TEST BANK FOR Introduction to Classical Mechanics By David Morin

Press
Chapter 1
Strategies for solving
problems
1.8. Pendulum on the moon
The only p way to get units of time from `, g, and m is through the combination
`=g. Therefore,
TM
TE
=
p
`=gM p
`=gE
=
r
gE
gM
=) TM ¼
p
6 TE ¼ 7:3 s: (1)
1.9. Escape velocity
(a) Using M = ½V , we have
v =
r
2G ¢ (4=3)¼R3½
R
=
p
(8=3)¼GR2½: (2)
(b) We see that v / R
p
½. Therefore,
vJ
vE
=
RJ
p
½J
RE
p
½E
= 11 ¢
1
p
4
= 5:5: (3)
1.10. Downhill projectile
The angle ¯ is some function of the form, ¯ = f(µ; m; v0; g). In terms of units, we
can write 1 = f(1; kg;m=s;m=s2). We can't have any m dependence, because there
is nothing to cancel the kg. And we also can't have any v0 or g dependence, because
they would have to appear in the ratio v0=g to cancel the meters, but then seconds
would remain. Therefore, ¯ can depend on at most µ. (And it clearly does depend
on µ, because ¯ = 90± for µ = 0 or 90±, but ¯ 6= 90± for µ 6= 0 or 90±.)
1.11. Waves on a string
The speed v is some function of the form, v = f(M; L; T). In terms of units, we can
write m=s = f(kg;m; kgm=s2). We need to get rid of the kg's, so we must use the
ratio T=M. We then quickly see that
p
LT=M has the correct units of m=s. Note
that this can also be written as
p
T=½, where ½ is the mass density per unit length.
1.12. Vibrating water drop
The frequency º is some function of the form, º = f(R; ½; S). In terms of units, we
can write 1=s = f(m; kg=m3; kg=s2). We need to get rid of the kg's, so we must use
the ratio S=½. We then quickly see that
p
S=½R3 has the correct units of 1=s. Note
that this can also be written as
p
S=M, where M is the mass of the water drop.
1
2 CHAPTER 1. STRATEGIES FOR SOLVING PROBLEMS
1.13. Atwood's machine
(a) This gives a1 = 0. (Half of m2 balances each of m1 and m3.)
(b) Ignore the m2m3 terms, which gives a1 = ¡g. (Simply in freefall.)
(c) Ignore the terms involving m1, which gives a1 = 3g. (m2 and m3 are in freefall.
And for every meter they go down, a total of three meters of string appears
above them, so m1 goes up three meters.)
(d) Ignore the m1m3 terms, which gives a1 = g. (m2 goes down at g, and m1 and
m3 go up at g.)
(e) This gives a1 = ¡g=3. (Not obvious.)
1.14. Cone frustum
The correct answer must reduce to the volume of a cylinder, ¼a2h, when a = b. Only
the 2nd, 3rd, and 5th options satisfy this. The correct answer must also reduce to
the volume of a cone, ¼b2h=3, when a = 0. Only the 1st, 3rd, and 4th options satisfy
this. The correct answer must therefore be the 3rd one, ¼h(a2 + ab + b2)=3.
1.15. Landing at the corner
The correct answer must go to in¯nity for µ ! 90±. Only the 2nd and 3rd options
satisfy this. The correct answer must also go to in¯nity for µ ! 45±. Only the 1st
and 2nd options satisfy this. The correct answer must therefore be the 2nd one.
1.16. Projectile with drag
Using the Taylor series for e¡®t, we have
y(t) =
1
®
³
v0 sin µ +
g
®
´ ³
1 ¡ (1 ¡ ®t + ®2t2=2 ¡ ¢ ¢ ¢)
´
¡
gt
®
¼
³
v0 sin µ +
g
®
´ ³
t ¡ ®t2=2
´
¡
gt
®
= (v0 sin µ)t ¡
1
2
gt2 ¡
1
2
®t2v0 sin µ: (4)
If ® ¿ g=(v0 sin µ), then the third term is much smaller than the second, and we
obtain the desired result. So ® ¿ g=(v0 sin µ) is what we mean by \small ®."
However, we also assumed ®t ¿ 1 in the expansion for e¡®t above, so we should
check that this doesn't necessitate a stricter upper bound on ®. And indeed, the total
time of °ight is less than 2v0 sin µ=g (because this t makes the above y(t) negative),
so the condition ® ¿ g=(v0 sin µ) implies ®t ¿ (g=v0 sin µ)(2v0 sin µ=g) = 2. So
®t ¿ 1 is guaranteed by ® ¿ g=(v0 sin µ).
1.17. Pendulum
Here is a Maple program that does the job:
q:=3.14159/2: # initial µ value
q1:=0: # initial µ speed
e:=.0001: # a small time interval
i:=0: # i will count the number of time steps
while q>0 do # run the program while µ > 0
i:=i+1: # increase the counter by 1
q2:=-(9.8)*sin(q)/1: # the given equation
q:=q+e*q1: # how q changes, by definition of q1
q1:=q1+e*q2: # how q1 changes, by definition of q2
end do: # the Maple command to stop the do loop
i*e; # print the value of the time
This yields a time of t = 0:5923 s. If we instead use a time interval of .00001 s, we
obtain t = 0:59227 s. And a time interval of .000001 s gives t = 0:592263 s.
3
1.18. Distance with damping
In the xÄ = ¡Ax_ case, we have the following Maple program:
x:=0: # initial x value
x1:=2: # initial x speed
T:=1: # the total time
e:=.001: # a small time interval
for i to T/e do # run the program for a time T
x2:=-(1)*x1: # the given equation
x:=x+e*x1: # how x changes, by definition of x1
x1:=x1+e*x2: # how x1 changes, by definition of x2
end do: # the Maple command to stop the do loop
x; # print the value of the position
To run the program for di®erent times, we simply need to change the value of T in
the 3rd line. Letting T equal 1 gives a ¯nal position of 1.264. Letting T equal 10 and
100 gives ¯nal positions of 1.99991 and 1.9999996, respectively. These approach 2.
In the xÄ = ¡Ax_ 2 case, the only change in the entire program is in the 6th line,
where we now have the square of x1:
x2:=-(1)*x1^2: # the given equation
Letting T equal 1, 10, 100, 1000, and 10000, gives ¯nal positions of 1.099, 3.044,
5.302, 7.600, and 9.903, respectively. Looking at the successive di®erences between
these values, we see that they approach roughly 2.3. This constant di®erence for
inputs of powers of 10 implies a log dependence on the time.
Chapter 2
Statics
2.20. Block under an overhang
Let's break up the forces into components parallel and perpendicular to the over-
hang. Let positive Ff point up along the overhang. Balancing the forces parallel
and perpendicular to the overhang gives, respectively,
Ff = Mg sin ¯ +Mg cos ¯; and
N = Mg sin ¯ ¡Mg cos ¯: (5)
N must be positive, so we immediately see that ¯ must be at least 45± if there is
any chance that the setup is static.
The coe±cient ¹ tells us that jFf j · ¹N. Using Eq. (5), this inequality becomes
Mg(sin ¯ + cos ¯) · ¹Mg(sin ¯ ¡ cos ¯) =)
¹ + 1
¹ ¡ 1
· tan ¯: (6)
We see that we must have ¹ > 1 in order for there to exist any values of ¯ that
satisfy this inequality. If ¹ ! 1, then ¯ can be as small as 45±, but it can't be any
smaller.
2.21. Pulling a block
The Fy forces tell us that N + F sin µ ¡ mg = 0 =) N = mg ¡ F sin µ. And
assuming that the block slips, the Fx forces tell us that F cos µ > ¹N. Therefore,
F cos µ > ¹(mg ¡ F sin µ) =) F >
¹mg
cos µ + ¹ sin µ
: (7)
Taking the derivative to minimize this then gives tan µ = ¹. Plugging this µ back
into F gives F > ¹mg=
p
1 + ¹2. If ¹ = 0, we have µ = 0 and F > 0. If ¹ ! 1, we
have µ ¼ 90± and F > mg.
2.22. Holding a cone
Let F be the friction force at each ¯nger. Then the Fy forces on the cone tell us
that 2F cos µ ¡ 2N sin µ ¡ mg = 0. But F · ¹N. Therefore,
2¹N cos µ ¡ 2N sin µ ¡ mg > 0 =) N ¸
mg
2(¹ cos µ ¡ sin µ)
: (8)
This is the desired minimum normal force. When ¹ = tan µ, we have N = 1. So
¹ = tan µ is the minimum allowable value of ¹.
2.23. Keeping a book up
The result of Problem 2.4 is F ¸ mg=(sin µ + ¹ cos µ), assuming that sin µ + ¹ cos µ
is positive (that is, tan µ > ¡¹). If it is negative, there is no solution for F. To ¯nd
the maximum force, consider two cases: