Subfield: Mechanics

$g=9.8 \mathrm{~N} / \mathrm{kg}$



$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$



$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$



$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$



$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$



$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$



$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$



$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$



$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$



$$

\begin{aligned}

& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

& e=1.602 \times 10^{-19} \mathrm{C} \\

& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

& \ln (1+x) \approx x \text { for }|x| \ll 1 \\

& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{aligned}

$$

$$

\begin{array}{lrl}

\text { velocity parameter } & \beta & =\frac{v}{c} \\

\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\

\text { relativistic momentum } & p & =\gamma \beta m c \\

\text { relativistic energy } & E & =\gamma m c^{2} \\

\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}

\end{array}

$$

Extra Supplementary Reading Materials:

Suppose a domino stands upright on a table. It has height $h$, thickness $t$, width $w$ (as shown below), and mass $m$. The domino is free to rotate about its edges, but will not slide across the table.

<img_4527>

Question: a. Suppose we give the domino a sharp, horizontal impulsive push with total momentum $p$.

i. At what height $H$ above the table is the impulse $p$ required to topple the domino smallest?



ii. What is the minimum value of $p$ to topple the domino?

Solution:
First we'll look for the height $H$ at which we should push to topple the domino with the least momentum. A convenient method is to look at the angular momentum in the domino because this is easy to calculate and describes rotational motion. Because the push is horizontal, the moment arm of the push (about the domino's rotational axis) is purely vertical. That means the angular momentum of the push is $p H$, where $p$ is the momentum imparted and $H$ is the height of the push. There is some minimum angular momentum $L_{\min }$ to topple the domino, so we set $L_{\min }=p H$. The bigger $H$, the smaller $p$, so we should choose the largest possible $H$. In other words, we should push at the very top of the domino, $H=h$. While we're on this part, note that if the push weren't constrained to be horizontal, the push could be a little bit smaller since the moment arm could be the entire diagonal of the thin edge of the domino.



Next we calculate the minimum $p$ using energy. As the domino rotates, it converts kinetic energy to potential energy, so we'll calculate both. The domino's potential energy is greatest when its center of mass is directly over the contact point of the domino and the table. That height is half the diagonal of the domino, so the distance from the contact point to the center of the domino is $\frac{1}{2} \sqrt{t^{2}+h^{2}}$. From the push until it reaches this point, the domino's potential energy increases by



$$

\Delta U=\frac{1}{2} m g\left(\sqrt{t^{2}+h^{2}}-h\right)

$$



Then the domino topples. By conservation of energy, $\Delta U$ is how much rotational kinetic energy the domino must have begun with.



Next we find the kinetic energy using the rotational kinetic energy formula, $T=L^{2} / 2 I$. The moment of inertia of the domino about its contact point with the table is $I=\frac{1}{3} m\left(h^{2}+t^{2}\right)$.







You can find this with an integral, or if you know the moment of inertia about the center $\left(\frac{1}{12} m\left(h^{2}+t^{2}\right)\right.$ ), you can use the parallel axis theorem. We know the momentum $L=p h$, so the kinetic energy is



$$

T=\frac{1}{2} \frac{L^{2}}{I}=\frac{1}{2} \frac{(p h)^{2}}{(1 / 3) m\left(h^{2}+t^{2}\right)}=\frac{3}{2} \frac{p^{2} h^{2}}{m\left(h^{2}+t^{2}\right)}

$$



Setting the initial kinetic energy equal to the gain in potential energy and solving for $p$,



$$

p_{\min }=\frac{1}{\sqrt{3}} \frac{m}{h} \sqrt{g\left(\sqrt{t^{2}+h^{2}}-h\right)\left(h^{2}+t^{2}\right)}

$$

Final answer: $H=h$ , $p_{\min }=\frac{1}{\sqrt{3}} \frac{m}{h} \sqrt{g\left(\sqrt{t^{2}+h^{2}}-h\right)\left(h^{2}+t^{2}\right)}$