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

MECHANICS
247
Fig. 370
From the ratio Ul =!i, where a2 is the acceleration of the second weight,
a2 '
we can find that
On the basis of Newton's second law, the tensions of the strings T 1 and Ta
are equal to:
, ,2
M 1 +m2 R+ R2(m2+ M2)
,a
ml+M1 +(ma+ M2) R2
ml+M1 +-k (m1+Mz-k)1
. ,2 .:
ml+Ml+(ms+Ms) RZ t
The force F which the system acts on the axis of the pulley with is
F=T 1+T.+(M1+Ms) g
219. Let the path travelled by the centre of gravity of the cylinder during
he time t be equal to s, and the velocity of the centre of gravity be v at
his moment (see Fig. 370).
Hence, from the law of conservation of energy we have
Mvl=Mgs sin ex,
Thus. the velocity is v=Vgs sin a and the acceleration a=g s~n a .
The velocity of the centre of gravity of the cylinder and the angular velocity
of its rotation will be
o=gs~nat and fJ)=g~~at