(^120) Aptitude Test Problems in Physics
upward direction is then
a. Vag ± va
— —
)2 a V109
r 3
1.15. Let tne velocity of the drops above the per-
son relative to the merry-go-round be at an angle a
to the vertical. This angle can be determined from
the viiodity triangle shown in Fig. 131.
Since in accordance with' the velocity composi-
on rule vo=vrel+vm.g•r, where vm.g.,. is the
velocity of the merry-go-round in the region of la
Fig. 131 Fig. 132
cation of the person, v„i = vo — vm.g. r• The ve-
locity of the merry-go-round is vm .g.r = O. Con-
sequently cot a = v 0 /(wr).
Therefore, the axis of the umbrella should be
tilted at the angle a = arccot [vo/(wr)] to the ver-
tical in the direction of motion of the merry-go-
round and perpendicular to the radius of the latter.
1.16•. Let the board touch the bobbin at point C
at a certain instant of time. The velocity of point
C is the sum of the velocity vo of the axis 0 of
the bobbin and the velocity of point C (relative to
point 0), which is tangent to the circle at point C
and equal in magnitude to vo (since there is no
slipping). If the angular velocity of the board at
this instant is w, the linear velocity of the point
of the board touching the bobbin will be
eiR tan-1 (a/2) (Fig. 132). Since the board remains
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