Solutions to Problems in Leighton and Vogt's Exercises in Introductory Physics

Solutions to problems in Leighton and Vogt’s textbook "Exercises in Introductory Introductory Physics" (Addison-Wesley, 1969) Pier F. Nali (Revised April 10, 2016)

CHAPTER 1 Atoms in Motion

B-3

1

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

nG ∼

a)

n L nG

 N 0  Molar_vol._at_STP

∼

∼

liquid _ air _ density air _ density

air _ density

6.025 ×10 23 molecules / mo mole

∼

3

3

22.41×10 cm / mole 1.0 g cm

−3

0.001 g cm

−3

∼ 10

0.001 g cm −3

3

19

2

3

∼ 10 molec olecul ules es/c /cm m .

⇒ n L ∼ 1022 molecules/cm 3 .

m∼

c)

Said σ    the (approx.) diameter of an air molecule the inverse of the mean free path is

1 

2

∼ σ 

nG

⋅ nG , where σ  ∼

3

∼

19

2.7 ×10 molecules / cm

1 n L

3

∼ 10

−23

b)

g/molecule .

−7

∼ 10 cm , assuming for a rough estimate that the molecules are condensed

in liquid air in such a way that each molecule occupies a cubic cell of side σ  . Thus 2 1 19 3 5 −7 −1 ∼ (10 cm ) × 10 molecules/cm ∼ 10 cm   or, calculating the reciprocal, the mean free path 

 ∼ 10

−5

cm .

From P : P

d)

P 

( STP )

( STP )

( STP )

( STP )

∼ nG : nG

∼

( STP )

:,

∼ 1 atm, ∼ 10

−5

cm,

 ∼ 1 m,

one gets P ∼ 10−7 atm .

B-4

Said P the pressure of argon (A) gas spread in a layer l ayer of volume V  (at  (at constant temperature), we have P P

( STP )

=

Molar_vol._at_STP V

= 2.839 ×1010 cm 3 /mole

⇒ V  =

P

( STP )

× Molar_vol._at_STP P

=

760 mm Hg × 22.41×10 3cm 3 /mole 6.0 ×10 −4 mm Hg

=

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

3

and number _ density ( A _ atoms _ per _ cm

3

)=

 N 0 V 

=

6.025 ×10 23 atoms/mole 10

3

2.839 ×10 cm /mole

= 2.123 ×1013 atoms / cm 3 .

Said A a constant of proportionality representing the effective target area per A atom and expressing (on experimental basis) the percent intensity reduction of the atomic beam of K atoms as  A × number _ density × thickness = 3.0 %  one gets

 A =

3.0 % number _ density × thickness

=

3.0 ×10 −2 13

−1

3

2.123 ×10 10 at atoms/cm ×10 cm

≈ 1.4 ×10 −14 cm 2 /atom .

B-5

2 x interatomic spacing s

Cl

Cl….. Cl…

Cl

Cl Na

The cubic lattice has  N × N × N   points and

Na

Na…. Na ….

Na

( N − 1) × ( N − 1) × ( N − 1)   spacings between nearest 3

3 3 3 neighbours (Cl and Na ions in alternance). Then lattice _ vol. = ( N − 1) ⋅ s ≈ N s   (for large  N  ) and

° 3   N 3 1 1 1 = number _ density  atoms / A  ≈ 3 3 = 3 = cm −3 . 3 − 24 ° s   22.4258 ×10    N s 2.820A    

Since the NaCl molecule is diatomic 1 number _ densi ty × molecular _ weight _ of   _ NaCl density _ of   _ NaCl = 2 ,  N 0

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

4

where: density _ of   _ Na NaCl = 2.165 g/cm 3 , molecular _ weight _ of   _ NaCl = 58.454 g/mole,

and hence  N 0 =

1

2

number _ density × molecular _ weight _ of   _ NaCl density _ of   _ NaCl

= 6.02 × 10 23 molecules/mo /mole .

B-6

The number of helium atoms expelled per second by one gram of radium in equilibrium with its disintegration products is 13.6 ×1010 atoms g −1 s −1 = 117.504 ×1014 atoms g −1 day −1 . The corresponding volume of helium produced per day by one gram of radium is 3 −3 −1 0.0824 ×10 cm day = 4.291 ×10 −4 cm 3 g −1 day −1  and thus the number density of helium atoms at STP −3 192 ×10 g is a)

( STP )

n He

=

117.504 ×1014 atoms g −1 day −1 −4

3

4.291×10 cm g

−1

day

−1

≈ 2.7 ×1019 atoms/cm 3

and the calculated Avogadro’s number b) ( STP )

 N 0 = n He

− × Molar_vol._at_STP Molar_vol._at_STP ≈ 2.7 ×1019 atoms cm 3 × 22.41 ×10 3 cm 3 /mole ≈ 6.1 ×10 23 atoms/mole .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

5

C-1

The molecular weight of H(CH 2)18COOH is approximately approximately 298 g/mole and the molar volume volume is molecular _ we weig igh ht  density

≈

298 g mole −1 0.8 g cm

−3

≈ 372. 372.5 5 cm cm 3 /mol /molee . −

The volume of oil is deduced by dividing 0.81 mg by its density, i.e.

The surface area over which this volume of oil is spread is

1 4

0.81×10 3 g 0.8 g cm

−3

≈ 1.0 ×10−3 cm 3 .

 × 842 cm 2 ≈ 5.542 ×103 cm 2 , so the π  ×

thickness of the oil film (calculated as if its density were the same as in non spread state) is 0.81×10 −3 g 0.8 g cm −3 1.0 ×10 −3 cm 3 −6 ≈ ≈ × l≈ 0 . 2 1 0 cm . 2 2 3 2 1 × × π  8 4 cm c m 5 . 5 4 2 1 0 cm c m 4 If the oil molecules were arranged in linear chains 20 atoms high ( ≈ thickness l of the oil film) × 4 atoms wide ( ≈ 15 × l ) so that a molecule occupies a volume roughly estimated (assuming in liquid state a rotational l × 15 l × 15 l =

degree l

motion

( 0 .2 × 1 0 − cm ) 6

3

25

of

≈

25

around

the

main

axis

of

the

chain)

as

3

≈ 3.2 × 10 −22 cm 3 .

The reciprocal of the latter ( 0.3 × 10 22 molecules/cm3 ) is the number density

 N 0  Molar _ vol.

. So

22 3 22 3 3 24  N 0 ≈ 0.3 × 10 molecules/cm × Molar _ vol . ≈ 0.3× 10 mol./cm × 372.5 cm /mole ∼ 10 molecules/mole.

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

6

C-2

. Said m   the mass of a gas molecule the gas density is  ρ ( gas ) = m ⋅ N g . If one assume for sake of simplicity that in a solid the molecules are tightly packed in cells of side σ   then  ρ ( solid ) ≈ m σ 3

(1)

.

Thus

2

 ρ ( gas )   ρ ( gas ) 1 ρ ( gas ) ⇒ N gσ 2 ≈ 3 N g ⋅  ≈  N g ⋅ σ 3 ⇒ σ ≈ 3 ⋅  .  ρ ( solid ) ρ   N g ρ ( solid ) s o l i d )   ( 3

(1)

 For our purposes here we may ignore that in closed packaging configuration the proportion of space occupied by spherical

particles is ∆ =

nr _ of _ particles _ per _ cell × V( particle   ) V ( unitcell ) V ( unitcell )

π 

≈ 74,048%   and therefore the number of particles per

3 2

3

σ  unit cell is ∆ ⋅ = ⋅ = 3 σ  V ( particle) 3 2 π  6 3

π

=

2 particles/cell particles/cell rather than 1 particle/cell,  ρ ( solid ) is

m σ  and so on, so the subsequent results should contain

 N g

3

2 ⋅ m σ  rather than

in locus of  N g . This is the densest packing of equal spheres 2 in three dimensions, as proved by b y Gauss in 1831. See Gauss, C. F. "Besprechung " Besprechung des Buchs von L. A. Seeber: Intersuchungen über die Eigenschaften der positiven ternären quadratischen Formen usw." Göttingsche Gelehrte Anzeigen (1831, July 9) 2, 188-196, 1876 = Werke, II (1876), 188-196. For a proof of this result see, e.g., J. W. S. Cassels, —An Introduction to the Geometry of Numbers, Springer-Verlag, 1971 [Chap. II, Th. III]

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

Comparing  = 1

 N gσ 2 = 1

(

(

)

2π N gσ 2  and η =

)

2π  = ρ ( gas ) ⋅ v

(

1

3

7

ρ ( gas ) ⋅ v ⋅   one also gets

)

2π ⋅ 3η .  

Comparing again the right-hand sides of latter two expressions for  N gσ 2   and cubing the second members one gets

2

  ρ ( gas )  3    ρ ≈ ⋅ v3  N g ⋅  g a s ( )     ρ ( solid ) 

(54

2π 3 ⋅η 3

and finally – assuming on experimental basis

(2)

)  ρ ( solid ) ≈ ρ ( liquid )  -

2

 N g ≈

(2)

 ρ ( gas )  ρ ( liquid )  ⋅ v3 3

3

54 2 ⋅ π ⋅η 

19

3

∼ 0.7 ×10 molecules/cm .

 This assumption is valid where we are dealing only whit orders of magnitude and is based on the observed condensation behaviour of many substances. Unfortunately, at the times of Loschimdt’s work (1865) was not yet known how to liquefy the air, so he could not directly observe condensation behaviour for air. However, the density of a substance in the condensed state can be estimated by its chemical composition, as Loschmidt did. See Porterfield, William W., and Walter Kruse. "Loschmidt and the Discovery of the Small.",  J. Chem. Educ. , 1995, 72 (10), p 870

Solutions to problems in Leighton and Vogt’s textbook "Exercises in Introductory Introductory Physics" (Addison-Wesley, 1969) Pier F. Nali (Revised April 10, 2016)

CHAPTER 2 Conservation of Energy, Statics

1

According to the principle of virtual work (PVW) “the sum of the heights times the weights does not change”, said h1   and h2   the heights of the weights W 1   and W 2   respectively, then

∆ (W1h1 + W2 h2 ) = W1∆h1 + W2 ∆h2 = 0 . If we assume that the weight W 1   goes down

∆h1 = −

1 2

∆h2 ⇒ W11 = W2 2 .  

1

∆h1   as the weight W 2   rises

∆h2 , then

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

2

2

The “s “sum of of th the he heigh ights tim times th the we weigh ights” ts” fo for an an ar arbitra trary number of of we weigh ights be becomes

∑ W ∆h = 0 i

i

i

and considering the distances positive or negative depending on the side of the fulcrum, so that ∆h j  W i  i = 0 . ∆hi = i ∆h j  (or ∆hi ∝  i  for any i, being  independent of j), it becomes thus   j

∑

  j

i

3

 n  The PVW can be expressed in a more general form as ∑  ∑ Fij i∆ri = 0 , where the sum over  j is i  j   N

extended to all the n forces acting on the mass i and the sum over i is extended to all the  N masses.

 n   = 1 , so the above formula reduces to  ∑ Fi i∆r = 0 . In the present case (a single body)  N  =  i  a) For n = 1  is

F i∆r

= 0 ; so, for the arbitrariness of the choice of the vector ∆r ,

F

=0.

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

b) For n = 2 is

( F1 + F2 )i∆r = 0   and, again for the arbitrariness of the choice of the vector ∆r , is

+ F2 = 0 , whence collinear. c) For n = 3  is, again, F1

F2 (apart

n = 2 ),

3

F1

= −F2 ; so, the two forces are equal in magnitude, opposite in direction and

F1

+ F2 + F3 = 0 . If we consider the plane on which lie the two vectors

F1  and

from the trivial case in which the two vectors are parallel, which degenerates to the case of

F3  also

vectors. Let two vectors

lies in the same pane, being expressible as a linear combination of the other two

X 0  the

vector joining the origin to the point of intersection of the lines of action of the

F1  and F2  and X  the

vector joining the origin to an arbitrary point of one or the other

line of action. Then the two lines of action are defined by the system of equations

 ( X − X 0 ) ∧ F1 = 0 .  ( X − X0 ) ∧ F2 = 0 By adding the latter two equation, immediately follows that X 0  defines

X 0  also

lies on the line of action of

F3 .

a single point (through which the three lines pass) as each line of action cannot intersect

each other in more than one point (provided the lines are not p arallel). d) From From

Fi i ∆r



n

∑ 

= F i ∆ r⋅ cos ∆ i   it follows that 

i =1



Fi  i∆r



n

arbitrariness of the choice of ∆r  -

∑ F  cos ∆ i

i =1

i

=0.





n

F cos ∆  ⋅ ∆ r = 0 , and – again for the ∑  

=

i

i =1

i

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

4

A-1

a’ b’

L

a b W

c’

c

45°

y

∆θ 

x

W

Suppose the weight W goes down  x as the centre of mass of the plank rises  y. Therefore (refer to the above figure)  y =  L2 sin ∆θ  ,  x = − ( c − c′ )  and a′ = a − L sin ∆θ , b′ = b + L (1 − cos ∆θ ) , c′ =

a ′2 + b′2 .

So – assuming ∆θ  small - c′ = c 1 − 2 acL2 sin ∆θ + 4

 L ( L +b ) c

2

sin 2

∆θ  2

≅ c (1 − acL sin ∆θ ) and  x ≅ − aLc sin ∆θ  , 2

where we have used the trigonometric double-angle formula 1 − cos α  = 2 sin 2 1 + x ≅ 1 +

 x

2

 for x small.

α  2

  and the approx.

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

Being

also

c = 2 ⋅ a   it

follows

W ⋅ ( − x ) = W( plank ) ⋅ y , whence W = −

 y

 x ≅ −

W ( plank ) =

 L 2

sin ∆θ  = − 2 ⋅ y .

W ( plank )

3

=

 x 2 2 (note: the length of the plank doesn't affect this derivation).

From

PVW

5 then

we

have

kg-wts .

A-2

Let W  the weight of the ball, F P  and F W   the forces exerted on the ball by the plane and by the wall n

respectively and made use of PVW in the form

∑ F  cos ∆ i

i

=0.

i =1

Thus, with respect to an horizontal line joining the point in which the ball touches the wall with the centre of mass of the ball, Fw cos ( 0 ) + F P cos ( π 2 + α ) = 0 , that is Fw − F P sin α  = 0 ; and, with respect to a vertical line lying in the plane of the sheet and passing through the centre of mass of the ball, FW cos π 2 + W cos π + F P cos α  = 0 , that is −W + F P cos α  = 0 . So F P =

W 

, cos α  FW  = W  tan α ,

and the forces on the planes are 1 F P =  kg-wt, cos α  FW  = tan α  kg-wt.

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

6

A-3

n

Again make use of PVW, this time in i n the form

∑W h = 0 . i i

i =1

-

Without the loads: (WC + WAA′CD ) ⋅ AP = WBB′GH ⋅ P   B , that is WC + WAA′CD = 2W  BB ′GH . 

-

With the loads, assume the weight W 1  goes down x as the weight W 2  rises y, being  x = −

Thus

that is (WC + WAA′CD + W1 ) ⋅ ( − x ) = (WBB ′GH + W   2)⋅ y,

WC + WAA′CD + W1 = 2 (WBB ′GH

subtracting to this equation the one without the loads above one gets

1

y. 2   + W2 ) , and

W1 = 2W 2 ,

W2 = 12 W 1 = 0.25 kg-wts .

A-4  L l1  x A

 y A

ϕ 

l2

−∆ y A

so

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

7

The weight A can move both vertically and horizontally. Vertical displacement of A (assumed downwards): W A ⋅ ( − ∆y A ) = W1 kg ⋅ ∆y1 kg + WB ⋅ ∆ yB . -

∆ y B   as A moves to the left):

Horizontal displacement of A (assuming B rises

(

)

W1 kg ⋅ − ∆y1 kg = W B ⋅ ∆yB

Case 1 – vertical displacement For the flattened triangle  (l1, ∆ y A , l1 + ∆l1 ), using Carnot’s theorem and assuming ∆ y A small, l1 + ∆l1 = l12 + 2l1∆ y A cos ( π 2 + ϕ ) + ∆ y A2 =

Neglecting

the

l1 + ∆l1 ≅ l1 1 − 2

 = sin ϕ  =

 y A l1 + ∆l1

≅

term

of

∆ y A

yA l1

∆y A

1

l1

∆ y A2   and

order

( ) sin ϕ ≅ l (1 −

. So ∆l1 ≅ −∆y A

l1

l12 + 2l1∆ y A (− sin ϕ ) + ∆ y A2 = l1 1− 2

)

sin ϕ   

 y A

using

the

∆ y A

∆y 2A

l1

l12

∆ y A l1

.

1 + x ≅ 1 + 12 x ,

approx.

∆l1 ≅ l1 (1 −

and

( ) sin ϕ +

)

sin ϕ − l1 ≅ − ∆ y A sin ϕ ,

then where

.

l1

∆l2 ≅ −∆y A

Similarly, for the triangle (not shown)  (l2, ∆ y A , l2 + ∆l2 ),

 y A l2

. But ∆l1 = ∆y1 kg and

∆l2 = ∆y , whence  B

W A ⋅ ( −∆y A ) ≅ W1 kg ⋅ ( −∆y A )

 y A l1

+ WB ⋅ ( − ∆y A )

= W1 kg ⋅ sin 30° + W B ⋅ sin 45° = 12 W1 kg +

1 2

yA l2

→ WA ≅ W1 kg ⋅

WB = 12 +

1 2

yA

+ WB ⋅

l1

WB .

yA l2

= 

 

Case 2 – horizontal displacement In the same way, for the triangle (not even shown as well)  ( l1 − ∆l1 l1 − ∆l1 = l12 + 2l1∆ x A cos 30° + ∆ xA2 = l1 1 + 2 l1 − ∆l1 ≅ l1 1 + 2

( ) cos 30° ≅ l (1 + ∆ x A

Substituting cos30° =   ( l2

1

l1

 x A l1

∆x A l1

( ) l1

cos 30° +

)

∆x2A l12

, and, with the above approximations,

(

cos 30° → ∆ l1 ≅ l1 − l1 1 +

∆x A l1

 L − x A l2

.

)

cos 30° ≅ − ∆ x A cos 30° .

 in the above expressions one gets ∆l1 ≅ − ∆x A

+ ∆l2  , ∆ x A , l2 ): ∆l2 ≅ ∆x A cos cos 45° = ∆x A

So from PVW

∆ x A

∆ x A , l1 ) with ∆ x A small,

 ,

 x A l1

, and similarly, for triangle

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

(

)

W1 kg ⋅ − ∆y1 kg = W B ⋅ ∆ yB → W1 kg ⋅ ( − ∆ l1 ) = WB ⋅ ∆ l2 → W1 kg ⋅ ∆ x A

→ W1 kg ⋅

 x A l1 3 2

→ W B =

= W B ⋅

L − xA l2

W  B =

(

1 2

+

3 2

3 2

→ W1 kg ⋅ cos 30° = WB ⋅ cos 45° → W1 kg ⋅

l1 3 2

= WB ⋅ ∆ xA = WB ⋅

2 2

L − xA l2  

⋅ W 1 kg .

Whence finally W A ≅ 12 + W  A =

 x A

8

1 2

WB = 12 +

3 2

W 1 kg .  In summary:

) kg-wts,

 kg-wts.

A-5

ϕ 

n

From the PVW (in the form

∑ F cos ∆ i

i

= 0 ):

i =1

-

Case 1 – vertical fixed line:

T1 cos ϕ + T2 cos ϕ + W cos π = 0 → (T1 + T2 ) cos ϕ = W → ( T1 + T2 ) 1 − sin 2 ϕ = W → 

(T1 + T2 ) 1 −

( ) AB

2

=W →

2 AC

1 2

(T1 + T2 ) = W .

 

-

Case 2 – horizontal fixed line: −T1 cos ( π2 − ϕ ) + T2 cos ( π2 − ϕ ) + W cos π 2 = 0 → T1 = T2 .

Thus T1 = T 2 =

W 

2

=

50 2

 lb-wts

 

 

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

9

A-7

V

ϕ   x

Let V the reaction force of the edge in contact with the wheel and assume null the reaction force of the n

wall at the ground soon as the force F is applied to the axle. Then from PVW

∑ F cos ∆ i

i

= 0:

i =1

 R − h

-

Case 1 – vertical fixed line: V cos ϕ + W cos π = 0 → V cos ϕ  = W → V

-

Case 2 – horizontal fixed line: F cos ( 0 ) + V cos ( π2 + ϕ ) + W cos π 2 = 0 → F = V sin ϕ  = V

where  x = Thus F = W

2

R 2 − ( R − h ) = h ( 2R − h ) .  x  R − h

= W 

h ( 2 R − h) ) R−h

.

 R

=W .

   x  R

,

 

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

10

A-9

F

 x

φ

F

F

Fx

n

Using

∑ F  cos ∆ i

i

= 0  we have:

i =1

horizontal line  AB : FO cos (π + 45° ) + FP cos π 2 + FQ cos 0 ° + Fx cos ( π + ϕ ) = 0 ⇒ − F cos 45 ° + F − Fx cos ϕ = 0 ,  

 

vertical line OP : FO cos ( π2 + 45° ) + FP cos 0° + FQ cos π2 + Fx cos ( π 2 + ϕ ) = 0 ⇒ − F sin 45 ° + F − Fx sin ϕ  = 0 ;   and confronting:  = 45° or ϕ  =  = 225° . cos ϕ = sin ϕ , whence ϕ  =

 = 45°  therefore it is parallel to, and has the same direction of, F O . Its magnitude F  x is positive when ϕ  =

(

is F x = F 

)

2 − 1 = 20.7 N , 45° down from  AB .

When the plate is in equilibrium the torque, t orque, with respect to the corner O, is F x ⋅ x ⋅ sin (π − ϕ ) + OP ⋅ FQ + 0 ⋅ FP + 0 ⋅ FO = 0 ⇒ F  x ⋅ x ⋅ sin ϕ + F ⋅ OP = 0 ⇒  x = −

F ⋅ OP F  x sin 45 45°

=−

i.e. .34 m left of O.

50 N × 0,100 m 20.7 N × 0.7071

= − 0,34 m,

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

A-10

Assume W1 > W 2 . Then as W 1 goes down − D sin θ  the weight W 2 rises + D sin θ . From the conservation of energy Thus v = 2 gD

W1 − W 2 W1 + W 2

1 2

W1 + W 2  2  g  ⋅ v + W1 ⋅ ( − D sin θ ) + W2 ⋅ D sin θ  = 0 .  

sin θ  .

A-11

See the previous problem. In the present case, 2W  2 1 ⋅ ⋅ v + W ⋅ D sin θ + W ⋅ ( − D sin φ ) = 0 ⇒ v = 2 g

gD ( sin φ − sin ϑ ) .

 

11

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

12

B-1

 y

 = 0 At t  =

( K .E.)1 = ( K .E.) 2 = 0, ( P.E.)1 = ( P.E.) 2 = Mg H 2  , being  M1 = M 2 = M 

 > 0 At t  >

( K .E.)1 = ( K .E.) 2 = 12 Mv 2 , ( P.E.)1 = Mgy, ( P.E.) 2 = Mgx

 x

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics  H 

(

13

)

1+ 2 2 Length of the cord on the side of M 2:  H − x  H  Length of the cord on the side of M 1: 2 − 1 + x 2 Height of M2:  x

Length of the cord:

(

Height of M1:  y =

 > 0 Then at t  >

 H 

1 

2 

2

1 −

−

)

x

2

 H  x  1  − − 1 t he conservation of energy   and from the   2 2 2    

( P.E.)1 = Mgy = Mg 

( K .E.)1,t =0 + ( K .E.) 2,t =0 + ( P.E.)1,t =0 + ( P.E.) 2,t =0 = ( K .E.)1,t >0 + ( K .E.) 2,t >0 + ( P.E.) 1,t >0 + ( P.E.) 2,t > 0 ⇒ H   x  + Mgx ⇒ 1 − 1  −  2 2 2 2 2 2 2   + Mg  H   1 − 1  . 2  MgH = Mv + Mgx  1 − 1    2 2 2  0 + 0 + Mg

 H

+ Mg

H

=

1

Mv 2 +

1

Mv 2 + Mg 

Taking the derivative with respect to t  of  of each member of the above expression dx dv 2 dv   1 = 2v = 2va . 0 = 2 Mva + Mgv 1 −  + 0 , where we used v =  and 2 dt  dt dt    a)

1 1   > 0  is a = − 1 − So the vertical acceleration of M 2 at t  > g . 2 2

b)

Being a < 0  the mass M 2 will move downward. The eight of M2 is  x =

 = 0  whereas  x = v ( 0 ) = 0  at t  = time t 1 = − H  = a

c)

2 H 

1 2

at 2 +

 H 

2

.

 H 

2

and its velocity

 > 0 ; the  at t  > then it will ill stri strik ke the the grou round ( x = 0 ) at

1   g 1 −  2   H  x  H   1  1  − Being  y = 1− the height of M 1, then for  x = 0  is y = 1−     < H ; so the 2  2 2 2 2  other mass (M1) will no strike the pulley.

 

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

14

B-2

 y

When the boom rotates an angle ∆θ    (assumed small, i.e. neglecting terms of the order if

2

( ∆θ ) or

higher) around the pivot the length  y of the horizontal cable will change of ∆ y  x cos θ∆θ . This can be seen using Carnot’s theorem and the approx.

1 + x ≅ 1 + 12 x  to find the length of the side opposite to

the angle θ :

∆ y = y′ − y = x 2 + x 2 cos 2 θ − 2 x 2 cos θ co c os (θ + ∆θ ) − x sin θ ≅    x sin θ

(

)

1 + 2 ctg θ∆θ − 1  x sin θ ⋅ ctg θ ∆θ = x cos θ ∆θ .  

Or, in a more straightforward  y′ ≅ x sin (θ + ∆θ ) = x sin θ cos ∆θ + x cos θ si s in ∆θ ≅ x sin θ + x cos θ ∆θ ⇒ 

way,

∆ y = y′ − y ≅ x cos θ .   the weight w of The height of the weight W changes of ∆ Z = L cos (θ + ∆θ ) − L cos θ ≅ − L sin θ ∆θ  and the boom (assumed applied to the centre of mass of the boom) moves of ∆ z =  L2 cos (θ + ∆θ ) − L2 cos θ ≅ − L2 sin θ ∆θ .   Then from PVW: T ∆y + W ∆Z + w∆z = 0 ⇒ T ⋅ x cos θ∆θ − W ⋅ L sin θ∆θ − w ⋅  L2 sin θ ∆θ ≅ 0 ⇒ T=

 L 

w W  +  tan θ .  2  x 

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

15

B-3

 L

r

s

Let L = 10 ft the length of the ladder, r  =  = 8 ft the initial height of the rollers rollers at the top end and and suppose that, slightly tilting the ladder leaning against the vertical wall, the rollers would go down ∆r . Then (assuming ∆r  small) 2

2

2 2 2 2 2 2 2  L = ( r + ∆r ) + ( s + ∆s ) ⇒ L ≅ r + 2r ∆r + s + 2s ∆s = (r + s ) + 2 ( r ∆r + s ∆s ) = L + 2 (r ∆r + s ∆s ) ⇒

⇒  r∆r + s∆s = 0. r  Thus ∆s = − ∆r . s Furthermore, said hW ,w  the heights of the weights W and w, hangings from rungs at  LW , w  feet from the

bottom end, then hW ,w =

 LW ,w  L

r  and ∆hW ,w =

 LW , w  L

∆r .

W is hung l = 2.5 feet from the top end and therefore LW  = L − l  whereas  Lw =

 L

2

.

Let R, H and V, respectively, the force of the roller on the wall, the horizontal and vertical forces of the ladder on the ground, then, from PVW,  L L  r    H ∆s + W ∆hW + w∆hw = 0 ⇒ H  − ∆r  + W W ∆r + w w ∆r = 0 ⇒ L L  s 

− H

r s

+W

 LW L

+w

Lw L

=0⇒ H =

s Lr

(WLW + wLw ) =

s  L − + W L l w . ( ) Lr    2 

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

16

n

Now, from PVW in the form

∑ F cos ∆ i

i

= 0  one gets  R = H  and V = W + w , whence

i =1

a)

 R = 45 lb-wts

b)

 H = 45 lb-wts, V  = 90 lb-wts

B-4

α  θ  α − θ 

Let  L = 3R  the length of the plank. Then we have sin α =

 L

2 R

=

3 2

⇒ α  = 60° .

( 2)

For the heights of the weights W and w one gets that hW  = R − R 2 −  L

2

cosθ ,

hw = R − R cos (α − θ )  and (for a small tilt ∆θ  )

∆hW  ≅

( 2)

R −  L 2

 = being cos α  =

1 2

2

(

sin θ∆θ = R 1 − L

2 R )

2

sin θ∆θ = R cos α sin θ∆θ =

 R

2

sin θ∆θ ,  

, ∆hw ≅ R sin (α − θ ) ∆θ  .

From PVW, remembering that ∆hW  and ∆hw  lie in opposite directions, we have W ∆hW = w∆hw , whence W

 R

2

sin θ∆θ = wR sin (α − θ ) ∆θ =

W 

2

R sin (α − θ ) ∆θ ⇒ sinθ = sin (α −θ ) ⇒ θ = α −θ ⇒ θ = 

α  2

= 30°.

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

17

B-5

 y  x

From PVW we have W ∆x sin 30° + W ∆y sin 60° = 0 . From the rectangular triangle shown (see problem B-3)  x∆x + y∆y = 0 ⇒ ∆y = −

 x  y

∆x = − tan α ∆x .

Then W ∆x sin 30° + W ∆y sin 60° = 0 ⇒ ∆x sin 30° + ∆y sin 60° = 0 ⇒ ∆x sin 30 ° − tan α ∆x sin 60 ° = 0 ⇒ sin 30° − tan α sin 60° = 0 ⇒ sin 30° − tan α cos 30 ° = 0 ⇒ tan α  =

sin30° cos30°

= tan 30 °,

 = 30° . and therefore α  =

B-6

h

α

 L  R r

Being  L  the length of the string and r   the radius of the sphere, then, from Carnot’s theorem,

( L + r ) + ∆L =

2

2 R + ( R − r ) − 2 R ( R − r ) cos ( α + ∆α ) .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

If ∆α  is small ( L + r ) + ∆L ≅

2

18

2

R + ( R − r ) − 2R ( R − r ) cos α + 2 R ( R − r ) sin α ∆α  .

2

2

2 But  R + ( R − r ) − 2 R ( R − r ) cos α  = ( L + r ) , thus

2

( L + r ) + ∆L ≅ ( L + r ) + 2R ( R − r ) sin α∆α ≅ ( L + r ) 1 + ( L + r ) +

 R ( R − r )

( L + r )

sin α∆α ⇒ ∆L 

R ( R − r )

( L + r )

sin α∆α .

2 R ( R − r )

( L + r )

2

sin α ∆α 

 

 

h  Furthermore, being the height of ∆h = ( R − r ) sin (α + ∆α ) − ( R − r ) sin α ≅ ( R − r ) cos α ∆α .

the

centre

of

the

sphere,

 

Let F  the strength of the string, from PVW: F ∆L = W ∆h ⇒ F = W

L + r  ∆h ≅ W   cot α . One also gets ∆ L R

 R − r  = 49 − 4.5 4.5 cm=44.5c 4.5cm m  L + r  = 40 + 4.5 cm cm=44.5cm  L + r  So the triangle is isosceles so that cot α  = 1 ⇒ F ≅ W  . Let h  the height of the isosceles, then 2 2h 2 44.5 22.25  L + r  2 =W = W  0.6W . = 44.52 − 24.52 = 1380  and F ≅ W h = ( L + r ) −  R 2 2h 2 × 1380 1380

( )

B-7

 

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

19

θ 

30°

θ 

The centres of the spheres lie on the vertices of a regular tetrahedron of edge a and height

 = cosθ  =

6a 3

 , then

6

. 3 Let wa   the component of the weight w  along each edge a. From symmetry considerations 3 × wa cos θ  = w , whence wa =

w

3cos θ 

w

=

6

= 2 ton-wts .

 = wa ⋅ Along the horizontal surface of the tetrahedron wa   has a component wa sin θ  =

3

  and the two 3 welds at the contact points of each sphere with the two other exert along the same direction a force 2T cos 30° = T  3 , being T  the  the tension of each weld. n

Then from PVW (in the form

∑ F  cos ∆ i

i

= 0 ) and fixing the horizontal direction, one gets

i =1

T 3 = wa ⋅

3 3

⇒ T  =

wa

3

, and thus, for a factor of safety of 3, T( 3) = 3 × T = wa = 2 ton-wts .

B-8

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

Let l the length length of the cord, cord, h1 ,

20

h2 the heights of the two masses when they lie at distances c1 ,

c2 from

the top in downwards direction along the two legs of the right triangle. Then from PVW T ∆l = m1∆h1 + m2 ∆h2  where:

∆h1 = ∆c1 sin30° = 12 ∆c1 ∆h2 = ∆c2 sin sin 60° =

3 2

∆c2

Being c1 = l cos α  c2 = l sin α 

one gets ∆c1 = −l sin α∆α + cos α ∆l ⇒ ∆h1 = − 2l sin α ∆α + 12 cos α ∆l , 

∆c2 = l cos α∆α + sin α∆l ⇒ ∆h2 =

3 2

l cos α ∆α +

3 2

sin α ∆ l  

Thus T ∆l = m1∆h1 + m2 ∆h2 =

(

1 2

m1 cos α +

3 2

)

m2 sin α ∆l −

(

1 2

m1 sin α −

3 2

)

m2 cos α l ∆α .

 

∆l , ∆α  being independent of one another, tacking ∆l = 0   one finds that the system is in static equilibrium when m1 sin α − 3m2 cos α = 0 ⇒ tan α  = 3 3 . Reading Reading

the

T = 12 m1 cos 79°.1 +

tables tables 3 2

one

gets

m2 sin 79°.1 = 265 g-wts .

α  =  = 79 °.1

and,

taking

∆α  = 0 ,

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

B-9

 L

r’

 L

∆ y

r

∆s ∆ x

Being ∆ L = 0 we have  x∆x + y ∆y = 0 ⇒ ∆y = −

 x  y

∆x = − cot 30 °∆x .

But ∆ x = r ′ cos 30° + ∆s − r cos 30 ° = ∆s − ( r − r ′ ) cos 30 °  . Assuming ∆s small we have r − r ′ ≅ ∆s cos30 cos 30 ° . Thus ∆ x ≅ ∆s (1 − cos 2 30 °) = ∆s sin 2 30 ° .

a)

from PVW

 cot 30° ⋅ sin 2 30°∆s  ∆ y  − cot 30°∆x  T ∆s + W ∆y = 0 ⇒ T = −W   = −W  ⇒  =W  ∆s ∆s ∆s     T = W cos 30°si s in 30° =

3 4

W .

21

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

b) normal

 N

T

horizontal  N

Using the method of the force components (FC) on cart we have: T cos 180° + N cos 300 ° = 0 ⇒ T =

3 2

N  (horizontal).  

Using FC on W:  N cos 0° + W cos 240° = 0 ⇒ N = So T =

3 4

1 2

W  .

W  .

B-10

Assume the centre of the bobbin goes down ∆ H and m rises ∆h  as the bobbin rolls downwards. Then from PVW:  M ∆H = m∆h .

22

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

23

For a rolling of the bobbin of ∆ L = R∆α  then M goes down ∆ H = ∆L sin θ  , whereas m rises (assuming

∆α small) ∆l ≅ r ∆α  =

r   R

∆L  and goes down ∆ H .

Thus

r   r   − sin θ  ⇒ M ∆L sin θ = m∆L  − sin θ  ⇒  R  R 

∆h = ∆l − ∆H = ∆L 

r r m r    M sin θ = m − m sin θ ⇒ ( M + m ) sin θ = m ⇒ sin θ = = 0, 5 ⇒  R R M +m R θ  = 30°.

 

B-11

∆h ∆ x

If the chain rises ∆h  resting in a horizontal circle, then its it s radius varies of ∆ x = of ∆l = 2π ∆x .

r Wh   ∆h = T ⋅ 2π∆x = T ⋅ 2π  ∆h ⇒ T =   From PVW: W ∆h = T ∆l ⇒ W . h 2π r 

r  h

∆h  and its length

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

24

B-12

lW 2

lW 3

lW 1

lw2 l w1

Assuming w rises ∆hw  whereas W goes down ∆hW  , being:

∆ ( lW + lW + lW  ) = 0 , ∆lW = ∆lW ⇒ ∆lW = − 12 ∆lW , ∆ ( lw  + lw 1

2

3

1

2

1

3

1

2

) = 0 , ∆l

W3

= − ∆lw = ∆lw , 1

2

then ∆hw = ∆lw2 + ∆lW3 = 2 ∆lw2 , ∆hW = ∆lW1 sin θ = − 12 ∆lW3 sinθ = − 12 ∆lw2 sin θ = − 14 ∆hw sin θ .

From PVW: w∆hw + W ∆hW  = 0 . So w∆hw − 14 sin θ ⋅W ∆hw = 0 ⇒ W =

 

4w   . sin θ 

B-13

FG  DF

 EF

FG

 x  EG

 y

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

25

From PVW: F1 x = F2 y , but  x : 2 = y : 1 ⇒ F2 = 2F 1  .

Using

∑ F cos ∆ i

i

= 0  then, with respect to the vertical direction, one gets

i

W  2W  F1 − W + F2 = 0 ⇒ F1 + F2 = W ⇒ 3F1 = W ⇒ F1 =   , F 2 =  . 3 3

Using FC at the joint G:

.

vertical components: F2 cos 0° + TFG cos 150 ° = 0 ⇒ F2 − TFG

3 2

= 0 ⇒ TFG =

2 F 2 3

. 

Using FC at the joint F: -

horizontal components: TFG cos120° + TEF cos 240 ° + F cos 0 ° = 0 ⇒ F =

-

vertical components: TFG cos 30° + TEF cos 150 ° = 0 ⇒

whence F = T FG =

2 F2 3

=

4W  3 3

.

3 2

1 2

(TFG + TEF )

(TFG − TEF ) 00 ⇒ TEF = TFG

 

 

 

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

26

B-14

a) members AB, BD, DF, FG should be rigid in order that A remains in fixed position; CD also rigid to fix C and D. The elements that could be flexible are therefore AC, CE, EG, BC, EF, ED. ˆ  = 3  , sin A= ˆ 4 . From b) meanwhile note that cosA From the resul resultt of the prec precedi eding ng probl problem em is then then 5 5 F 1 =

W 

3

 , F 2 =

2W  3

 .

Using FC at the joint A: -

horizontal:

ˆ AC ABcos ABcos A=AC A=

-

vertical:

ˆ =F  = W ⇒ AB = 5W  AB sin A 1 3 12

Using FC at the joint B: -

horizontal:

ˆ BD osA= ( AB+BC ) cosA=B

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

-

vertical:

27

ˆ = W  AB=BC ⇒ BD=2ABcosA 2

Using FC at the joint C: -

vertical:

BC = CD

Using FC at the joint D:

-

vertical:

So BD=

W 

2

DE = CD ⇒ DE=CD=BC=AB=

 , DE=

5W  12

5W  12

 .

B-15

At its maximum swing the pendulum has zero velocity. Then from the conservation of energy ∆h W ∆hW + w∆hw = 0 ⇒ W = −w w . At the maximum swing (pendulum at the ceiling) ∆hw = 3 , ∆hW 

∆hW  = −4 . Thus W = 34 w .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

28

B-16

As the third mass reaches the table top, the two equal masses rise 50

(

)

2 − 1 cm ; when the third mass

2   ∆ h   has descended a distance ∆h  the two equal masses have risen by 50 1 +  −1 .     50   

Then, said V   and v  the final velocities (in module) of the third mass and of the two equal masses respectively, v =

V 

2

.

The kinetic energy varies of 2 ( mg ) ⋅ 50

(

1 2

  the potential energy of ( 2m ) V 2 + 2 ⋅ 12 mv 2 = m (V 2 + v 2 ) = 23 mV 2   and

)

(

)

2 − 1 − ( 2mg ) ⋅ 50 = −2mg ⋅ 50 2 − 2 .

From the conservation of energy:

3 2

(

)

mV 2 = 2 mg mg ⋅ 50 2 − 2 ⇒ V =

With g = 981 cm s −2  thus one obtains V  = 196 cm s −1 .

(

)

200 g 2 − 2   . 3

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

29

B-17

Let’s consider a small layer of liquid of thickness ∆ z  a distance h up from the hole. Its potential energy of position is  ρ gA∆z ⋅ h .

The internal tension force due to the weight of the overlying liquid,  ρ gA ( H − h ) , corresponds to a pressure  p = ρ g ( H − h )  and an internal energy  pA∆z = ρ gA ( H − h ) ∆z .

The total potential energy is the sum of the position energy and of the internal energy:

 ρ gA∆z ⋅ h + ρ gA ( H − h ) ∆z = ρ gAH ∆z (independent of

h,

and

is

converted

hence

also

valid

for

h = 0

corresponding to the hole position).

Soon

as

the

hole

is

open

the

 ρ gAH ∆z = 12 ( ρ A∆z ) v 2 ⇒ v = 2 gH .

above  

energy

to

kinetic

energy,

so

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

30

C-1

W W

T

X W

α  T

 β   β + 60° + θ = 90° ⇒ β = 30° − θ  60°

α = 60° − β = 30° + θ 

θ 

From FC: T cos α + X cos β  = W 

-

vert.

-

horiz. T sin α = X sin β 

-

 = T cos 60 obliq. W sin θ  = 60° =

T 

2

(Note that in the shown arrangement T( limit) = 2W sin θ  ).

In terms of cot β  :  X cos β = X sin β cot β = T sin α cot β ⇒ T cos α + T sin α cot β  = W  .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

31

But

cos α = cos (30 30 ° + θ ) = cos 30 °co cos θ − sin 30 °si s in θ =

3 2

cos θ − 12 sin θ

sin α = sin ( 30° + θ ) = sin 30° co c os θ + cos 30 °si sin θ = 12 cos θ + cot β = ( cot 30° co cot θ + 1) ( cot θ − cot 30°) =

(

3 2

 

sin θ

 

) ( cot θ  − 3 )

3 cot θ + 1

Thus we have



T ( cos α + sin α cot β ) = W ⇒ T 

3 2



cos θ − 12 sin θ +

(

⇒ 3 cos θ si sin θ − sin 2 θ + ( cos θ si sin θ + 3 sin 2 θ ) ⋅  

1 2

cos θ +

3 2

3 cot θ  + 1

cot θ  − 3

3 cot θ  + 1  T  sin θ ⋅    =W = 2sin θ  cot θ  − 3 

)

=1.

In terms of cot θ  :

cos θ sin θ  = sin 2 θ  =

Thus

cot θ  1 + cot 2 θ  1

1 + cot 2 θ 

3 cot θ − 1 1 + cot 2 θ

finally θ  = tan −1

+

1 3 3

cot θ + 3 1 + cot 2 θ 

= 10°, 9 .

⋅

3 cot θ  + 1 cot θ − 3

= 1⇒

2 3 cot θ  − 3

= 1 ⇒ co cot θ = 3 3 ⇒ tan θ  =

1 3 3

 

and

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

32

C-2

As w moves of a (small) distance ∆hw  the spool rotates an angle ∆α  ≅

∆hw

 and W moves in the same r  direction by the same distance ∆hw  due to the vertical displacement of the spool, and in the opposite direction of a distance ∆hW  ≅ R∆α  due to the rotation of the spool. Then from PVW:

  w∆hw = W ( ∆hW − ∆hw ) = W ( R ∆α − r ∆α ) = W ( R − r ) ∆α ⇒ w  r ∆α = W ( R − r ) ∆α  , whence W  =

wr   R − r 

.

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

C-3

T

θ ′

T ′

T ′′

θ ′′

Define t ′ = tan θ ′ =

B-A 9m

, t ′′ = tan θ ′′ =

From FC at the ends of B, A, C: B)

Horiz. 1.

Vert. 2.

A)

T 

2 T

2

= T ′

= T′

1 1 + t ′2 t ′

1 + t ′

2

+w

A-C 9m

 and w =

4.8 ×103 kg-wts 6

.

C

T ′′′

33

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

Horiz. 3.

T′

Vert. 4.

T′

1

= T ′′

1 + t ′2 t′

1 + t′

= T ′′

2

1 1 + t ′′2 t ′′

1 + t ′′

2

+w

C)

Horiz. 5.

T ′′

Vert. 6.

T ′′

1 1 + t ′′2 t ′′

1 + t ′′2

= T ′′′

=w

t′

From 3. and 4.: T ′′

1 + t ′′2

From 6. and the latter: T ′′

t ′′

= T ′′

1 + t ′′2

t ′′

1 + t ′′

2

= T ′′

+ w ⇒ T ′′

( t ′ − t ′′) 1 + t ′′

2

( t ′ − t ′′ ) 1 + t ′′2

= w.

.

t ′ Thus t ′′ = t′ − t ′′ ⇒ t ′′ =    . 2 t ′

From 6. and 4.: T ′

1 + t ′

2

= 2w , whence from 2.

Finally, from 1., 3wt ′ = 2 w ⇒ t ′ = 23 ⇒ t ′′ = 13  . Thus

But But C = 2.00 2.00 m, so A=5 m, B= B=11 11 m. To determine T max let’s return back to 1. – 6.: T

1’.

2’.

3’.

2 T

2 3T ′ 13

=

=

=

3T ′ 13

2T ′ 13

,

whence T > T ′

+w

3T ′′ 10

,

whence T ′ > T ′′

T 

2

= 3w .

B-A 9m

=

2 3

,

A-C

1 = . 9m 3

34

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

4’.

5’.

6’.

2T ′ 13 3T ′′ 10 T ′′

10

=

T ′′

10

+ w = 2w using 6’. below, from which T ′ = 13w

= T ′′′ ,

whence T ′′ > T ′′′

=w

Finally, therefore, Tmax = T =

3 2 13

T ′ = 3 2w =

4.8 ×10 1 03 kg-wts 2

C-4

FA

≈ 34 × 103 kg-wts .

35

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics 

a) Given the condition F   B = 0 , from PVW (in the form

∑W  i

i

= 0 ), with respect to A, one gets

i

 L Fh = WL ⇒ F = W   . h

b) From FC: 

h. F = F  A cos θ  

v. W = F  A sin θ 

whence tan θ =



W F

=

h ⇒ θ  = tan −1     and L L h

2

2

h L =W 1 + tan θ  = W 1 +   = W 1 +   .  F A = W ⋅ h cos θ  h h L h  L

1

L

2

L

36

Solutions to problems in Leighton and Vogt’s textbook "Exercises in Introductory Introductory Physics" (Addison-Wesley, 1969) Pier F. Nali (Revised April 10, 2016)

CHAPTER 3 Kepler’s Laws and Gravitation

1

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

In general, for a curve having parametric equations  x = ϕ ( t ) ,

2

y = ψ ( t ) , the curvilinear trapezoid area

in the figure below

a

b

 x

t 2

where a = ϕ (t1 ) ,

∫

b = ϕ ( t 2 )  and   ψ ( t ) ≥ 0  in the interval ( t1 , t 2 ) , is A = ψ ( t ) ⋅ ϕ ′ ( t ) dt .   t 1

We can use this formula to calculate the area of an ellipse having parametric equations  x = a cos θ , y = b sin θ  . From the symmetry of the ellipse we can calculate the area of a quarter ellipse between the limits  x = 0 (ϑ  = π 2 ) ,  x = a (ϑ  = 0 ) . b

Hence

A 4

0

π  2

∫

∫

π 

0

= b sin θ ⋅ ( − a sin θ ) dθ = ab sin 2 θ dθ = π 4 ab ⇒ A=π ab .  2

A-1

 x=0

a

x=a

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

For the moon a =

1 2

(r

a

+ r p

)=

405,500 km + 363,300 km 2

3

= 384,500 km

For the Earth sat.

a =

1 2

(r

a

+ rp ) = 12 ( r⊕ + 710 km + r⊕ + 225 km ) = 12 ( d ⊕ + 935 k m ) =

3 2

12,756 km + 935 km 2

= 6, 845.5 km

3 2

a  6,845.5 km  × 27 d .322 ≈ 0 d .065 ≈ 1.6 hr Thus t =   t  =   384, 500 km   384, a 

A-2

From the Kepler’s second law ra va = rp v p ⇒

v p va

=

ra rp

=

A-3

a+c a−c

=

1 + e⊕ 1 − e⊕

=

1 + 0.0167 1 − 0.0167

=

1.0167 0.9833

= 1.033 .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

g⊕ = G

 M ⊕  R⊕2

1.6 m s −2 

,

M  R⊕ 

2

M   R⊕ 

4

2

2

1.000  6, 37 378 km  = × × 9.80 m s −2 = g =G 2 ⇒   ⇒g =   g⊕ =  81.3  1, 73 738 km  R g⊕ M ⊕  R  M⊕  R   M

g

g⊕

6

A-4

2

a)

2 3

3

 Tcomet   acomet   T comet   =  ⇒ acomet =   ⋅ a⊕ . Thus T a T   ⊕   ⊕   ⊕ 

From the Kepler’s third law  2 3

acomet

 (1986 − 1456 ) / 7  =  × 1 A.U. ≈ ( 75.7 ) A.U. ≈ 17.9 A.U.  and 1 yr y r   2 3

racomet = 2acomet − r pcomet ≈ 2 × 17.9 A.U. - 0.60 A.U. ≈ 35.8 A.U. - 0.60 A.U. ∼ 35.2 A.U.

b)

From the Kepler’s second law vmax r pcomet = vmin racomet  ⇒

vmax vmin

=

r acomet r  pcomet

=

35.2 A.U. 0.60 A.U.

∼ 59  .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

5

A-5

From the Kepler’s third law 2

3

2 3

2 3

 T24 h   R24 h   T 24 h   24 × 60 min  ⇒  R = =      ⋅ R E =   24 h  ⋅ RE = (14.4 ) ⋅ RE T R T  100 min    100 min    E    100 min 

B-1

2 3

∼ 5.9 RE  .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

6

 = 0   (equator). Although a synchronous orbit a) The geographic latitude of P is restricted to λ  = doesn’t need to be equatorial, a body in a synchronous non equatorial orbit could not remain in a fixed position with respect to P: it would appear to oscillate between the north and south of the Earth's equator. Moreover, a body in a synchronous non circular (elliptical) orbit would appear to oscillate East-and-West. The combination of the two latter (synchronous non-circular nonequatorial) would appear as a path in the shape of eight. 3 2

2 3

 r   1  1  r⊕ = r⊕ . b) From the Kepler’s third law t s (= 1 day)=  s  T ( = 27 days) ⇒ rs =   9  27   r ⊕ 

B-2

a) From the Kepler’s third law

T2 a⊕ 3

=

4π 2 GM ⊕

,

T ⊕ 2 a⊕ 3

 T   = ⇒  GM   T⊕ 

From Earth and Moon data:

d 

rP⊕ = 1.47 ×1011 m,

rA⊕ = 1.52 ×1011 m, T ⊕ = 365 .2 .242

rP = 3.63 × 108 m, m,

rA = 4.06 ×1011 m,

d 

T  = 27 .3 .322

4π 2

2

 a⊕  M  .  = a M   ⊕  ⊕ 3

⋅

 

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

a⊕ = a =

(r (r

1 2 1 2

+ r A⊕ ) = 1.50 × 1011 m

P⊕

+ r A

P

) = 3.84 ×10

 27 d .3 . 322  = Thus  d   M ⊕  365 .242   M 

b)

T2 a

3

=

4π 2 GM

,

a = 421, 800 km,

m

3

 1.50 ×1011 m  5 ⋅  = 3.33 × 10 8  3.84 ×10 m  2

T   a  M = ⇒   ⋅  = GM ⊕  T   a⊕ 3 M ⊕

T2

4π 2

a⊕3

   

d  . 769 a⊕ = 384, 000km, T  =1 .7

 27 d .322  =  d  Thus   M ⊕  1 .769   M 

2

8

2

3

 421, 800 km  ⋅  = 318 . 3 8 4 , 0 0 0 k m  

B-3

For elliptic orbits ma + mb =

4π 2 R 3 GT 2

.

For the Earth-Sun system m + m⊕ ( ≅ m ) =

4π 2 (1 A.U.) G (1 yr )

3

.

7

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

Thus ma + mb =

 R3 T 2

8

m  with R expressed in A.U. and T  in  in years.

B-4

II: unchanged.

The Kepler’s second low says that the areal velocity is constant, i.e., said d   the arc length of the R

trajectory trajectory of m and R the distance distance from M : dA dt

=

1 2

R

d dt

=

1 2

Rv =

1 2m

R ( mv ) =

L

2m

d 

= const. , being L the orbital angular angular momentum momentum of m .

But L remains constant for central forces, so the areal velocity also doesn’t change. Indeed, 

  GMm    GMm   = R × F = R × − (3+ a ) R  = − ( 3+ a ) ( R × R ) = 0 , being  R × R ≡ 0 . dt  R   R 

dL







Thus also  L cost. and

d A dt  

so well.

2



 = III: For a circular orbit F = −mω  R , where ω  =

2π  T 

. Hence

 GMm  GM GM  2 4π 2 ( 3+ a ) 4π 2 2 2 − (3+ a )  R = −mω R ⇒ ω  = ( 3+ a ) ⇒ 2 = ( 3+a) ⇒ T = R . T GM   R R R

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

9

C-1

Ignoring the moon’s gravitation one can assume g =

Moon at zenith: g z = g −

GM 

( R⊕ − R⊕ )

2

GM ⊕  R⊕2

, at nadir: g n = g +

.

GM 

( R⊕ + R⊕ )

2

.

    M   ∆g g n − g z 1 1  = = + Thus 2 2   . g g M ⊕   R  ⊕ + 1  R⊕ − 1     R⊕   R⊕  

But

 R⊕  R⊕

=

3.84 ×10 108 m 6.38 × 10 106 m

,

R⊕ R⊕

+1 =

3.90 ×10 10 8 m 6.38 ×10 10 6 m

,

R⊕ R⊕

−1 =

3.78 ×10 10 8 m

 M 

7.34 ×10 22 kg

= , , so 6.38 ×10 10 6 m  M ⊕ 5.98 ×10 24 kg

2 2 7.34  6.38   6.38   −6 −6 −6 = ×  +  × 10 = 1.23 × ( 2.68 + 2.85 ) ×10 = 1.23 × 5.53 ×10 =   g 5.98  3.90   3.78  

∆g

= (approx.) 7 × 10 −6 .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

10

C-2

 M =

4π 2 a 3 GT 2

,

M =

4π 2 a ⊕3 GT⊕2



 a  ⇒ M  =   a  ⊕ 

3

2

 T   ⋅  ⊕   .   T 

-1

Being T in days and V in km s  , V =

2π a [ km ]

  seconds   = × T T  s d a y s [ ]    day    

(

86, 400 km km Thus  M = 86,400

=

2π a [ km ] T × 86, 400 [s ]

3

2π ×1.50 ×10

8

⇒a=

86, 86, 400 2π 

× ( 365 days ) × TV km ) 2

3

TV .

 

⋅ M  = 1.02 ×10 −7 TV 3 M ⊕ .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

C-3

For the ellipse the radius of curvature is  Rc =

where  Ra = a (1 + e ) >

( R p Ra )

3 2

ab

 ,

R p = a (1 − e ) , b = a 1 − e 2 .

Thus  Rc = R p (1 + e ) = Ra (1 − e )  and

1 r

=

1 Rc

(1 − e cos ϕ ) .

11

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

1

a) From the conservation of energy:

2

v2 −

GM   R p

=

1 2

va2 −

12

GM     (simplifying the comet mass), Ra

where v is the velocity at perihelion, and from the Kepler’s second law vR p = va Ra .

1

Thus

2

va = v

so

1 2

 1 GM  1  − = 2e ,   R p Ra  R c  

( v 2 − va2 ) = GM  

 R p  Ra

=v

1− e 1+ e

⇒ v − va = 2v

( v 2 − va2 ) = 2v 2

e

(1 + e )

2

e

1+ e

= 2e

,

GM   Rc

v + va = 2v

1 1+ e

 ,

, whence 2

2

2

( vR p ) ( 500.0 km s −1 ×1.00 ×106 km ) GM    Rc  ⋅ = = 1.88 ×106 km .  Rc =  ⇒ Rc = 2 11 3 −2  v GM    1.33 ×10 km s  R p  b) 1 + e =

 Rc  R p

⇒e=

Rc Rp

− 1,

1.00 ×106 km 10 6 km = = = = 8.33 × 106 a=  R 1.88 1− e 0.12 2− 2− c 1.00  R p  R p

2 c

c) From the Kepler’s third law T =

Rp

4π 2 a3 GM 

⇒ T c =

2π a a GM 

km  .

 .

From a) GM  =

vR p  Rc

⇒ T c =

2π a aRc vR p

=

2π × 8.33 ×106 km 8.33 ×10 6 km × 1.88 ×10 6 km 500.0 Km s −1 ×1.00 ×10 6 km −1

(  ) 414,000 s  seconds   days  = × = ≈ 4.8 days . Tc [days ] = Tc [s ] ×  T  s [ ] c  day  −1 8 6 , 4 0 0 s d a y  s   

 414,000 s