Physicsguy51 ha scritto: ↑9 set 2023, 5:36
According to the condition of the problem, the ball rolls without slipping, therefore, the speeds of those points of the ball that at a given moment of time touch the metal at AB (Fig.) are equal to zero. Considering the ball to be an absolutely rigid body (that is, the distance between any two points of the ball is unchanged), we conclude that at a given moment in time all points of the ball lying on the segment AB are motionless. And this means that at each moment of time the movement of the ball is a rotation about the axis AB. (It is clear that points A and B - the points where the ball touches the metal - are moving with the speed
.)
The instantaneous velocity of any point of the ball is
, where
is the angular velocity of rotation,
is the distance from the point to the axis AB. The speed of the center of the ball (point O in the figure) is equal to
; the distance from point O to axis AB is
. Hence,
It is clear that the points of the ball most distant from the axis AB have the maximum speed. From geometric considerations it is clear that at any moment there is only one point that is maximally distant from the axis - in the figure this is point Q. The distance from point Q to the axis of rotation is
)
, and speed of Q is:
=v \frac{2+\sqrt{3}}{\sqrt{3}}
)
Sorry, but I disagree with this solution. In the diagrams attached by @Pigkappa, which coincides with that of AAPT in
this document it is not entirely clear where points
and
are. Are they
)
respectively the lowest and highest points of the sphere, or
)
the point nearest where the plates meet and the point furthest from it?
The right-hand diagram appears to show the plates end-on, so is looking up the slope. That means
is the correct interpretation. This makes the answer '
' correct. The text is wrong to refer to them as
.
Under interpretation
, the speed of
is
multiplied by the ratio of distances from
to
and
to
, namely,
, so it is
v)
.
Under interpretation
,
is not the fastest point, and its speed is indeterminate because we do not know the angle of the slope. As to how the AAPT answer came to be wrong, it was a rehash of an old question and they forgot to change everything consistently. It happens, and to note, this is not the first time this has occurred in the USAPhO.
It sometimes helps to consider an extreme case. Suppose the plates are vertical, which they could be, as far as we know. The point at the top of the sphere is
from
, so moves at speed
. Meanwhile, the point furthest from the plate join still moves at
)
. I have tried to verify your solution, graphically, using all kinds of software and graphing calculator, but it is in no way possible to obtain an equilateral triangle
, with all three angles of 60 degrees, while keeping the two metal planes orthogonal to each other (as explicitly required by the text). It is not really possible. There is a huge discrepancy between the text and the proposed graphical scheme. I cannot understand why in your opinion the
triangle is equilateral in the case
and
are not at the same level.
in the problem I posted, 
and the result is
, isn't it?