Chapter 8 Even Problem Solutions
38.
Because the track is frictionless, conservation of energy can be used. Let's
dene (naturally) 0 gravitational potential energy at a height 0. Because
the particle was initially at rest, it's initial energy is:
(1)
Ei = mghi = mg(3.8m)
a.
At the point B, it's energy must be conserved, thus (hB is the height
at point B):
1
mv 2 + mghB = Ei = mghi
2
1
⇒ mv 2 = mg(hi − hB )
2
p
m
⇒ v = 2g(hi − hB ) = 4.8
s
EB =
b.
40.
p
2g(hi − hC ) = 7.0
m
s
(4)
(5)
The particle will turn around when it reaches a point with the same
height as A. This is because at that point, all of the particle's energy
must be potential energy (by conservation of energy) and therefore
it must have 0 kinetic energy and therefore 0 velocity at that point.
Looking on the graph, it appears this point is around 11.5m.
The turning points are when the total energy is completely in potential
energy, as argued above. Thus (we assume the constants in the potential
energy function are in the correct units):
7.0 − 8.0xturn + 1.7x2turn = 3.5
⇒ xturn = .48m
48.
(3)
We do the same thing for point C:
v=
c.
(2)
or
4.2m
(6)
(7)
We know that the force is the negative of the spatial derivative of the
potential energy:
dU
Fx = −
(8)
dx
Thus:
Fx (x = 10nm)
d Fx (x = .10nm) = − U dx x=.10nm
=−
=
(9)
d
[286(x − xe )2 − 6.22 ∗ 1012 (x − xe )3 ].10nm
(10)
dx
−[572(x − xe ) − 1.87 ∗ 1013 (x − xe )2 ].10nm (11)
1
We know that xe = .0741nm, so:
x − xe = .1nm − .0741nm = .0259nm
(12)
Therefore:
Fx
= − 572(.0259 ∗ 10−9 ) + 1.87 ∗ 1013 (.0259 ∗ 10−9 )2
= − 2.27 ∗ 10−9 N
64.
Using Einstein's formula:
(13)
(14)
E = mc2
(15)
3.85 ∗ 1026 W
∆E
P
∆m
9 kg
= 2
= 2 =
m 2 = 4.3 ∗ 10
8
∆t
c ∆t
c
(3 ∗ 10 s )
s
(16)
We immediately get:
This is about 400 million tons of mass which the Sun loses every second.
Luckily, the Sun has a total mass about 2 ∗ 1021 million tons, so it's not
in danger of radiating away most of its mass anytime soon.
78.
Let θ denote the angle of the displacement vector of the bug as measured
from the vertical, and let r denote the radius of the spherical head. The
forces on the bug are the normal and the gravitational force (draw a picture
for this). The gravitational force is −mg going down, while the normal
force acts only in the perpendicular direction going outwards. From drawing a gure, we can deduce the total forces in the radial direction (positive
being outwards):
Fr = −mg cos(θ) + N
(17)
In order for the y to trace a circular trajectory (i.e. it doesn't leave the
head), it must have a total force acting on it in the radial direction which
is equal to the centripetal force:
Fc = −m
v2
r
(18)
We have the negative sign there because the centripetal force must point
inwards. We now set the two equal:
−mg cos(θ) + N = −m
v2
r
(19)
Notice that the normal force N can only be positive (the normal force,
by denition, cannot act inward, i.e. it cannot act to attract the y to
the head). As the y moves down the head, θ gets larger and larger, and
therefore cos(θ) gets smaller and smaller. At the same time, the velocity
of the y will increase. The normal force is the force trying to keep the
two sides of the equation equal; however, at some point, the normal force
2
will have to turn negative in order for the above equation to be true, and
this cannot happen. Therefore, when the normal force turns from positive
to negative, that is when the y will leave the head (and no longer trace a
circular trajectory). Thus, the y will detach from the head at the point
where:
v2
(20)
−mg cos(θ) = −m
r
Let y be the vertical distance that the y has dropped (i.e. the y starts
at y = 0). From conservation of energy, we know:
1
mv 2 = mgy
2
(21)
(22)
v 2 = 2gy
Also note that by basic trigonometry cos(θ) =
equation (20):
g
2gy
r−y
=
r
r
r
⇒y=
3
3
r−y
r .
Thus, plugging into
(23)
(24)