Subfield: Mechanics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1

\end{array}

$$

Extra Supplementary Reading Materials:

## Pitfall



A bead is placed on a horizontal rail, along which it can slide frictionlessly. It is attached to the end of a rigid, massless rod of length $R$. A ball is attached at the other end. Both the bead and the ball have mass $M$. The system is initially stationary, with the ball directly above the bead. The ball is then given an infinitesimal push, parallel to the rail.



<img_4549>



Assume that the rod and ball are designed in such a way (not shown explicitly in the diagram) so that they can pass through the rail without hitting it. In other words, the rail only constrains the motion of the bead. Two subsequent states of the system are shown below.

<img_4509>
Context question:
a. Derive an expression for the force in the rod when it is horizontal, as shown at left above.
Context answer:
\boxed{$T=M g$}


Context question:
b. Derive an expression for the force in the rod when the ball is directly below the bead, as shown at right above.
Context answer:
\boxed{$T=9 Mg$}

Question: c. Let $\theta$ be the angle the rod makes with the vertical, so that the $\operatorname{rod} \operatorname{begins}$ at $\theta=0$. Find the angular velocity $\omega=d \theta / d t$ as a function of $\theta$.

Solution:
The height of the ball is $R \cos \theta$, so the released gravitational potential energy is



$$

M g R(1-\cos \theta)

$$



The kinetic energy may be decomposed into rotation about the center of mass and translation of the center of mass; this translation is purely vertical by conservation of momentum. The vertical speed of the center of mass is $(R / 2) \omega \sin \theta$, so the kinetic energy is



$$

\frac{1}{2} I \omega^{2}+\frac{1}{2}(2 M) v_{\mathrm{CM}}^{2}=\frac{1}{4} M R^{2} \omega^{2}+\frac{1}{4} M R^{2} \omega^{2} \sin ^{2} \theta

$$



Equating these two expressions and simplifying gives



$$

\omega^{2}=\frac{4 g}{R} \frac{1-\cos \theta}{1+\sin ^{2} \theta}

$$



Some students instead treated the motion of the bead and ball separately; this can lead to the correct answer, but it is easy to make a mistake. Another common route was to apply Lagrangian mechanics, solving the Euler-Lagrange equations, or equivalently to solve the $F=m a$ equations. These are quite complicated, and nobody managed to integrate them to get the correct answer.



Final answer: $\omega^{2}=\frac{4 g}{R} \frac{1-\cos \theta}{1+\sin ^{2} \theta}$