USAPhO 2019, Полуфинал — Задача 6. Question B3

a.

Since this problem is subtle, we will present multiple solutions.

First 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 $MgR$, the speed $v$ of the ball obeys

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,

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 is also associated with a vertical acceleration of the ball and hence irrelevant. Thus the force in the rod is just a tension

A common pitfall to think of the motion of the system as pure rotation about the bead, giving the incorrect answer $F=Mv^2/R=2Mg$. The reason this is incorrect is that, while it correctly describes the instantaneous velocity, to compute the acceleration one must also account for the acceleration of the instantaneous pivot point, which is not located at the bead for general times. It is possible to do this, but the computation is much more involved.

Second solution: While the first solution required only two lines of algebra, it also required careful reasoning. 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

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

Therefore, the position of the ball is

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

Differentiating this twice, the acceleration of the ball is

In this case, $\theta=90^\circ$ and we are only interested in $a_x$, giving

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 $Mg$ to the left, and hence a tension in the rod. Incidentally, yet another solution is to use $F=Mv^2/R$ where $R$ is the instantaneous radius of curvature of the ellipse, which may be computed directly.

Third solution: 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

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 $T=Mg$.

b.

First solution: At this point the released gravitational potential energy is $2MgR$, and both masses are moving horizontally with speed $v$, where

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

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 $Mg$ due to gravity, the force in the rod is a tension

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

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

which, as before, indicates a tension of $T=9Mg$.

Third solution: 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

This makes answering $5Mg$ 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

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

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

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

c.

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

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

Equating these two expressions and simplifying gives

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 $F=ma$ equations. These are quite complicated, and nobody managed to integrate them to get the correct answer.