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# TEST BANK FOR Introduction to Classical Mechanics By David Morin

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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:

## [Solved] TEST BANK FOR Introduction to Classical Mechanics By David Morin

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