(^172) Aptitude Test Problems in Physics
1.70. The steady-state motion of the system in air
will be the falling of the balls along the vertical at
a constant velocity. The air drag F acting on the
lower (heavier) and the upper ball is the same
since the balls have the same velocity and size.
Therefore, the equations of motion for the balls
can be written in the form
?nig — T — F = 0, rn 2 g T — F = O.
Solving this system of equations, we obtain the
tension of the thread:
T = (m1 — ma) g
2 •
En*. At each instant of time, the instantaneous
axis of rotation of the ball passes through the point
of contact between the thread and the cylinder.
This means that the tension of the thread is per-
pendicular to the velocity of the ball, and hence
it does no work. Therefore, the kinetic energy of
the ball does not change, and the magnitude of its
velocity remains equal to v.
In order to determine the dependence 1 (1), we
mentally divide the segment of the thread un-
wound by the instant t into a very large number N
of small equal pieces of length Al = //N each.
Let the time during which the nth piece is un-
wound be At,. During this time, the end of the
thread has been displaced by a distance v At„, and
the thread has turned through an angle Aq:) =
v AtAn Al) (Fig. 173). The radius drawn to the
point of contact between the thread and the
cylinder has turned through the same angle, i.e.
Al
AWn= --=–T
whence
n (A/) 2
Atn
VT
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