)^2 \\ &=& \frac{2}{3} M L^2 - \frac{1}{2} M L^2 \\ &=& (\frac{4}{6} - \frac{3}{6}) M L^2 \\ &=& \frac{1}{6} M L^2 \\ \end{eqnarray} $$ This is the same result as in part (b).

Solution

$I$ through the origin

We use $I = \int r^2 dm$, where $r$ is the distance of each mass element from the axis of rotation ($z$-axis in this case). The total area of the triangle is $A = \frac{1}{2} L^2$. We have $dm = \frac{M}{A} dA = \frac{M}{L^2/2} dA = \frac{2 M}{L^2} dA$ (note that $\frac{M}{A}$ is the mass per unit area) and $dA = dx dy$ is the area element. Therefore, we have: $$ \begin{eqnarray} I &=& \int r^2 dm \\ &=& \int (x^2 + y^2) \frac{2 M}{L^2} dA \\ &=& \frac{2 M}{L^2} \int (x^2 + y^2) dA \\ &=& \frac{2 M}{L^2} \left[ \int x^2 dA + \int y^2 dA \right] \end{eqnarray} $$ The two integrals are equal due to symmetry (if you switch $x$ and $y$ the triangle and the area element remain unchanged), so we only need to calculate one of them and then double the result. We choose to calculate $\int x^2 dA$. The limits of integration for $x$ is from $0$ to $L$, and for each fixed $x$, the limits of integration for $y$ is from $0$ to $L - x$. Therefore, we have: $$ \begin{eqnarray} \int x^2 dA &=& \int_{x=0}^L \left( \int_{y=0}^{L-x} x^2 dy \right) dx \\ &=& \int_{0}^{L} x^2 (L-x) dx \\ &=& \int_{0}^{L} (L x^2 - x^3) dx \\ &=& \left[ \frac{L x^3}{3} - \frac{x^4}{4} \right]_0^L \\ &=& \frac{L^4}{3} - \frac{L^4}{4} \\ &=& \frac{1}{12} L^4 \end{eqnarray} $$ Therefore, we have: $$ \begin{eqnarray} I &=& \frac{M}{L^2} \left[ \int x^2 dA + \int y^2 dA \right] \\ &=& \frac{2 M}{L^2} \left[ 2 \int x^2 dA \right] \\ &=& \frac{2 M}{L^2} \left[ 2 \cdot \frac{1}{12} L^4 \right] \\ &=& \frac{1}{3} M L^2 \end{eqnarray} $$ Integration by brute force

As $1/4$ of a square

We can view the triangle as $1/4$ of a square of side length $L$ as shown in the figure. The mass of the square is $M_{square} = 4 M_{triangle} = 4 M$, and the length of the square is $L_{square} = \sqrt{2} L$. The moment of inertia of the square about its center is: $$ \begin{eqnarray} I_{square} &=& \frac{1}{6} M_{square} L_{square}^2 \\ &=& \frac{1}{6} (4 M) (\sqrt{2} L)^2 \\ &=& \frac{1}{6} (4 M) (2 L^2) \\ &=& \frac{4}{3} M L^2 \end{eqnarray} $$ Since the triangle is $1/4$ of the square, its moment of inertia about the same axis is: $$ I_{triangle} = \frac{1}{4} I_{square} = \frac{1}{4} \cdot \frac{4}{3} M L^2 = \frac{1}{3} M L^2 $$ This is the same result as in part (a).

Solution

$I_{zz}$

We use $I_{zz} = \sum_i m_i (x^2 + y^2)$, where $i$ sums over the six masses. Note that $x^2 + y^2$ is the square of distance of each mass from the axis of rotation ($z$-axis in this case). $$ \begin{eqnarray} I_{zz} &=& \sum_i m_i (x^2 + y^2) \\ &=& 2 m (L^2 + 0^2) + 2 m (0^2 + L^2) + 2 m (0^2 + 0^2) \qquad \text{Contribution from masses on the $x$, $y$, and $z$ axes respectively} \\ &=& 4 m L ^2 \\ \end{eqnarray} $$ Note that there are two masses on each axis, hence the factor of 2 in front of each term. The two masses on the $z$-axis makes no contribution because their distance to the $z$-axis is zero (i.e., $x = 0$ and $y = 0$) since they are already on the $z$-axis.

$I_{xx}$ and $I_{yy}$

By symmetry, we have: $$ I_{xx} = I_{yy} = I_{zz} = 4 m L^2 $$

Off-diagonal terms

$$ I_{xy} = I_{yx} = - \sum_i m_i x_i y_i = 0 $$ It is zero because for the masses at $x = \pm L$, $y = 0$. For the masses at $y = \pm L$, $x = 0$. Hence, all terms in the sum are zero.
Similar argument means all off-diagonal terms are zero: $$ \begin{eqnarray} I_{xy} &=& I_{yx} = - \sum_i m_i x_i y_i = 0 \\ I_{xz} &=& I_{zx} = - \sum_i m_i x_i z_i = 0 \\ I_{yz} &=& I_{zy} = - \sum_i m_i y_i z_i = 0 \\ \end{eqnarray} $$

Moment of inertia tensor

Therefore, the moment of inertia tensor is given by: $$ \mathbf{I} = \begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \\ \end{bmatrix} = \begin{bmatrix} 4 m L^2 & 0 & 0 \\ 0 & 4 m L^2 & 0 \\ 0 & 0 & 4 m L^2 \\ \end{bmatrix} = 4 m L^2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = 4 m L^2 \mathbf{I_3} $$ where $\mathbf{I_3}$ is the $3 \times 3$ identity matrix. This means the moment of inertia tensor is proportional to the identity matrix, and is therefore independent of the orientation of the object. This explains why in a previous exercise, the moment of inertia about the $z$-axis remained the same after rotation about the $y$-axis.

Solution

$I_{zz}$

We use $I_{zz} = \int (x^2 + y^2) dm$, where $x^2 + y^2$ is the square of the distance of each mass element from the axis of rotation ($z$-axis in this case). We have $dm = \frac{M}{L^2} dA$ (note that $\frac{M}{L^2}$ is the mass per unit area) and $dA = dx dy$ is the area element. Therefore, we have: $$ \begin{eqnarray} I_{zz} &=& \int (x^2 + y^2) \frac{M}{L^2} dA \\ &=& \frac{M}{L^2} \int_0^L \int_0^L (x^2 + y^2) dx dy \\ &=& \frac{M}{L^2} \left[ \int_0^L x^2 dx \int_0^L dy + \int_0^L y^2 dy \int_0^L dx \right] \\ &=& \frac{M}{L^2} \left[ \frac{L^3}{3} \cdot L + \frac{L^3}{3} \cdot L \right] \\ &=& \frac{2}{3} M L^2 \\ \end{eqnarray} $$

$I_{xx}$

We use $I_{xx} = \int (y^2 + z^2) dm$: $$ \begin{eqnarray} I_{xx} &=& \int (y^2 + z^2) dm = \int (y^2 + 0^2) dm = \int y^2 dm \\ &=& \int y^2 \frac{M}{L^2} dA \\ &=& \frac{M}{L^2} \int_0^L \int_0^L y^2 dx dy \\ &=& \frac{M}{L^2} \left[ \int_0^L y^2 dy \int_0^L dx \right] \\ &=& \frac{M}{L^2} \left[ \frac{L^3}{3} \cdot L \right] \\ &=& \frac{M}{L^2} \left[ \frac{L^3}{3} \right] \\ &=& \frac{1}{3} M L^2 \\ \end{eqnarray} $$

$I_{yy}$

By symmetry, we have: $$ I_{yy} = I_{xx} = \frac{1}{3} M L^2 $$

Off-diagonal terms

Because both $I_{xz} = I_{zx} = - \int xz dm $ involves $z = 0$ (the plane of the square), we have: $$ I_{xz} = I_{zx} = 0 $$ For $I_{xy} = I_{yx} = - \int xy dm$, we have: $$ \begin{eqnarray} I_{xy} &=& - \int xy dm \\ &=& - \frac{M}{L^2} \int_0^L \int_0^L xy dx dy \\ &=& - \frac{M}{L^2} \int_0^L x dx \int_0^L y dy \\ &=& - \frac{M}{L^2} \frac{L^2}{2} \cdot \frac{L^2}{2} \\ &=& - \frac{1}{4} M L^2 \end{eqnarray} $$

Moment of inertia tensor

Therefore, the moment of inertia tensor is given by: $$ \mathbf{I} = \begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \\ \end{bmatrix} = \begin{bmatrix} \frac{1}{3} M L^2 & - \frac{1}{4} M L^2 & 0 \\ - \frac{1}{4} M L^2 & \frac{1}{3} M L^2 & 0 \\ 0 & 0 & \frac{2}{3} M L^2 \\ \end{bmatrix} = M L^2 \begin{bmatrix} \frac{1}{3} & - \frac{1}{4} & 0 \\ - \frac{1}{4} & \frac{1}{3} & 0 \\ 0 & 0 & \frac{2}{3} \\ \end{bmatrix} $$

Solution

$F(m, n)$

$$ \begin{eqnarray} F(m, n) &=& \int x^m y^n dV \\ &=& \int_{x=0}^1 \int_{y=0}^{1-x} \int_{z=0}^{1-x-y} x^m y^n dz dy dx \\ &=& \int_{x=0}^1 \int_{y=0}^{1-x} x^m y^n (1-x-y) dy dx \\ &=& \int_{x=0}^1 \int_{y=0}^{1-x} x^m ((1-x) y^n- y^{n+1}) dy dx \\ &=& \int_{x=0}^1 x^m \left[ (1-x) \frac{(1-x)^{n+1}}{n+1} - \frac{(1-x)^{n+2}}{n+2} \right] dx \\ &=& \int_{x=0}^1 x^m (1-x)^{n+2} \left[ \frac{1}{n+1} - \frac{1}{n+2} \right] dx \\ &=& \frac{1}{(n+1) (n+2)}\int_{x=0}^1 x^m (1-x)^{n+2} dx \\ &=& \frac{1}{(n+1) (n+2)} B(m+1, n+3) \\ &=& \frac{1}{(n+1) (n+2)} \frac{m! (n+2)!}{(m+n+3)!} \\ &=& \frac{m! n!}{(m+n+3)!} \end{eqnarray} $$

Volume of the tetrahedron

The volume of the tetrahedron is given by: $$ \begin{eqnarray} V &=& \int dV = F(0,0) = \frac{0! 0!}{(0+0+3)!} = \frac{1}{6} \end{eqnarray} $$

Density of the tetrahedron

The density of the tetrahedron is given by: $$ \begin{eqnarray} \rho &=& \frac{M}{V} = \frac{1}{1/6} = 6 \end{eqnarray} $$

$I_{zz}$

We use $I_{zz} = \int (x^2 + y^2) dm$, where $x^2 + y^2$ is the square of the distance of each mass element from the axis of rotation ($z$-axis in this case). We have $dm = \rho dV$ and $dV = dx dy dz$ is the volume element. Therefore, we have: $$ \begin{eqnarray} I_{zz} &=& \int (x^2 + y^2) \rho dV \\ &=& \rho \int (x^2 + y^2) dV \\ &=& \rho \left[ \int x^2 dV + \int y^2 dV \right] \\ &=& \rho \left[ F(2,0) + F(0,2) \right] \\ &=& 6 \left[ \frac{2! 0!}{(2+0+3)!} + \frac{0! 2!}{(0+2+3)!} \right] \\ &=& 6 \left[ \frac{2}{5!} + \frac{2}{5!} \right] \\ &=& 6 \left[ \frac{4}{120} \right] \\ &=& \frac{1}{5} \end{eqnarray} $$

$I_{xx}$ and $I_{yy}$

Note that if we relabel the axes (x, y, and z), the tetrahedron stays the same. So the integrals are symmetric. Therefore, we have: $$ I_{xx} = I_{yy} = I_{zz} = \frac{1}{5} $$

Off-diagonal terms

For $I_{xy} = I_{yx} = - \int xy dm$, we have: $$ \begin{eqnarray} I_{xy} &=& - \int xy dm \\ &=& - \rho \int xy dV \\ &=& - 6 \cdot F(1,1) \\ &=& - 6 \cdot \frac{1! 1!}{(1+1+3)!} \\ &=& - 6 \cdot \frac{1}{5!} \\ &=& - \frac{1}{20} \end{eqnarray} $$ By symmetry, we have: $$ \begin{eqnarray} I_{xy} &=& I_{yx} = - \frac{1}{20} \\ I_{xz} &=& I_{zx} = - \frac{1}{20} \\ I_{yz} &=& I_{zy} = - \frac{1}{20} \\ \end{eqnarray} $$

Moment of inertia tensor

Therefore, the moment of inertia tensor is given by: $$ \mathbf{I} = \begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \\ \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & - \frac{1}{20} & - \frac{1}{20} \\ - \frac{1}{20} & \frac{1}{5} & - \frac{1}{20} \\ - \frac{1}{20} & - \frac{1}{20} & \frac{1}{5} \\ \end{bmatrix} $$

Restoring $L$ and $M$

To restore the original values of $L$ and $M$, we note that the moment of inertia scales as $M L^2$. Therefore, we have: $$ \mathbf{I} = M L^2 \begin{bmatrix} \frac{1}{5} & - \frac{1}{20} & - \frac{1}{20} \\ - \frac{1}{20} & \frac{1}{5} & - \frac{1}{20} \\ - \frac{1}{20} & - \frac{1}{20} & \frac{1}{5} \\ \end{bmatrix} = \frac{M L^2}{20} \begin{bmatrix} 4 & -1 & -1 \\ -1 & 4 & -1 \\ -1 & -1 & 4 \\ \end{bmatrix} $$

Solution

By the cylindrical symmetry of the object, we can assume both $\vec L$ and $\vec \omega$ lie in the $xz$-plane without loss of generality. Therefore, we have: $$ \vec \omega = \omega_x \hat i + \omega_z \hat k $$ From the definition of the moment of inertia tensor, we have: $$ \begin{eqnarray} \vec L &=& \mathbf{I} \vec \omega \\ &=& \begin{bmatrix} I_s & 0 & 0 \\ 0 & I_s & 0 \\ 0 & 0 & I_n \\ \end{bmatrix} \begin{bmatrix} \omega_x \\ 0 \\ \omega_z \\ \end{bmatrix} \\ &=& \begin{bmatrix} I_s \omega_x \\ 0 \\ I_n \omega_z \\ \end{bmatrix} \end{eqnarray} $$ Therefore, we have: $$ \vec L = I_s \omega_x \hat i + I_n \omega_z \hat k $$ In components, we have: $$ \begin{eqnarray} L_x &=& I_s \omega_x \\ L_z &=& I_n \omega_z \\ \end{eqnarray} $$ From the definition of $\theta$ and $\phi$, we have: $$ \begin{eqnarray} \tan \theta &=& \frac{L_x}{L_z} = \frac{I_s \omega_x}{I_n \omega_z} \\ &=& \frac{I_s}{I_n} \frac{\omega_x}{\omega_z} \\ &=& \frac{I_s}{I_n} \tan \psi \\ \end{eqnarray} $$ Rearranging, we have: $$ \begin{eqnarray} \tan \psi &=& \frac{I_n}{I_s} \tan \theta \end{eqnarray} $$ Therefore, we have: $$ \begin{eqnarray} \tan \phi &=& \tan (\theta - \psi) \\ &=& \frac{\tan \theta - \tan \psi}{1 + \tan \theta \tan \psi} \\ &=& \frac{\tan \theta - \frac{I_n}{I_s} \tan \theta}{1 + \tan \theta \frac{I_n}{I_s}} \\ &=& \frac{\tan \theta \left(1 - \frac{I_n}{I_s}\right)}{1 + \tan \theta \frac{I_n}{I_s}} \\ &=& \frac{(I_s - I_n) \tan \theta}{I_s + I_n \tan^2 \theta} \end{eqnarray} $$ Hey, it works!