Aptitude Test Problems in Physics Science for Everyone by S Krotov ( PDFDrive.com )

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Solutions (^189)
the equilibrium can be only neutral, i.e. the cen-
tre of mass of the rod must be on the same level
for any position of the rod. If the end of the rod
leaning against the surface has an abscissa (x) the
ordinate (y(1) of its other end touching the vertical
wall can be found from the condition
/2 = Ey (x)._ yol2 + x2, (^) yo y (X) ± V- 1 2 — x 2.
Since the rod is homogeneous, its centre of mass is
at the midpoint. Assuming for definiteness that the
ordinate of the centre of mass is zero, we obtain
Yo+Y (x) __
2
whence
1//2
x2
Y (x)= ± 2
Only the solution with the minus sign has the
physical meaning. Therefore, in general, the cross
section of the surface is described by the function
V 1 2 —x 2
y (x) -_ a
2
where a is an arbitrary constant.
1.90. In the absence of the wall, the angle of de-
flection of the simple pendulum varies harmoni-
cally with a period T and an angular amplitude cc.
The projection of the point rotating in a circle of
radius a at an angular velocity w = 23t/T per-
forms the same motion. The perfectly elastic colli-
sion of the rigid rod with the wall at an angle of
deflection corresponds to an instantaneous jump
from point B to point C (Fig. 185). The period is
reduced by At = 2y/co, where y = arccos (13/a).
Consequently,
2g 2y
'

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