Bukhovtsev-et-al-Problems-in-Elementary-Physics

MECHANICS
Fig. 368
two sections, we can write
241
v and in rotational motion with the
velocity VI =roR. The resultant velocity
V2 of the upper section of the hoop
can be found as the geometrical sum of
the velocities v and VI (Fig. 368):
v:=vr+v 2 +2wI cos ex
For a symmetrical section
v:=vl+v~-2vvl cos ex
The total kinetic energy of both sections
is
~E,,=
A I I
= mvs +~= Amv 2 + f1m O)' R2
2 2
Since this expression is valid for any
for the entire hoop
Mvs MR20)2
EII=-2-+-2-
If the hoop rolls without slipping, v=roR and, therefore, Ek=Mv l •
2Gv 2
208. Ek=-g- (",,+1).
209. The cylinder made of a denser material will obviously be hollow. At
the same translational velocities without slipping the kinetic energy of rotation
will be greater in the hollow cylinder, since the particles of its mass are
further from the centre and, therefore, have higher velocities.
For this reason the hollow cylinder will roll down an inclined plane without
slipping slower than the solid one. At the end of the plane, the total
kinetic energies of both cylinders are the same. This is possible only when
the velocities are different, since when the velocities are the same, the energies
of translational motion are identical, while the energy of rotational motion
of a solid cylinder will always be smaller than that of a hollow one.
210. When the drum moves, the force of friction performs no work, since
the cable and the drum do not slip. Hence, the energy of the system does
not change:
!!-v'+GR=G-px u2+