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

224 ANSWERS AND SOLUTIONS
171. The motion of the body can be considered as superrosftion of movement
along a circumference With a radius R in a horizonta plane and vertical
falling. Accordingly, the velocity of the body v at the given moment
can be represented as the geometrical sum of two components: VI =v cos a
directed horizontally and v 2=v sin a. directed vertically (Fig. 342). Here a
is the angle formed by the helical line of the groove with the horizon.
In curvilinear motion the acceleration of a body is equal to the geometrical
sum of the tangential and normal accelerations. The normal acceleration
that corresponds to movement along the circumference is
v~ v 2 cost ex
al n = 7[= R
The vertical motion is rectilinear, and therefore a2n = O.
The sought acceleration a = V Q~"t+a:'t+ a~n, where a 1t and a2't are the
tangential accelerations that correspond to motion along the circumference
and along the vertical. The total tangential acceleration Q't Is obviously
equal to a't=V a~1'+a:1'..
The value of a't can be found by mentally developIng the surface of the
cylinder with the helical groove into a plane. In this case the groove will
become an inclined plane with B height nh and a length of Its base 2nRn.
h
Apparently, a~=g sin a.=g y .
h 2+4n2 R2
To determine DIn' let us find v from the law of conservation of energy:
mv 2 2 Bn"nhgR .
T=mghn. Consequently, () =2ghn and a 1n h 2+4n2
R" Upon Inserting
Fig. 343