This preview shows page 1. Sign up to view the full content.
Unformatted text preview: re the ball is moving perpendicular to the radial line from
This checks in the special cases where C = 0 or D = 0.
you to the ball, so therefore the velocity is never perpendicular to the slope of
4.16. Angledhill, which means that such a θ isn’t possible in the setup of Exercise 3.51,
a rails
as the position of each mass along that they could produce an angle position.
Let x be we found. (If someone claimed the rail, relative to the equilibrium θ < θ0 in
Then Exercise 3.51, then thetime reversed motion would yield an anglesin < θ0 for
the spring stretches a distance 2x sin θ, yielding a force of 2kx θ θ. The
which the distance decreases at is 2kx sin2 in the ﬂight, namely just after the
component of this force along the rail some point θ. So F = ma along the rail gives
ball passes ¨
through the “hill”, in contradiction with the result of this problem.)
−2kx sin2 θ = mx. Hence, ω = 2k/m sin θ. 30 4.17. Eﬀective spring constant
3.53. Projectile with drag
(a) Let the mass move a distance x to the right. Then the two springs pull to the
(a) left= maforces −k= −αx −k2 x.y The−g − αy . Using the initial −(k1 + thexx
F with gives x 1 x and˙ and ¨ = total force is therefore F = speed, k2 ) .
¨
˙
equationeﬀ = k1 + k2 . Note that if k1 = 0, then keﬀ = k2 , as expected. And if
integrates to
Hence, k
k1 = ∞, then keﬀ = ∞, as expected.
x = Ae−αt =⇒ x = v0 cos θ e−αt .
˙
˙
(53)
(b) Let the mass move a distance x to the right. How much does each spring
stretch? The initial position of springs must exert the same force, otherwise
Assuming an key is that both zero, this then integrates to
there would be a nonzero net force on some part of the massless springs, and
this part= −(v0 cos θ/undergo+ B =⇒ x = (v0 cosLetαthe − e−αt ).stretch(54)
by
x would then α)e−αt inﬁnite acceleration. θ/ )(1 springs
x1 and x2 . Then we have k1 x1CHAPTER 4. also x1 + x2 = x, of course.
= k2 x2 . And OSCILLATIONS 29 Solving this system of two equations gives x1 = k2 x/(k1 + k2 ) and x2 =
k1 x/(k1 + k2 ). The force in each spring (which is the force that the mass feels)
is therefore k1 x1 = k2 x2 = k1 k2...
View Full
Document
 Fall '13
 Brandenberger
 mechanics, Force

Click to edit the document details