Solutions 145
otherwise, the "support" would acquire an infinitely
large acceleration:
N 1 sin a = N2 sin a, N^1 = N2.
Moreover, since the lower ball does not ascend,
the normal components of the accelerations of the
balls relative to the right inclined plane must be
equal (there is no relative displacement in this
direction). Figure 152 shows that the angle between
the direction of the normal reaction N2 of the sup-
port and the right inclined plane is It/2 — 2a, and
hence the latter condition can be written in the
form
mg cos a — m 2 g cos a — N2 cos 2a
T711 (^) m2
whence m 2 = m 1 cos 2a. Thus, the lower ball will
"climb" up if the following condition is satisfied:
m 3 < ml cos 2a.
1.45. As long as the cylinder is in contact with
the supports, the axis of the cylinder will be exact-
ly at the midpoint between the supports. Conse-
quently, the horizontal component of the cylinder
velocity is v/2. Since all points of the cylinder axis
move in a circle with the centre at point A , the
total velocity u of each point on the axis is perpen-
dicular to the radius OA =r at any instant of time.
Consequently, all points of the axis move with
a centripetal acceleration ac = u 2 /r.
We shall write the equation of motion for point
0 in terms of projections on the "centripetal" axis:
2
mg cos a— N mac mu
(1)
where N is the normal reaction of the stationary
support. The condition that the separation between
the supports is r implies that the normal reac-
tion of the movable support gives no contribution
to the projections on the "centripetal" axis. Accord-
ing to Newton's third law, the cylinder exerts
10-0771