(^146) Aptitude Test Problems in Physics
the force of the same magnitude on the stationary
support. From Eq. (1), we obtain
mug
= mg cos a
r •
At the moment when the distance between points
A and B of the supports (see Fig. 18) is AB =
, —
r V 2, we have
cos a=
(^1)
=,
2r -VI •
The horizontal component of the velocity of point 0
is u cos a = vI2, whence u= v1/- 2. Thus, for AB =
rii27 the force of normal pressure exerted by the
cylinder is
mg mv 2
N=
2r •
For the cylinder to remain in contact with the sup-
ports until AB becomes equal to ri/2, the condi-
tion el( 2 > v 2 1(2r) must be satisfied, i.e. v <
gr /2.
1.46. The cylinder is acted upon by the force of
gravity mig, the normal reaction NI of the left
Fig. 153
inclined plane, and the normal reaction N3 of the
wedge (force N 3 has the horizontal direction). We
shall write the equation of motion of the cylinder
in terms of projections on the zraxis directed
along the left inclined plane (Fig. 153):
meal = mig sin a — Ns cos a, (1)
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