Preparation problems

8-68. An 840-kg roller-coaster car is launched from a giant spring of constant k = 31 kN/m into a frictionless loop-the-loop track of radius 6.2 m, as shown in Figu. 8-41. What is the minimum amount that the spring must be compressed if the car is to stay on the track?

Solution:

The first thing we must determine in solving this problem is the condition which results in the car just barely staying on the track after it is launched by the spring. Let's consider a few options. First, suppose the cart were going fast enough to just make it to the top of the loop with a speed of 0 m/s at the top. This clearly cannot work, as the cart would then fall straight down from the track. A better situation would be to be going fast enough that the acceleration the cart experiences would be enough to pull it into a circular path with the same radius as the track. In this situation, the cart is experiencing zero g's at the top of the track. We can express this condition mathematically as:

ar { top }
=
vf2
r
 = g
vf2
=
rg
(1)
This will allow us to determine the kinetic energy at the top of the track, and we can use conservation of energy to solve the problem.

We can call the initial condition the time when the car is against the compressed spring, and the final condition when the car is moving with the velocity given by Eq. (1) at the top of the loop. Our statement of the conservation of energy then looks like

DK + DU
=
0
[ Kf - Ki ] + [ Uf - Ui ]spring + [ Uf - Ui ]gravity
=
0
é
ê
ë 		1
2
 mvf2 - 0 ù
ú
û 		+ é
ê
ë 		0 - 1
2
 kx2 ù
ú
û 		+ [ mg(2r) - 0 ]
=
0
é
ê
ë 		1
2
 mvf2 - 0 ù
ú
û 		+ é
ê
ë 		0 - 1
2
 kx2 ù
ú
û 		+ [ mg(2r) - 0 ]
=
0
1
2
 mvf2 - 1
2
 kx2 + mg(2r)
=
0
1
2
 kx2
=
1
2
 mvf2 + 2mgr
Substituting our velocity from Eq. (1) into this equation, we get our solution
1
2
 kx2
=
1
2
 m(gr) + 2mgr
1
2
 kx2
=
5
2
 mgr
x2
=
5mgr
k
x
=
  æ
Ö
5mgr
k
 
x
=
2.87  m



Follow-up problems

8-17. A particle moves along the x-axis under the influence of a force F = ax2 + b, where a and b are constants. Find its potential energy as a function of position, taking U = 0 at x = 0.

Solution:

We begin with the definition of potential energy

DUA ® B = - ó
õ 		B

A 
 F(x) dx
We therefore must perform the integration between the origin and another point x to find the potential energy as a function of position. Taking the integral from x = 0 allows us to ensure that U = 0 at x = 0, as desired. The integration gives
U(x)
=
- ó
õ 		x

0 
 F(x) dx
=
- ó
õ 		x

0 
 ( ax2 + b ) dx
=
- é
ê
ë 		1
3
 ax3 + bx ù
ú
û 		x

0 
=
- 1
3
 ax3 - bx

8-38. A particle slides along the frictionless track shown in Fig. 8-30 (p.198), starting at rest from point A. Find (a) its speed at B, (b) its speed at C, and (c) the approximate location of its right-hand turning point.

Solution:

(a) We can use the principle of conservation of energy to find the particle's velocity at point B. The loss in the potential energy of the particle will be made up by its gain in kinetic energy. Thus

DK + DU
=
0
[ KB - KA ] + [ UB - UA ]
=
0
é
ê
ë 		0 - 1
2
 mvB2 ù
ú
û 		+ [ mgyB - mgyA ]
=
0
1
2
 mvB2
=
mgyA - mgyB
1
2
 vB2
=
gyA - gyB
vB2
=
2g( yA - yB )
vB
=
  ___________
Ö 2g( yA - yB )
 
vB
=
4.85  m/s

(b) We can again use the principle of conservation of energy to find the particle's velocity at point C. Here we have

DK + DU
=
0
[ KC - KA ] + [ UC - UA ]
=
0
é
ê
ë 		0 - 1
2
 mvC2 ù
ú
û 		+ [ mgyC - mgyA ]
=
0
1
2
 mvC2
=
mgyA - mgyC
1
2
 vC2
=
gyA - gyC
vC2
=
2g( yA - yC )
vC
=
  ___________
Ö 2g( yA - yC )
 
vC
=
7  m/s

(c) The right-hand turning point is where we expect the particle to stop moving and turn around to the other direction. We would expect this to happen when all of its energy were potential energy, or its height were 3.8 m. Looking at the figure, this occurs at about x = 11 m.

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Solutions translated from TEX by TTH, version 1.57.