Download full text - ELSA - Europa

TEST14
Same as TEST13 but uses same
expected.
∆ t as TEST01. This calculation is unstable, as
TEST21
To conclude this exercise, let us consider a similar problem whereby we replace the
gravity g by an initial velocity v
0
directed downwards.
The initial kinetic energy of the system is:
1 2
EK
0
= Mv0
2
The mass will stop when all this energy has been transformed into elastic energy in
the cable or bar. Assuming linear elastic behaviour and Poisson’s coefficient ν = 0
the resistance force of the bar is:
ES
R= σS = EεS
= λ
L0
where λ is the elongation:
λ L−
L 0
The elastic energy is:
2
∆L
ES ES ( ∆L)
EE
= ∫ λdλ
=
0
L0 L0
2
By posing EE
= EK0
one obtains from the above expressions:
LM
0
∆ L=
v0
ES
In order to obtain an elongation of, say, 20 % of the initial length:
∆ L= 0.2L0
= 0.2 m
the initial velocity should be:
11 −8
ES 2.0× 10 ⋅ 2.5×
10
v0
= ∆ L= = 1.414 m/s
LM
0
1⋅100
Let us now compute the time τ to reach the maximum elongation. The equation of
motion of the system may be written as:
ES
ES
M λ =−
L
λ → λ =−
0
ML
λ
0
The solution of this equation, for the particular initial conditions considered in this
problem, is the following harmonic motion:
λ() t = sin( ωt)
with the pulsation given by:
ES
−1
ω = = 7.071 s
ML0
i.e. the same value as in the case considered previously, with gravity and zero initial
velocity. The desired time τ is clearly ¼ of an oscillation period T , and therefore is
given by:
T 2π
τ = = = 0.222 s
4 4ω
14