Solutions (^163)
must be equal to u = v tan a (At is so small that
the angle a practically remains unchanged). Con-
sequently, the linear velocity of all points of the
hoops must have the same magnitude. According
to the energy conservation law, we have
mv 2
mg Ax= 2Mu 2 + —
2
= 2Mv 2 tang a+
mv
2
2
where Mug is the kinetic energy of each hoop at a
given instant. From this equality, we obtain
v 21
2Ax = 4M tan^2 a m g 1+4 (M/m) tan^2 a g.^
As Ax 0, we can assume that v^2 = 2a Ar, where
a is the acceleration of the ring at the initial in-
stant of time. Consequently,
1
a — (^) g.
1+4 (M/m) tang a
1.60. Let the rope move over a distance Al during
a small time interval At after the beginning of mo-
tion and acquire a velocity v. Since At is small,
we can assume that
v 2 = 2a Al, (1)
where a is the acceleration of all points of the rope
at the initial instant.
From the energy conservation law (friction is
absent), it follows that
Mv 2
2
where M is the mass of the rope, and A Wp is the
change in the potential energy of the rope during
the time interval At. Obviously, A Wp corresponds
to the redistribution of the mass of the rope, as a
result of which a piece of the rope of length Al
"passes" from point A to point B (see Fig. 31).
Therefore,
AWp = gh Al.
(3)
=. Awp,^ (2)
11•
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