2 CHAPTER 1. STRATEGIES FOR SOLVING PROBLEMS
1.13. Atwood’s machine
(a) This gives a
1
= 0. (Half of m
2
balances each of m
1
and m
3
.)
(b) Ignore the m
2
m
3
terms, which gives a
1
= −g. (Simply in freefall.)
(c) Ignore the terms involving m
1
, which gives a
1
= 3g. (m
2
and m
3
are in freefall.
And for every meter they go down, a total of three meters of string appears
ab ove them, so m
1
go es up three meters.)
(d) Ignore the m
1
m
3
terms, which gives a
1
= g. (m
2
go es down at g, and m
1
and
m
3
go up at g.)
(e) This gives a
1
= −g/3. (Not obvious.)
1.14. Cone frustum
The correct answer must reduce to the volume of a cylinder, πa
2
h, when a = b. Only
the 2nd, 3rd, and 5th options satisfy this. The correct answer must also reduce to
the volume of a cone, πb
2
h/3, when a = 0. Only the 1st, 3rd, and 4th options satisfy
this. The correct answer must therefore be the 3rd one, πh(a
2
+ ab + b
2
)/3.
1.15. Landing at the corner
The correct answer must go to infinity for θ → 90
◦
. Only the 2nd and 3rd options
satisfy this. The correct answer must also go to infinity for θ → 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
−αt
, we have
y(t) =
1
α
³
v
0
sin θ +
g
α
´³
1 − (1 −αt + α
2
t
2
/2 − ···)
´
−
gt
α
≈
³
v
0
sin θ +
g
α
´³
t − αt
2
/2
´
−
gt
α
= (v
0
sin θ)t −
1
2
gt
2
−
1
2
αt
2
v
0
sin θ . (4)
If α ¿ g/(v
0
sin θ ), then the third term is much smaller than the second, and we
obtain the desired result. So α ¿ g/(v
0
sin θ ) is what we mean by “small α.”
However, we also assumed αt ¿ 1 in the expansion for e
−αt
ab ove, so we should
check that this doesn’t necessitate a stricter upper bound on α. And indeed, the total
time of flight is less than 2v
0
sin θ /g (because this t makes the above y(t) negative),
so the condition α ¿ g/(v
0
sin θ ) implies αt ¿ (g/v
0
sin θ )(2 v
0
sin θ /g) = 2. So
αt ¿ 1 is guaranteed by α ¿ g/(v
0
sin θ ).
1.17. Pendulum
Here is a Maple program that does the job:
q:=3.14159/2: # initial θ value
q1:=0: # initial θ 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 θ > 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.
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