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>

Question: a. Derive an expression for the force in the rod when it is horizontal, as shown at left above.

Solution:
The motion of the rod can be decomposed as a superposition of rotation about the center of mass, and translation of the center of mass. We claim that the bead is stationary at this moment. To see this, note that the bead cannot have a vertical velocity component since it is fixed to the rod. Moreover, it cannot have a horizontal velocity component, because since the system experiences no external forces in the horizontal direction, the motion of the center of mass has no horizontal component.



Therefore, the bead is stationary, and since the released gravitational potential energy is $M g R$, the speed $v$ of the ball obeys



$$

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

$$



The motion of the system consists of a downward speed $v / 2$ of the center of mass, superposed







on a rotation about the center of mass which gives the masses speed $v / 2$.



Now consider the forces on the ball. The only horizontal force on the ball is the force in the rod, so we need to find the horizontal acceleration of the ball. Part of this is due to the centripetal force associated with the rotational motion,



$$

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

$$



Another part of the ball's acceleration is due to the acceleration of the center of mass, but this is purely vertical and hence irrelevant here. A third part is due to the angular acceleration about the center of mass, but this also is associated with a vertical acceleration of the ball and hence irrelevant. Thus the force in the rod is just a tension



$$

T=M g

$$



A common pitfall to think of the motion of the system as pure rotation about the bead, giving the incorrect answer $F=M v^{2} / R=2 M g$. The reason this is incorrect is that, while it correctly describes the instantaneous velocity, to compute the acceleration one must also account for the rate of change of the pivot point. It can be easy to miss this, because the point at which the pivot is currently located (the bead, in this case) always has zero instantaneous velocity. Yet the pivot point itself does have an instantaneous velocity, which must be accounted for. It is possible to solve the problem this way, but it's subtle.
Here we present a more complicated, but more straightforward alternative. Let $x$ be the horizontal position of the bead. The position of the ball is



$$

\mathbf{r}=(x+R \sin \theta, R \cos \theta)

$$



By conservation of momentum, the center of mass must be at the same horizontal position as it started, so



$$

2 x+R \sin \theta=0

$$



Therefore, the position of the ball is



$$

\mathbf{r}=((R / 2) \sin \theta, R \cos \theta)

$$



That is, it follows an ellipse whose major axis is twice the minor axis.



Differentiating this twice, the acceleration of the ball is



$$

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

$$



In this case, $\theta=90^{\circ}$ and we are only interested in $a_{x}$, giving



$$

a_{x}=-\frac{R \omega^{2}}{2}

$$



At this point, one needs to find $\omega(\theta)$ using conservation of energy. Plugging in the result from either part (c) below or the first solution above gives $a_{x}=-g$, which indicates a force of $M g$ to the left, and hence a tension in the rod. Incidentally, yet another solution is to use $F=M v^{2} / R$ where $R$ is the instantaneous radius of curvature of the ellipse, which may be computed directly.
We can work in the noninertial reference frame of the bead. In this frame, the ball simply rotates around the bead, with the naive centripetal force



$$

F=\frac{M v^{2}}{R}=2 M g

$$



However, since the tension accelerates the bead to the right, there is a fictitious force of magnitude $T$ pulling the ball to the left in this frame. The real tension force and the fictitious force add up to the centripetal force, so $T+T=F$, giving $T=M g$.

Final answer: $T=M g$