Download Lecture 5 Forces that are a function of Position

Lecture 5
Forces that are a function of Position
Beginning with the expression that we derived from last time,
F (x) =
dT
dx
(5.1)
where T = 1/2mv 2 is the kinetic energy, we may “solve” for T given a particular force F (x).
That is,
Z
Z
x
T − To =
x
dT =
F (x)dx
(5.2)
where To is the kinetic energy evaluated at V (xo ). Note that,
Z x
F (x)dx
W =
(5.3)
xo
xo
xo
is just the work done on the particle by the applied force F (x).
• This is an important result which states that the work is just the change in kinetic
energy.
• Let us define a function V (x) such that,
−
then it is clear that,
Z x
Z
F (x)dx = −
xo
dV (x)
= F (x)
dx
(5.4)
x
dV
= −V (x) + V (xo ) = T − To
xo
T + V (x) = To + V (xo ) = constant = E
Thus,
1 2
mv + V (x) = E
2
(5.5)
which is just the Energy Equation where we identify V (x) as the potential energy
of the particle.
LECTURE 5. FORCES THAT ARE A FUNCTION OF POSITION
18
Important
• ∴ if the applied force is only a function of position, then the total energy remains
constant throughout the motion.
• Such a force is called a conservative force and these forces play an important role
in many areas of physics.
• Note that from the energy equation we can solve for the motion of the particle as
follows:
r
dx
2
v=
=±
[E − V (x)]
dt
m
• in integral form, this becomes
Z
x
t − to =
xo
dx
q
± m2 [E − V (x)]
which gives us the time as a function of x!
So what are the allowed values of x?
• we are restricted to values of x for which E ≥ V (x).
• also note that the velocity,v, vanishes when E = V (x). Here, the particle comes to
rest and reverses direction and these points are called turning points of the motion:
V(x)
Allowed Region
E
Turning Points
x
Example 5.1 Variation of Gravity with Height
a.
LECTURE 5. FORCES THAT ARE A FUNCTION OF POSITION
19
• Newton’s Law of Gravity states that,
Fr = −
GM1 M2
r2
(5.6)
where the force is directed along the line connecting the two bodies and only depends on the separation between the two bodies.
• Consider now a mass m. The force due to gravity at the surface of the earth is:
−mg
• Take the mass of the earth to be M , then
GM m
re2
GM
g =
re2
mg =
• If the mass is a distance z above the earth, then relative to the center of the earth
r = re + z and,
F (z) = −
GM m
mgre2
GM m
=
−
=
−
= mz̈
r2
(re + z)2
(re + z)2
where we have used GM m = mgre2 .
• Now write,
z̈ =
dv dz
dv
dv
=
= v
dt
dz dt
dz
and therefore,
Z
−mgre2
z
zo
dz
=
(re + z)2
Z
Z
v
v
mvdv =
vo
¡
¢
d ø12mv 2
vo
Solving the above integrals we find,
µ
¶
1
1
1
1
2
mgre
−
= mv 2 − mvo2
r e + z r e + zo
2
2
• This is again just the energy equation, where we identify the potential energy as,
·
¸
re2
V (z) = −mg
(re + z)
which is in a different form than encountered before for the potential energy
(V (z) = mgz)
b. Escape Speed
LECTURE 5. FORCES THAT ARE A FUNCTION OF POSITION
20
• if the mass, m, is projected upward with velocity vo at the surface of the earth,
zo = 0, then the energy equation becomes:
µ
¶
−z
2
2
2
v = vo + 2gre
r (r + z)
Ãe e !
2gre
−z
= vo2 +
re
1 + rze
µ
¶−1
z
2
2
v = vo − 2gz 1 +
re
(Note:if z << re , then we obtain the standard result for a body moving in a
uniform gravitational field)
What is the turning point of the motion?
• this is just the point for which v = 0 which occurs at the maximum height zmax = h.
vo2 zmax
= 2gzmax
re
vo2 vo2 zmax
+
= zmax
2g
2gre
µ
¶
vo2
v2
zmax 1 −
= o
2gre
2g
µ
¶−1
vo2
vo2
zmax = h =
1−
2g
2gre
∴ vo2 +
To obtain the escape speed, we need to obtain the value of h that is infinite. We
can get this by setting the expression above in brackets to zero,
2
vO
= 2gre → vescape = (2gre )1/2
for g = 9.8m/s2 and re = 6.4 × 106 m we find,
vescape ≈ 11km/sec
(5.7)
Questions:
i. Why does the earth retain its atmosphere where as the moon does not have an
atmosphere?
ii. What would happen if the earth’s atmosphere were composed of hydrogen?
5.1
Force as a function of time
We consider now the case where F = F (t) such that we have:
F (t) = m
dv
dt
(5.8)
LECTURE 5. FORCES THAT ARE A FUNCTION OF POSITION
which may be integrated to give,
Z t
F (t0 )dt0 = mv(t) − mv(to ) = ∆p
21
(5.9)
o
where ∆p is the change of momentum.
• This is just the impulse: change of momentum of a body due to a force F (t) acting
over a given time interval.
Z t
dx
F (t0 ) 0
= v(t) = v(to ) +
dt
(5.10)
dt
m
o
¸
Z t
Z t ·Z t
F (t0 ) 0
0
0
dt dt
(5.11)
x − xo =
v(t )dt = v(to )t +
m
o
o
o
Example 5.2 Consider a force of the following form:
F (t) = ct
(think about driving a car and pressing down increasingly on the accelerator). Hence,
dv
= ct
dt
Z t
ct
ct2
∴v =
dt =
m
2m
Zo t 2
ct
ct3
x =
dt =
6m
o 2m
m
• v increases quadratically with t
• x increases as the cube of t
• note that the “jerk”, defined as,
d3 x
c
da
= 3 =
= constant
dt
dt
m
5.2
Velocity-Dependent Forces
• There are many cases where the force on a body depends on the velocity F = F (v).
• If we also include a constant force that does not depend on either v or t, then
Fo + F (v) = m
dv
dv
= mv
dt
dx
(5.12)
• Such forces arise, for example, in the case of viscous resistance exerted on a body
moving through a fluid.
LECTURE 5. FORCES THAT ARE A FUNCTION OF POSITION
22
• This is not necessarily some simple function, but we can make a first approximation
as,
F (v) = −c1 v − c2 v|v|
(5.13)
where the negative signs are because the force is opposite to v. The constants c1 and
c2 depend on the size and shape of the body.
a.
Example 5.3
• Consider a block of mass m which is initially projected with velocity
vo along some smooth surface (i.e. ignore friction) and there is air resistance
(linear). Then, taking Fo = 0 in 5.12 and using the leading term in 5.13, we
have,
µ ¶
Z v
dv
mdv
−mdv
−m
v
−c1 v = m , dt = −
→t=
=
ln
dt
cv
cv
c1
vo
vo
we can now invert this to find v:
v = vo e−c1 t/m
and then integrate to find:
Z t
¢
mvo ¡
1 − e−c1 t/m
x=
vo e−c1 t/m dt =
c1
o
• Note that the velocity of the block decreases exponentially with time and reaches
a limiting position of:
·
¸
¢
mvo ¡
mvo
−c1 t/m
xlim = lim x = lim
1−e
=
t→∞
t→∞
c1
c1
b. Terminal Velocity
• We consider now vertical fall through a fluid so that our constant force, Fo , is just
the force due to gravity, Fo = −mg, and thus,
−mg − c1 v = m
then,
Z
v
t=
vo
hence,
or
dv
dt
mdv
m
= − ln
(−mg − c1 v)
c1
µ
(5.14)
mg + c1 v
mg + c1 vo
¶
mg + c1 v
−tc1
ln
=
mg + c1 vo
m
µ
¶
−mg
mg
v=
+
+ vo e−c1 t/m
c1
c1
¶
µ
(5.15)
LECTURE 5. FORCES THAT ARE A FUNCTION OF POSITION
23
for t À c1 /m, the velocity of the body reaches the limiting value,
vterm =
−mg
c1
(5.16)
referred to as the terminal velocity
• this is the condition where the total force on the body is zero → so there is no
acceleration.
• the above equation can be slightly modified as,
v = −vt + (vt + vo ) et/τ
(5.17)
where vt is the terminal velocity and τ = m/c1 is the characteristic time.
• for a body falling from rest (i.e. at time t = 0, vo = 0) we have,
¡
¢
v = −vt 1 − e−t/τ
(5.18)
c. Some Comments
• For spherical objects, the constants c1 and c2 are:
c1 = 1.55 × 10−4 D, C2 = 0.22D2
where D is the diameter of the sphere in meters.
• if we take the ratio of the two terms in 5.13:
0.22v|v|D2
= 1.4 × 103 |v|D
1.55 × 10−4 vD
• hence, for objects the size of a basketball (D ∼ 0.07m), the quadratic term will
dominate when,
1.4 × 103 |v|D ≥ 1, v ∼ 0.01m/s
(5.19)
• for speeds less than this, the linear term will dominate.
¨