Machine Dynamics Problems

74 B. Dyniewicz, Cil. Bajer
y(r) =~r(r -1)I,fI (a + i -1)~b+ i-I) rk
a-I k=1i=1 C + I -1 k!
(42)
where t: = vt / I > 0 is time parameter, a = Nl /(2mv2) > 0 determines the
dimensionless parameter. Parameters a, band c are given below:
a _3±~ b _3+~ c=2
1,2 - 2 1,2 - 2 ' (43)
In the case of a = 1 the initial problem has a closed solution. Here we consider the
case of aot:1.
Proof
In (42) we will include the term r(t: - t) into the sum. Thus the equation can be
reduced to the following form:
(l_r)i;(ak)(bk) rk =abr + i;(ak-1)(bk-1)((a+k-l)(b+k-l) _l)_r_k_ (44)
k=1 (ck) k! C k=2 (C k - 1 ) k(c+k-l) (k-l)!
In the above relation (a k ) = a(a + 1)...(a + k -1), (b k ) = b(b + 1)...(b + k -1) and
(c k ) = cCc + 1)...(c + k -1). By using Rabbe criterion one can show that for
a + b < c + 2 the limit
1· [4a (1 )~l1k (a+i-l)(b+i-l)rk ]. fi .
un --r - t: L" IS Illite.
r~l a-I k=1i=1 C + i-I k!
Now we can estimate the value of the sum (44). The sum of the first two-three
terms, depending on parameters, including abilc, is positive. Next terms are all
positive (remember, that P is negative in our numerical example). This proves that
the sum (44) is finite and is greater than O.
·0.5
0
.§
·1
·1.5
Fig. 2. Comparison of particle's trajectory moving on massless and inertial string
v1IL