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{$T=M g$}


Question: b. Derive an expression for the force in the rod when the ball is directly below the bead, as shown at right above.

Solution:
At this point the released gravitational potential energy is $2 M g R$, and both masses are moving horizontally with speed $v$, where



$$

2 \frac{1}{2} M v^{2}=2 M g R \quad \Rightarrow \quad v=\sqrt{2 g R}

$$



Work in the frame moving to the right with speed $v$. In this frame the bead is stationary and the ball has velocity $2 v$ and is instantaneously rotating about the bead, so it must be experiencing a centripetal force



$$

\frac{M(2 v)^{2}}{R}=8 M g

$$



Unlike in part (a), there are no additional contribution from the acceleration of the rotation center, because the bead can only ever accelerate horizontally, and the force in the rod at this moment is vertical. Since the ball also experiences a downward force of $M g$ due to gravity, the force in the rod is a tension



$$

T=9 M g

$$
We can also solve the problem using the algebraic approach. Following what we found in the second solution to part (a),



$$

a_{y}=-R\left(\alpha \sin \theta+\omega^{2} \cos \theta\right)=R \omega^{2}

$$



Plugging in the result for $\omega^{2}$, one finds



$$

a_{y}=8 g

$$



which, as before, indicates a tension of $T=9 \mathrm{Mg}$.
Work in the original frame, where the masses are instantaneously rotating about the midpoint of the rod. Because of this motion, the ball must be experiencing a







centripetal force



$$

\frac{M v^{2}}{R / 2}=4 M g

$$



This makes answering $5 \mathrm{Mg}$ a tempting pitfall; to get the right answer, we must also account for the vertical acceleration of the center of mass. The height of the center of mass is $y_{\mathrm{CM}}=(R / 2) \cos \theta$, so



$$

a_{\mathrm{CM}}=\frac{d}{d t}\left(-\frac{R}{2} \omega \sin \theta\right)=-\frac{R}{2}\left(\omega^{2} \cos \theta+\alpha \sin \theta\right)

$$



At this point, $\sin \theta=0$ and $\cos \theta=-1$, so



$$

a_{\mathrm{CM}}=\frac{R}{2} \omega^{2}=\frac{v^{2}}{R / 2}=4 g

$$



where we used $v=\omega(R / 2)$. Therefore, the total upward acceleration of the ball is $8 g$, as we found earlier, so the force in the rod is a tension $T=9 \mathrm{Mg}$.



In summary, there are many valid ways to arrive at the correct answers of $M g$ and $9 M g$, but all of them require careful thought. When this exam was given, only about $2 \%$ of students successfully found these answers.

Final answer: $T=9 Mg$