Solutions (^195)
of rotation of the hoop about its centre of mass.
Since the velocity of point A of the hoop is always
equal to zero, the two kinetic energy components
are equal (the velocity of the centre of mass is
equal to the linear velocity of rotation about the
centre of mass). Therefore, the total kinetic energy
of the hoop is mv 2 (m is its mass, and v is the ve-
locity of the centre of mass). According to the
energy conservation law, mv 2 = mg (r — hA),
where hA is the height of the centre of mass of the
hoop above point A at each instant of time. Con-
sequently, the velocity of the centre of mass of the
hoop is v = j/g(r — hA ). On the other hand, the
velocity of the pendulum B at the moment when
it is at a height hA above the rotational axis A is
v = /2g (r — hA), i.e. is VT times larger. Thus,
the pendulum attains equilibrium -CT times sooner
than the hoop, i.e. in
t ==
112
0.35 s.
1.98. It should be noted that small oscillations of
the load occur relative to the stationary axis AB
Fig. 189
Fig. 189). Let DC I AB. Then small oscillations
of the load are equivalent to the oscillations of a
simple pendulum of the same mass, but with the
13'
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