Solutions
153
rotation. When co = Q, the relative velocities of
the regions with the centres of mass at points Al
and A 2 are perpendicular to the segment 00' (di-
rected along the segment A 1 C in Fig. 157), and the
frictional torque about the axis of the smaller disc
is zero. Consequently, the smaller disc will rotate
at the steady-state angular velocity Q.
For co = Q, all the forces of friction acting on
similar pairs of regions of the smaller disc will be
equal in magnitude and have the same direction,
viz, perpendicular to the segment 00'. According
to Newton's third law, the resultant of all the
forces of friction acting on the larger disc will be
applied at the point of the larger disc touching the
centre 0' of the smaller disc and will be equal to
pmg. In order to balance the decelerating torque of
this force, the moment of force
oft = pmgd
must be applied to the axis of the larger disc.
1.50. After the translatory motion of the system
has been established, the ratio of the forces of fric-
tion Ffri and Ffr2 acting on the first and second
rods will be equal to the ratio of the forces of pres-
sure of the corresponding regions: Ffri/Ff r 2
Ni/N2. Since each force of pressure is proportional
to the mass (N 1 = mg and N2 = m 2 g), the ratio
of the forces of friction can be written in the formFfrl (^) — m1 (1)
Ffr2 m2
On the other band, from the equality of the mo-
ments of these forces about the vertex of the right
angle (Fig. 158, top view), we obtain
lFfrl cos q:$= 'Fin sin (I), (2)
where l is the distance from the vertex to the cen-
tres of mass of the rods. From Eqs. (1) and (2), we
obtain tan cp = m 1 /m 2 , where (p = a — W2. Con-
sequently, a = n/2 arctan (ml/m2).
1.51. If the foot of the football player moves at a
velocity u at the moment of kick, the velocity of