Unformatted text preview: 2 +(v sin θ t − gt2 /2)2 = v 2 t2 − vg sin θ t3 + g 2 t4 /4. (51)
4.13. kx force
Trying a solution ofderivative x(t) =and αt in kx) = mx gives αless ± k/m, so The
We want the the form of ( Ae thus 2 to never be = than zero. the
most general solution isequals zero if
derivative d /dt
x(= 2v 2 t − 3vg sin θBe− gk/m t .
+ t 2 + 2 t3
0 t) = Ae 22
=⇒ otherwise 3vg sin θ t + 2 2
We want A = 0, because 0 = g t −the ﬁrst term vwould become large for √
− k/m t
2 g 2 sin2 = − 8v 2 g 2k /m e− k/m t .
So x(t) = Be
=⇒which gives 3vg= xθ .± Hence, v (t) θ −x0
t = 2 B sin 0
Therefore, v (0) = −x0 k/m.
4.14. RopeA solution does not exist for t if the discriminant is less than zero, that is, if
on a pulley
Let x sin θthe 2 2/3 =each end is5above if θ is less thanaverage height. .5 , then
be < distance ⇒ θ < 70. . So and below the or equal to 70 Then the
never decreases during the ﬂight.
net force along the rope is σ (2x)g , so F = ma gives 2σ gx = σ Lx. So we essentially
¨ have the Exercise.5◦ . Ifwith k/m → 2g/LatYouangle θ larger than θ0 ,the higher end
(b) Let θ0 ≡ 70 4.13 you throw a ball . an should therefore pull then there is a
down point a speed v (0) =points)g/Lthe ﬂight where d /dt = 0. This means that at
with (actually two x0 2 in .
this point the ball is moving in the direction perpendicular to the radial line
from you to the ball. So if this radial line is considered to be the slope of a
Taking the derivative to ﬁnd the max (or min) yields tan ω t = D/C . At this time
hill, then the ball at this point has a velocity that is perpendicular to the hill.
The time-reversed motion of the ball therefore satisﬁes the setup in Exercise
x(t) = C cos ω t + D sin ω t = C · √
= C 2 + D2 . (89)
C 2 than
Conversely, if you throw a ball C 2 an D2 θ smaller + D2 θ0 , then there doesn’t
at + angle
exist a point whe...
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