p08_018.pdf

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Chapter 8 - 8.18
18. We use Eq. 8-18, representing the conservation of mechanical energy (which neglects friction and other
dissipative effects). The reference position for computing U (and height h ) is the lowest point of the
swing; it is also regarded as the “final” position in our calculations.
L cos θ when the angle
is measured from vertical as shown. Thus, the gravitational potential energy is U = mgL (1
−
−
cos θ 0 )
at the position shown in Fig. 8-32 (the initial position). Thus, we have
K 0 + U 0 = K f + U f
1
2 mv 0 + mgL (1
−
cos θ 0 )= 1
2 mv 2 +0
which leads to
v =
2
m
1
2 mv 0 + mgL (1
−
cos θ 0 ) = v 0 +2 gL (1
−
cos θ 0 ) .
(b) We look for the initial speed required to barely reach the horizontal position – described by v h =0
and θ =90 ◦ (or θ =
90 ◦ , if one prefers, but since cos(
−
φ )=cos φ , the sign of the angle is not a
concern).
K 0 + U 0 = K h + U h
1
2 mv 0 + mgL (1
−
cos θ 0 )=0+ mgL
which leads to v 0 = √ 2 gL cos θ 0 .
(c) For the cord to remain straight, then the centripetal force (at the top) must be (at least) equal to
gravitational force:
mv t
r
= mg =
⇒
mv t = mgL
wherewerecognizethat r = L . We plug this into the expression for the kinetic energy (at the top,
where θ = 180 ◦ ).
K 0 + U 0 = K t + U t
1
2 mv 0 + mgL (1
−
cos θ 0 )= 1
2 mv t + mg (1
−
cos180 ◦ )
1
2 mv 0 + mgL (1
cos θ 0 )= 1
−
2 ( mgL )+ mg (2 L )
which leads to v 0 = gL (3+2cos θ 0 ).
(d) The more initial potential energy there is, the less initial kinetic energy there needs to be, in order
to reach the positions described in parts (b) and (c). Increasing θ 0 amounts to increasing U 0 ,sowe
see that a greater value of θ 0 leads to smaller results for v 0 in parts (b) and (c).
(a) Careful examination of the figure leads to the trigonometric relation h = L
−
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