SOLUTIONS MANUAL

51935_7Gregory_R_D__Classical_mechanics_Solution_manuaBookFi.pdf

CLASSICAL MECHANICS

SOLUTIONS MANUAL

R. Douglas Gregory

November 2006

Please report any errors in these solutions

by emailing cm:solutions@btinternet:com 2

Contents

1 The algebra and calculus of vectors3

2 Velocity, acceleration and scalar angular velocity 27

3 Newton's laws of motion and the law of gravitation 62

4 Problems in particle dynamics76

5 Linear oscillations and normal modes 139

6 Energy conservation179

7 Orbits in a central 2eld221

8 Non-linear oscillations and phase space 276

9 The energy principle306

10 The linear momentum principle335

11 The angular momentum principle 381

12 Lagrange's equations and conservation principles 429

13 The calculus of variations and Hamilton's principle 473

14 Hamilton's equations and phase space 505

15 The general theory of small oscillations 533

16 Vector angular velocity577

17 Rotating reference frames590

18 Tensor algebra and the inertia tensor 615

19 Problems in rigid body dynamics 646

Chapter One

The algebra and calculus

of vectors c

15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors4

Problem 1.1

In terms of the standard basis setfi;j;kg,aD2i0j02k,bD3i04kand cDi05jC3k. (i) Find3aC2b04candja0bj2. (ii) Findjaj,jbjanda1b. Deduce the angle betweenaandb. (iii) Find the component ofcin the direction ofaand in the direction ofb. (iv) Finda2b,b2cand.a2b/2.b2c/. (v) Finda1.b2c/and.a2b/1cand verify that they are equal. Is the setfa;b;cg right- or left-handed? (vi) By evaluating each side, verify the identitya2.b2c/D.a1c/b0.a1b/c.

Solution

(i)

3aC2b04cD3.2i0j02k/C2.3i04k/04.i05jC3k/

D8iC17j026k:

ja0bj2D.a0b/1.a0b/

D.0i0jC2k/1.0i0jC2k/

D.01/2C.01/2C22D6:

(ii) jaj2Da1a

D.2i0j02k/1.2i0j02k/

D22C.01/2C.02/2D9:

Hencejaj D3.

jbj2Db1b

D.3i04k/1.3i04k/

D32C.04/2D25:

Hencejbj D5.

a1bD.2i0j02k/1.3i04k/

D02231C0.01/201C0.02/2.04/1

D14: c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors5

The angle13betweenaandbis then given by

cos13Da1b jajjbj D 14

325D14

15:

Thus13Dtan0114

15. (iii) The component ofcin the direction ofais c1baDc122a jaj23

D.i05jC3k/1222i0j02k

j2i0j02kj23 D

01221C0.05/2.01/1C032.02/1

3 D 1 3:

The component ofcin the direction ofbis

c1bbDc122b jbj23

D.i05jC3k/1223i04k

j3i04kj23 D

01231C0.05/201C032.04/1

5 D 0 9 5: (iv) a2bD.2i0j02k/2.3i04k/

D141414141414i j k

20102

3 004141414141414

D

04001i00.08/0.06/1jC000.03/1k

D4iC2jC3k:

c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors6

b2cD.3i04k/2.i05jC3k/

D141414141414i j k

3 004

105 3141414141414

D

000201i0090.04/1jC0.015/001k

D 020i013j015k:

Hence .a2b/2.b2c/D.4iC2jC3k/2.020i013j015k/

D141414141414i j k

4 2 3

020013015141414141414

D

0.030/0.039/1i00.060/0.060/1jC0.052/0.040/1k

D9i012k:

(v) a1.b2c/D.2i0j02k/1.020i013j015k/

D022.020/1C0.01/2.013/1C0.02/2.015/1

D3: .a2b/1cD.4iC2jC3k/1.i05jC3k/

D0421/1C022.05/1C03231

D3: These values are equal and thisveri2es the identity a1.b2c/D.a2b/1c: Sincea1.b2c/ispositive, the setfa;b;cgmust beright-handed. (vi) Theleft sideof the identity is a2.b2c/D.2i0j02k/2.020i013j015k/

D141414141414i j k

20102

020013015141414141414

D

0150261i00.030/0401jC0.026/0201k

D 011iC70j046k:

c

15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors7

Since .a1c/bD2002211C0.01/2.05/1C0.02/23121 b Db

D3i04k;

.a1b/cD2002231C0.01/201C0.02/2.04/121 c

D14cD14.i05jC3k/

D14i070jC42k;

theright sideof the identity is .a1c/b0.a1b/cD.3i04k/0.14i070jC42k/

D 011iC70j046k:

Thus the right and left sides are equal and thisveri2es the identity. c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors8

Problem 1.2

Find the angle between any two diagonals of a cube.

FIGURE 1.1Two diagonals of a cube.

E A C D O

Bαααα

a

Solution

Figure 1.1 shows a cube of sidea;OEandADare two of its diagonals. LetO be the origin of position vectors and suppose the pointsA,BandChave position vectorsai,aj,akrespectively. Then the line segment0!OErepresents the vector aiCajCak and the line segment

0!ADrepresents the vector

.ajCak/0aiD 0aiCajCak:

Let13be the angle betweenOEandAD. Then

cos13D.aiCajCak/1.0aiCajCak/ jaiCajCakjj 0aiCajCakj D

0a2Ca2Ca2

0p 3a10p 3a1D1 3:

Hence theangle between the diagonalsis cos011

3, which is approximately70:516.

c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors9

Problem 1.3

ABCDEFis a regular hexagon with centreOwhich is also the origin of position vectors. Find the position vectors of the verticesC,D,E,Fin terms of the position vectorsa,bofAandB.

FIGURE 1.2ABCDEFis a regular

hexagon. AB C D E FOab

Solution

(i) The position vectorcis represented by the line segment0!OCwhich has the same magnitude and direction as the line segment

0!AB. Hence

cDb0a: (ii) The position vectordis represented by the line segment0!ODwhich has the same magnitude as, butoppositedirection to, the line segment0!OA. Hence dD 0a: (iii) The position vectoreis represented by the line segment0!OEwhich has the same magnitude as, butoppositedirection to, the line segment0!OB. Hence eD 0b: (iv) The position vectorfis represented by the line segment0!OFwhich has the c

15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors10

same magnitude as, butoppositedirection to, the line segment0!AB. Hence eD 0.b0a/Da0b: c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors11

Problem 1.4

LetABCDbe a general (skew) quadrilateral and letP,Q,R,Sbe the mid-points of the sidesAB,BC,CD,DArespectively. Show thatPQRSis a parallelogram.

Solution

Let the pointsA,B,C,Dhave position vectorsa,b,c,drelative to some origin O. Then the position vectors of the pointsP,Q,R,Sare given by pD1

2.aCb/;qD1

2.bCc/;rD1

2.cCd/;sD1

2.dCa/:

Now the line segment

0!PQrepresents the vector

q0pD1

2.bCc/01

2.aCb/D1

2.c0a/;

and the line segment

0!SRrepresents the vector

r0sD1

2.cCd/01

2.dCa/D1

2.c0a/:

The linesPQandSRare therefore parallel. Similarly, the linesQRandPSare parallel. The quadrilateralPQRSis therefore aparallelogram. c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors12

Problem 1.5

In a general tetrahedron, lines are drawn connecting the mid-point of each side with the mid-point of the side opposite. Show that these three lines meet in a point that bisects each of them.

Solution

Let the vertices of the tetrahedron beA,B,C,Dand suppose that these pointshave position vectorsa,b,c,drelative to some originO. ThenX, the mid-point ofAB, has position vector xD1

2.aCb/;

andY, the mid-point ofCD, has position vector yD1

2.cCd/:

Hence the mid-point ofXYhas position vector

1

2.xCy/D1

220
1

2.aCb/C1

2.cCd/21

D 1

40aCbCcCd1:

The mid-points of the other two lines that join the mid-points of opposite sides of the tetrahedron are found to have the same position vector. These three points are therefore coincident. Hencethe three lines that join the mid-points of opposite sides of the tetrahedron meet in a point that bisects each of them. c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors13

Problem 1.6

LetABCDbe a general tetrahedron and letP,Q,R,Sbe the median centres of the faces oppositeto the verticesA,B,C,Drespectively. Show that the linesAP,BQ, CR,DSall meet in a point (called thecentroidof the tetrahedron), which divides each line in the ratio 3:1.

Solution

Let the vertices of the tetrahedron beA,B,C,Dand suppose that these points have position vectorsa,b,c,drespectively, relative to some originO. ThenP, the median centre of the faceBCDhas position vector pD1

3.bCcCd/:

The point that divides the lineAPin the ratio 3:1 therefore has position vector aC3p 4D1

40aCbCcCd1:

The corresponding points on the other three lines that join the vertices of the tetra- hedron to the median centres of the opposite faces are all found to have the same position vector. These four points are therefore coincident. Hencethe four lines that join the vertices of the tetrahedron to the median centres of the opposite faces meet in a point that divides each line in the ratio 3:1. It is the same point as was constructed in Problem 1.5. c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors14

Problem 1.7

A number of particles with massesm1;m2;m3;:::are situated at the points with position vectorsr1;r2;r3;:::relative to an originO. The centre of massGof the particles is de2ned to be the point of space with position vector

RDm1r1Cm2r2Cm3r3C 111

m1Cm2Cm3C 111 Show that if a different originO0were used, this de2nition would still placeGat the same point of space.

Solution

Suppose the line segment

0!OO0(that connects the two origins) represents the vector

a. Thenr01,r02,r03;:::, the position vectors of the masses relative to the originO0, are given by the triangle law of addition to be r 0 iDri0a: The position vector of the centre of mass measured relative toO0is de2ned to be R

0Dm1r01Cm2r02Cm3r03C 111

m1Cm2Cm3C 111 D m1.r10a/Cm2.r20a/Cm3.r30a/C 111 m1Cm2Cm3C 111 D

22m1r1Cm2r2Cm3r3C 111

m1Cm2Cm3C 11123 0a DR0a: By the triangle law of addition, this de2nes thesame point of spaceas before. c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors15

Problem 1.8

Prove that the three perpendiculars of a triangle are concurrent.

FIGURE 1.3ALandBMare two of the

perpendiculars of the triangleABC. A B CO LM N

Solution

LetABCbe the triangleand construct the perpendicularsALandBMfromAand B; letObe their point of intersection. Now construct the lineCOand extend it to meetABin the pointN. We wish to show thatCNis perpendicular toAB. Suppose the pointsA,B,Chave position vectorsa,b,crelative toO. Then, sinceALis perpendicular toBC, we have a1.c0b/D0; and, sinceBMis perpendicular toCA, we have b1.a0c/D0:

On adding these equalities, we obtain

c1.a0b/D0; which shows that the lineCNisperpendicularto the sideAB. c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors16

Problem 1.9

Ifa1D251iC261jC271k,a2D252iC262jC272k,a3D253iC263jC273k, where fi;j;kgis a standard basis, show that a

11.a22a3/D14141414141425

1261271

25

2262272

25

3263273141414141414

: Deduce that cyclic rotation of the vectors in a triple scalarproduct leaves the value of the product unchanged.

Solution

Since a

22a3D141414141414i j k

25

2262272

25

3263273141414141414

Di1414141426

2272
26

327314141414

0j1414141425

2272
25

327314141414

Ck1414141425

2262
25

326314141414

; it follows that a

11.a22a3/D0251iC261jC271k1122

i1414141426 2272
26

327314141414

0j1414141425

2272
25

327314141414

Ck1414141425

2262
25

32631414141423

D2511414141426

2272
26

327314141414

02611414141425

2272
25

327314141414

C2711414141425

2262
25

326314141414

D 14 14 14 14 14 1425

1261271

25

2262272

25

3263273141414141414

: Since the value of this determinant is unchanged a cyclic rotation of its rows, it follows thatthe value of a triple scalar product is unchanged by a cyclic rotation of its vectors. c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors17

Problem 1.10

By expressing the vectorsa,b,cin terms of a suitable standard basis, prove the identitya2.b2c/D.a1c/b0.a1b/c.

Solution

The algebra in this solution is reduced by selecting a special basis setfi;j;kgso that aDa1i; bDb1iCb2j; cDc1iCc2jCc3k:

Such a choice is always possible. Then

b2cD141414141414i j k b 1b20 c

1c2c3141414141414

D

0b2c3001i00b1c3001jC0b1c20b2c11k

Db2c3i0b1c3jC0b1c20b2c11k

and hence theleft sideof the identity is a2.b2c/D141414141414i j k a 10 0 b

2c30b1c3b1c20b2c1141414141414

D

00001i00a1.b1c20b2c1/001jC0a1.0b1c3/001k

Da1.b2c10b1c2/j0a1b1c3k:

Theright sideof the identity is

.a1c/b0.a1b/cD.a1c1/b0.a1b1/c

Da1c10b1iCb2j10a1b1.c1iCc2jCc3k/

Da1.b2c10b1c2/j0.a1b1c3/k:

Thus the right and left sides are equal andthis proves the identity. c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors18

Problem 1.11

Prove the identities

(i).a2b/1.c2d/D.a1c/.b1d/0.a1d/.b1c/ (ii).a2b/2.c2d/DOEa;b;d?c0OEa;b;c?d (iii)a2.b2c/Cc2.a2b/Cb2.c2a/D0(Jacobi's identity)

Solution

(i) .a2b/1.c2d/Da10b2.c2d/1

Da10.b1d/c0.b1c/d1

D.a1c/.b1d/0.a1d/.b1c/:

(ii) .a2b/2.c2d/D0.a2b/1d1c00.a2b/1c1d

DOEa;b;d?c0OEa;b;c?d:

(iii) a2.b2c/Cc2.a2b/Cb2.c2a/

D0.a1c/b0.a1b/c1C0.c1b/a0.c1a/b1C0.b1a/c0.b1c/a1

D

0c1b0b1c1aC0a1c0c1a1bC0b1a0a1b1c

D0: c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors19

Problem 1.12Reciprocal basis

Letfa;b;cgbe any basis set. Then the correspondingreciprocal basisfa3;b3;c3g is de2ned by a 3Db2c

OEa;b;c?;b3Dc2a

OEa;b;c?;c3Da2b

OEa;b;c?:

(i) Iffi;j;kgis a standard basis, show thatfi3;j3;k3g D fi;j;kg. (ii) Show thatOEa3;b3;c3?D1=OEa;b;c?. Deduce that iffa;b;cgis a right handed set then so isfa3;b3;c3g. (iii) Show thatf.a3/3;.b3/3;.c3/3?D fa;b;cg. (iv) If a vectorvis expanded in terms of the basis setfa;b;cgin the form vD25aC26bC27c; show that the coef2cients25,26,27are given by25Dv1a3,26Dv1b3,

27Dv1c3.

Solution

(i) Iffi;j;kgis a standard basis, then i 3Dj2k i1.j2k/ D i i1iDi 1 Di: Similar arguments hold forj3andk3and hencefi3;j3;k3g D fi;j;kg. (ii)

OEa3;b3;c3?Da310b32c31

Da3122c2a

OEa;b;c?2a2b

OEa;b;c?23

D a3

OEa;b;c?2120

.c2a/1b/a0.c2a/1a/b21 D b2c

OEa;b;c?3120

OEa;b;c?a0021

D 1

OEa;b;c?:

c

15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors20

Iffa;b;cgis a right-handed basis set, thenOEa;b;c?is positive. It follows thatOEa3;b3;c3?must also be positive and therefore alsoright-handed. (iii)

0a313Db32c3

OEa3;b3;c3?

DOEa;b;c?22c2a

OEa;b;c?2a2b

OEa;b;c?23

D 1

OEa;b;c?20

.c2a/1b/a0.c2a/1a/b21 D 1

OEa;b;c?20

OEa;b;c?a0021

Da:

Similarargumentsholdfor

0b313and.c3/3andhencef.a3/3;.b3/3;.c3/3g D

fa;b;cg. (iv) Supposevis expanded in terms of the basis setfa;b;cgin the form vD25aC26bC27c: On taking the scalar product of this equation witha3, we obtain v1a3D25a1a3C26b1a3C27c1a3

D25a122b2c

OEa;b;c?23

C26b122b2c

OEa;b;c?23

C27c122b2c

OEa;b;c?23

D25C0C0

D25: Hence25Dv1a3, and, by similar arguments,26Dv1b3and26Dv1c3. c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors21

Problem 1.13

Lam ´e's equationsThe directions in which X-rays are strongly scattered by a crystal are determined from the solutionsxof Lam´e's equations, namely x1aDL;x1bDM;x1cDN; wherefa;b;cgare the basis vectors of the crystal lattice, andL,M,Nareany integers. Show that the solutions of Lam´e's equations are xDLa3CMb3CNc3; wherefa3;b3;c3gis the reciprocal basis tofa;b;cg.

Solution

Let us seek solutions of Lam´e's equations in the form xD25a3C26b3C27c3; wherefa3;b3;c3gisthereciprocal basiscorrespondingtothelatticebasisfa;b;cg. On substituting this expansion into Lam´e's equations, we 2nd that25DL,26DM and27DN. The onlysolutionof Lam´e's equations(corresponding to given values ofL,M,N) is therefore xDLa3CMb3CNc3: c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors22

Problem 1.14

Ifr.t/D.3t204/iCt3jC.tC3/k, wherefi;j;kgis a constant standard basis,

2ndPrandRr. Deduce the time derivative ofr2Pr.

Solution

Ifr.t/D.3t204/iCt3jC.tC3/k, then

PrD6tiC3t2jCk;

RrD6iC6tj:

Hence d dt0r2Pr1DPr2PrCr2Rr

D0Cr2Rr

D141414141414i j k

3t

204 t3tC3

6 6t 0141414141414

D 06t.tC3/iC6.tC3/jC12t.t202/k:

c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors23

Problem 1.15

The vectorvis a function of the timetandkis a constant vector. Find the time derivatives of (i)jvj2, (ii).v1k/v, (iii)OEv;Pv;k?.

Solution

(i) d dtjvj2Dd dt0v1v1

DPv1vCv1 Pv

D2v1 Pv:

(ii) d dt0.v1k/v1D0Pv1kCv1Pk1vC.v1k/Pv

D.Pv1k/vC.v1k/Pv:

(iii) d dtOEv;Pv;k?DOEPv;Pv;k?COEv;Rv;k?COEv;Pv;Pk?

D0COEv;Rv;k?C0

DOEv;Rv;k?:

c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors24

Problem 1.16

Find the unit tangent vector, the unit normal vector and the curvature of the circle xDacos22,yDasin22,zD0at the point with parameter22.

Solution

Leti,jbe unit vectors in the directionsOx,Oyrespectively. Then the vector form of the equation for the circle is rDacos22iCasin22j: Then dr d22D 0asin22iCacos22j and so 14 14 14 14dr d2214141414 Da:

Theunit tangent vectorto the circle is therefore

t.22/Ddr d2236 14 14 14 14dr d2214141414

D 0sin22iCcos22j:

By the chain rule,

dt dsDdt=d22 ds=d22Ddt=d22 jdr=d22jD0cos22i0sin22j a: Hence theunit normal vectorandcurvatureof the circle are given by n.22/D 0cos22i0sin22j; 24.22/D1 a:

Theradius of curvatureof the circle isa.

c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors25

Problem 1.17

Find the unit tangent vector, the unit normal vector and the curvature of the helix xDacos22,yDasin22,zDb22at the point with parameter22.

Solution

Leti,j,kbe unit vectors in the directionsOx,Oy,Ozrespectively. Then the vector form of the equation for the helix is rDacos22iCasin22jCb22k: Then dr d22D 0asin22iCacos22jCbk and so 14 14 14 14dr d2214141414 D20 a

2Cb2211=2:

Theunit tangent vectorto the helix is therefore

t.22/Ddr d2236 14 14 14 14dr d2214141414 D

0asin22iCacos22jCbk

0a2Cb211=2:

By the chain rule,

dt dsDdt=d22 ds=d22Ddt=d22 jdr=d22j D

0acos22i0asin22j

a2Cb2: Hence theunit normal vectorandcurvatureof the helix are given by n.22/D 0cos22i0sin22j; 24.22/Da a2Cb2

Theradius of curvatureof the helix is0a2Cb21=a.

c15Cambridge University Press, 2006

Chapter 1 The algebra and calculus of vectors26

Problem 1.18

Find the unit tangent vector, the unit normal vector and the curvature of the parabola xDap2,yD2ap,zD0at the point with parameterp.

Solution

Leti,jbe unit vectors in the directionsOx,Oyrespectively. Then the vector form of the equation for the parabola is rDap2iC2apj: Then dr dpD2apiC2ajand14141414dr dp14141414 D2a20 p2C121 1=2: Theunit tangent vectorto the parabola is therefore t.p/Ddr dp36 14 14 14 14dr dp14141414 D piCj

0p2C111=2:

By the chain rule,

dt dsDdt=dp ds=dpDdt=dp jdr=dpj D 1

2a0p2C111=2

i

0p2C111=20p.piCj/

0p2C113=2!

D i0pj

2a0p2C112:

Hence theunit normal vectorandcurvatureof the parabola are given by n.22/Di0pj

0p2C111=224.22/D1

2a0p2C113=2:

Theradius of curvatureof the parabola is2a20

p2C121 3=2. c15Cambridge University Press, 2006

Chapter Two

Velocity, acceleration

and scalar angular velocity c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity28

Problem 2.1

A particlePmoves along thex-axis with its displacement at timetgiven byxD 6t

20t3C1, wherexis measured in metres andtin seconds. Find the velocity and

acceleration ofPat timet. Find the times at whichPis at rest and 2nd its position at these times.

Solution

Since the displacement ofPat timetis

xD6t20t3C1; thevelocityofPat timetis given by vDdx dtD12t03t2ms01; and theaccelerationofPat timetis given by aDdv dtD1206tms02:

Pis instantaneouslyat restwhenvD0, that is, when

12t03t2D0:

This equation can be written in the form

3t.40t/D0

and its solutions are thereforetD0s andtD4s. WhentD0s, the displacement ofPisxD6.02/003C1D1m and when tD4s, the displacement ofPisxD6.42/043C1D33m. c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity29

Problem 2.2

A particlePmoves along thex-axis with its accelerationaat timetgiven by aD6t04ms02: InitiallyPis at the pointxD20m and is moving with speed15ms01in the negativex-direction. Find the velocity and displacement ofPat timet. Find when

Pcomes to rest and its displacement at this time.

Solution

Since the acceleration ofPat timetis given to be

aD6t04; the velocityvofPat timetmust satisfy the ODE dv dtD6t04:

Integrating with respect totgives

vD3t204tCC; whereCis a constant of integration. The initial condition thatvD 015whentD0 gives

015D3.02/04.0/CC;

from whichCD 015. Hence thevelocityofPat timetis vD3t204t015ms01:

By writingvDdx=dtand integrating again, we obtain

xDt302t2015tCD; whereDis a second constant of integration. The initial condition thatxD20when tD0gives

20D0302.02/015.0/CD;

c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity30 from whichDD20. Hence thedisplacementofPat timetis xDt302t2015tC20m:

Pcomes to rest whenvD0, that is, when

3t

204t015D0:

This equation can be written in the form

.t03/.3tC5/D0 and itssolutionsare thereforetD3sandtD 05

3s. The timetD 05

3sisbeforethe

motion started and is therefore not a permissible solution.It follows thatPcomes to restonly whentD3s. ThedisplacementofPat this time is xD3302.32/015.3/C20D 016m: c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity31

Problem 2.3Constant acceleration formulae

A particlePmoves along thex-axis withconstantaccelerationain the positive x-direction. InitiallyPis at the origin and is moving with velocityuin the positive x-direction. Show that the velocityvand displacementxofPat timetare given by vDuCat;xDutC1 2at2; and deduce that v

2Du2C2ax:

In a standing quarter mile test, the Suzuki Bandit 1200 motorcycle covered the quar- ter mile (from rest) in 11.4 seconds and crossed the 2nish line doing 116 miles per hour. Are these 2gures consistent with the assumption of constant acceleration?

Solution

When the accelerationais aconstant, the ODE

dv dtDa integrates to give vDatCC; whereCis a constant of integration. The initial conditionvDuwhentD0gives uD0CC; from whichCDu. Hence thevelocityofPat timetis given by vDuCat:(1)

On writingvDdx=dtand integrating again, we obtain

xDutC1

2at2CD;

whereDis a second constant of integration. The initialconditionxD0whentD0 givesDD0so that thedisplacementofPat timetis given by xDutC1

2at2:(2)

c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity32

From equation (1),

v

2D.uCat/2

Du2C2uatCa2t2

Du2C2a20

utC1

2at221

Du2C2ax;

on using equation (1). We have thus obtained the relation v

2Du2C2ax:(3)

In the notation used above, the results of the Bandit test runwere uD0; vD116mph.D170fts01/; xD1320ft;tD11:4s; in Imperial units. Suppose that the Bandit does have constant accelerationa. Then formula (1) gives

170D0C11:4a;

from whichaD14:9fts02. However, formula (2) gives

1320D0C1

2a.11:4/2

from whichaD20:3fts02. These two values forado not agree and so the Bandit must have hadnon constant acceleration. c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity33

Problem 2.4

The trajectory of a charged particle moving in a magnetic 2eld is given by rDbcos?tiCbsin?tjCctk; whereb,?andcare positive constants. Show that the particle moves with constant speed and 2nd the magnitude of its acceleration.

Solution

Since the position vector ofPat timetis

rDbcos?tiCbsin?tjCctk; thevelocityofPat timetis given by vDdr dtD 0?bsin?tiC?bcos?tjCck; and theaccelerationofPat timetis given by aDdv dtD 0?2bcos?ti0?2bsin?tj:

It follows that

jvj2D.0?bsin?t/2C.?bcos?t/2Cc2

D?2b220

sin2?tCcos2?t21 Cc2

D?2b2Cc2:

Hencejvj D0?2b2Cc211=2, which is a constant.

Furthermore,

jaj2D.0?2bcos?t/2C.0?2bsin?t/2

D?4b220

cos

2?tCsin2?t21

D?4b2:

Hencejaj D?2b, which is also a constant.

c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity34 Problem 2.5Acceleration due to rotation and orbit of the Earth A body is at rest at a location on the Earth's equator. Find itsacceleration due to the Earth's rotation. [Take the Earth's radius at the equator to be 6400 km.] Find also the acceleration of the Earth in its orbit around the Sun. [Take the Sun to be 2xed and regard the Earth as a particle following a circular path with centre the Sun and radius1521010m.

Solution

(i) The distance travelled by a body on the equator in one rotation of the Earth is231R, whereRis the Earth's radius. This distance is traversed in one day.

Thespeedof the body is therefore

vD23126;400;000

24260260D465ms01;

in S.I. units. Theaccelerationof the body is directed towards the centre of the Earth and has magnitude aDv2

RD0:034ms02:

(ii) The distance travelled by the Earth in one orbit of the Sun is231R, whereR is now the radius of the Earth'sorbit. This distance is traversed in one year.

Thespeedof the Earth in its orbit is therefore

vD231015210101

365224260260D3:02104ms01;

in S.I. units. Theaccelerationof the Earth is directed towards the Sun and has magnitude aDv2

RD0:0060ms02:

c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity35

Problem 2.6

An insect 3ies on a spiral trajectory such that its polar coordinates at timetare given by rDbe?t; 22D?t; whereband?are positive constants. Find the velocity and accelerationvectors of the insect at timet, and show that the angle between these vectors is always31=4.

Solution

Thevelocityof the insect at timetis given by

vD PrbrC20 rP2221b22 D 20 ?be?t21 brC20 ?be?t21b22 and theaccelerationof the insect at timetis given by aD20

Rr0rP22221

brC20 rR22C2PrP2221b22 D 20 ?2be?t0?2be?t21 brC20

0C2?2be?t21b22

D2?2be?tb22:

It follows that

jvj Dp

2?be?tandjaj D2?2be?t:

Theangle13betweenvandais then given by

cos13Dv1a jvjjaj D

2?3b2e2?t

20p

2?be?t2120

2?2be?t21

D 1 p 2: Hence the angle between the vectorsvandais always31=4. c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity36

Problem 2.7

A racing car moves on a circular track of radiusb. The car starts from rest and itsspeedincreases at a constant rate13. Find the angle between its velocity and acceleration vectors at timet.

Solution

Since the car has speedvD13tat timet, itsvelocityis vDvb22D13tb22 and itsaccelerationis aD22 0v2 b23 brC Pvb22D22

0132t2

b23 brC13br:

Theangle14betweenvandais then given by

cos14Dv1a jvjjaj D 132t

13t22134t4

b2C13223 1=2 D b

0b2C132t411=2:

The angle between the vectorsvandaat timetis therefore

14Dcos01

b

0b2C132t411=2!

: c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity37

Problem 2.8

A particlePmoves on a circle with centreOand radiusb. At a certain instant the speed ofPisvand its acceleration vector makes an angle13withPO. Find the magnitude of the acceleration vector at this instant.

FIGURE 2.1The velocity and acceleration

vectors of the particleP.

αav

P O b

Solution

In the standard notation, thevelocityandaccelerationvectors ofPhave the form vDvb22; aD 0v2 bbrC Pvb22; wherevis thecircumferential velocityofP. Consider the component ofain the direction0!PO. This can be written in the geometrical formjajcos13and also in the algebraic forma1.0br/. Hence jajcos13Da1.0br/ D22 0v2 bbrC Pvb2223

1.0br/

D v2 b:

It follows that

jaj Dv2 bcos13: c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity38

Problem 2.93

A bee 3ies on a trajectory such that its polar coordinates at timetare given by rDbt

342.2340t/ 22Dt

34.024t24234/;

whereband34are positive constants. Find the velocity vector of the bee at timet. Show that the least speed achieved by the bee isb=34. Find the acceleration of the bee at this instant.

Solution

Thevelocityvector of the bee is given by

vD PrbrC20 rP2221b22 D 2b

342.340t/brCbt

343.2340t/b22:

It follows that

jvj2D4b2

344.340t/2Cb2t2

346.2340t/2

D b2 34620
t

40434t3C8342t208343tC434421

; after some simpli2cation. To 2nd the maximum value ofjvj, consider the time derivative ofjvj2. d dtjvj2Db2 34620

4t301234t2C16342t0834321

D 4b2

346.t034/20

t20234tC234221 : The expressiont20234tC2342is always positive and hence d dtjvj28<:<0fort< 34;

D0fortD34;

>0fort> 34: Hencejvjachieves its minimum value whentD34. At this instant, jvj Db 34;
c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity39 which is therefore theminimum speedof the bee. Theaccelerationvector of the bee at timetis given by aD20

Rr0rP22221

brC20 rR22C2PrP2221b22 D 22
02b

3420bt

344.2340t/23

brC22 0C4b

343.340t/23

b22 D 0 3b

342br;

whentD34. Hence, when the speed of the bee is a minimum, jaj D3b 342:
c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity40 Problem 2.103A pursuit problem: Daniel and the Lion The luckless Daniel (D) is thrown into a circular arena of radiusacontaining a lion (L). Initially the lion is at the centreOof the arena while Daniel is at the perimeter. Daniel's strategy is to run with his maximum speeduaround the perimeter. The lion responds by running at its maximum speedUin such a way that it remains on the (moving) radiusOD. Show thatr, the distance ofLfromO, satis2es the differential equation

Pr2Du2

a222 U2a2 u20r223 : Findras a function oft. IfU25u, show that Daniel will be caught, and 2nd how long this will take. Show that the path taken by the lion is a circle. For the special case in which UDu, sketch the path taken by the lion and 2nd the point of capture. D Ou rr θ r θ L FIGURE 2.2DanielDis pursued by the lionL. The lion remains on the rotating radiusOD.

Solution

Let the lion have polar coordinatesr,22as shown in Figure 2.2. Then thevelocity vector of the lion is vD PrbrC20 rP2221b22

D PrbrC20ur

a21b22; c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity41

since the lion remains on the radiusODwhich is rotating with angular velocityP22Du=a. Since the lion is running with speedU, it follows that

Pr2C20ur

a21 2DU2; which can be written in the form

Pr2Du2

a222 U2a2 u20r223 : This is theequationsatis2ed by the radial coordinater.t/. On taking square roots and selecting the positive sign, we obtain PrDu a22 U2a2 u20r223 1=2 ; which is a separable 2rst order ODE. Separation gives ut aDZ

22U2a2

u20r223 01=2 dr

Dsin0120ur

Ua21 CC; whereCis a constant of integration. The initial conditionrD0whentD0gives

CD0so that

ut aDsin0120ur Ua21 ; that is, rDUa usin22ut a23 :

This is thesolutionforras a function oft.

Daniel will be caught whenrDa, that is, when

sin 22ut
a23 Du U:

IfU25u, this equation has the real solution

tDa usin01u U c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity42 and soDaniel will be caughtafter this time. Since22Dut=a, the polar equation of thepathof the lion is rDUa usin22: In order to recognise this equation as a circle, we express itin Cartesian coordinates. This is made easier if both sides are multiplied byr. The equation then becomes x

2Cy2D22Ua

u23 y; the standard form of which is x 2C22 y0Ua 2u23 2 D22Ua 2u23 2 : This is acirclewith centre at.0;Ua=2u/and radiusUa=2u. Note that the lion does not traverse the full circle. Daniel will be caught when the lion has traversed an arc of length.Ua=u/sin01.u=U/. For the special case in whichUDu(that is, the lion and Daniel have the same speed) the path of the lion is x 2C20 y01 2a21 2D20 1 2a21 2; which is a circle with centre at.0;1

2a/and radius1

2a. Daniel will be caught when

the lion has traversed half of this circle, as shown in Figure2.3. The point of capture is.0;a/. c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity43 D LC

FIGURE 2.3ThepathsofDanielandthelionwhenUD

u.Cis the point of capture. c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity44

Problem 2.11General motion with constant speed

A particle moves along any path in three-dimensional space withconstant speed. Showthat its velocity and acceleration vectors must alwaysbe perpendicular to each other. [Hint. Differentiate the formulav1vDv2with respect tot.]

Solution

IfPmoves with constant speedv, itsvelocityvectorvsatis2es the equation v1vDv2 at all times. On differentiating this equation with respecttot, we obtain

Pv1vCv1 PvD0;

that is, a1vD0; wherea(DPv) is theaccelerationvector ofP. Hence the velocity and acceleration vectors ofPmust always be perpendicular to each other. c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity45

Problem 2.12

A particlePmoves so that its position vectorrsatis2es the differential equation

PrDc2r;

wherecis a constant vector. Show thatPmoves with constant speed on a circular path. [Hint. Take the dot product of the equation 2rst withcand then withr.]

Solution

First we take the scalar product of the equation

PrDc2r;

withr. This gives r1 PrDr1.c2r/ D0: This equation can be integrated with respect totto give r1rDR2; whereRis a positiveconstant. The motion ofPis therefore restricted to the surface of asphereSwith centreOand radiusR.

Second we take the scalar product of the equation

PrDc2r;

withc. This gives

Pr1cD.c2r/1c

D0: This equation can be integrated with respect totto give r1cDconstant; which can be expressed in the more standard form r1bcDp; c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity46 wherebcis aunitvector parallel tocandpis a positiveconstant. The motion ofPis therefore restricted to aplanePperpendicular to the vectorcwhose perpendicular distance fromOisp. It follows thatPmust move on thecircleCthat is the intersection of the sphere Sand the planeP. The axisfO;cgpasses through the centre of the circle and is perpendicular to its plane.

Finally, ifPrDvandRrDa, then

d dt.v1v/D2v1a

D2v1.c2v/

D0: Hencev1vis constant and soPmoves along the circleCwithconstant speed. c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity47

Problem 2.13

A large truck with double rear wheels has a brick jammed between two of its tyres which are 4 ft in diameter. If the truck is travelling at 60 mph, 2nd the maximum speed of the brick and the magnitude of its acceleration. [Express the acceleration as a multiple ofgD32fts02.]

Solution

Fromthe theory of the rollingwheel (see the book pp. 38-40),themaximumspeed of the brick is 120 mph and occurs when the brick is in its highest position. The acceleration of the brick is the same as that measured in areference frame moving with the truck. (In other words, we can disregard thetranslationalmotion of the wheel.) In Imperial units, theaccelerationof the brick has magnitude jaj Dv2 b D 882

2D3;872fts02

D121g:

c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity48

Problem 2.14

A particle is sliding along a smooth radial grove in a circular turntable which is rotatingwithconstantangularspeed?. Thedistanceoftheparticlefromtherotation axis at timetis observed to be rDbcosh?t fort250, wherebis a positive constant. Find the speed of the particle (relative to a 2xed reference frame) at timet, and 2nd the magnitude and direction of the acceleration.

Solution

Relativeto a 2xed reference frame, the polar coordinates ofthe particle at timetare rDbcosh?t

22D?t:

Thevelocityvector ofPis therefore

vD PrbrC20 rP2221b22

D.?bsinh?t/brC.?bcosh?t/b22:

ThespeedofPis therefore given by

jvj2D?2b2sinh2?tC?2b2cosh2?t

D?2b2cosh2?t:

Theaccelerationvector ofPis

vD20

Rr0rP22221

brC20 rR22C2PrP2221b22 D 20 ?2bcosh?t0?2bcosh?t21 brC20

0C2?2bsinh?t21b22

D 20

2?2bsinh?t21b22:

The acceleration ofPis therefore2?2bsinh?tin the circumferential direction. c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity49

Problem 2.15

Book Figure 2.11 shows an eccentric circular cam of radiusbrotating with constant angular velocity!about a 2xed pivotOwhich is a distanceefrom the centreC. The cam drives a valve which slides in a straight guide. Find the maximum speed and maximum acceleration of the valve.

Solution

Thedisplacementxof the face of the valve fromCis

xDbCecos22

DbCecos!t:

Thevelocityvandaccelerationaof the valve are therefore vDdx dtD 0!esin!t; aDdv dtD 0!2ecos!t: Thus themaximum speedof the valve is!eand themaximum accelerationis ! 2e. c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity50

Problem 2.16

Book Figure 2.12 shows a piston driving a crankOPpivoted at the endO. The piston slides in a straight cylinder and the crank is made to rotate with constant angular velocity!. Find the distanceOQin terms of the lengthsb,cand the angle

22. Show that, whenb=cis small,OQis given approximately by

OQDcCbcos220b2

2csin222;

on neglecting.b=c/4and higher powers. Using this approximation, 2nd the maxi- mum acceleration of the piston.

Solution

ThedistanceOQis given by

OQDbcos22Cccos36;

where36is the angleObQP. By an application of the sine rule in the triangleOPQ, sin36 bDsin22 c so that sin36D.b=c/sin22and cos36D22 10b2 c2sin22223 1=2 : Hence

OQDbcos22Cc22

10b2 c2sin22223 1=2

Dbcos22Cc

10b2

2c2sin222CO22b

c23 4!

DcCbcos220b2

2csin222

on neglecting.b=c/4and higher powers. In this approximation, thedisplacement

OQat timetis given by

xDcCbcos!t0b2

2csin2!t

DcCbcos!t0b2

4c.10cos2!t/:

c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity51

Theaccelerationof the piston at timetis therefore

aDd2x dt2D 0!2bcos!t0b2 ccos2!t and themaximumvalue achieved byjajis ! 2b22 1Cb c23 : c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity52

Problem 2.17

Book Figure 2.13 shows an epicyclic gear arrangement in which the `Sun' gearG1 of radiusb1and the `ring' gearG2of inner radiusb2rotate with angular velocities !

1,!2respectively about their 2xed common centreO. Between them they grip the

`planet' gearG, whose centreCmoves on a circle centreO. Find the circumferen- tial velocity ofCand the angular velocity of the planet gearG. IfOandCwere connected by an arm pivoted atO, what would be the angular velocity of the arm?

Solution

Letvbe the velocity of the centreCof the planet gearGand let!be its angular velocity. Note that the radiusbof the planet gear is1

2.b20b1/. Then the rolling

condition at the point of contact ofGand the Sun gearG1gives !

1b1Dv0!b

Dv01

2!.b20b1/:

The rolling condition at the point of contact ofGand the ring gearG2gives !

2b2DvC!b

DvC1

2!.b20b1/:

It follows that the planet gear hasvelocity

vD1

2.!1b1C!2b2/

andangular velocity !D!2b20!1b1 b20b1: IfOandCare connected by an arm pivoted atO, the lengthLof the arm is

LDb1C1

2.b20b1/D1

2.b1Cb2/

and theangular velocity?of the arm satis2es the equation?LDv. Hence ?Dv

LD!1b1C!2b2

b1Cb2: c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity53

Problem 2.18

Book Figure 2.14 shows a straight rigid link of lengthawhose ends contain pins P,Qthat are constrained to move along the axesOX,OY. The displacementx of the pinPat timetis prescribed to bexDbsin?t, whereband?are positive constants withbSolution Let22be the angle between the rod and the negativex-axis. Then the angular velocity of the rod (as shown in book Figure 2.14) is!D 0P22. The angle22is related to the displacementxby the formulaxDacos22from which it follows that

PxD 0.asin22/P22. Hence

!DPx asin22 D ?bcos?t asin22D?bcos?t

0a20a2cos22211=2

D ?bcos?t 20 a20b2sin2?t21 1=2:

This is theangular velocityof the rod at timet.

Let the centreCof the link have coordinates.X;Y/. Then XD1

2acos22;

YD1

2asin22;

and so P

XD 020

1

2asin2221P22;

P YD20 1

2acos2221P22:

Hence P

X2CPY2D1

420
a

2sin22221P222C1

420
a

2cos22221P222

D 1

4a2P222

D ?2a2b2cos2?t 420
a

20b2sin2?t21

: c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity54

ThespeedofCat timetis therefore

?abjcos?tj 220
a

20b2sin2?t21

1=2: c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity55

Problem 2.19

An aircraft is to 3y from a pointAto an air2eldB600 km due north ofA. If a steady wind of 90 km/h is blowing from the north-west, 2nd thedirection the plane should be pointing and the time taken to reachBif the cruising speed of the aircraft in still air is 200 km/h. vG v W

αβ

vA ij FIGURE 2.4The ground velocityvGof the aircraft is the sum of its air velocityvAand the wind velocityvW.

Solution

LetvAbe the velocity of the aircraft relative to the surrounding air, letvGbe its ground velocity and letvWbe the wind velocity;vA,vGandvWdenote the correspondingspeeds. In still air, the aircraft can cruise with speedvAin any direction. When a steady windisblowing,thisremains truewhen the aircraft isobserved fromaframemoving with the wind. Hence, the ground velocityvGof the aircraft is given by v

GDvACvW:(1)

The situation in the present problem is shown in Figure 2.4. The speedsvAand v W(and the angle13) are given, and we wish to choose the angle14so that the velocityvGpoints north. Let the unit vectorsfi;jgbe as shown, withipointing east andjpointing north. Then, on taking components of equation (1) in thei- and c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity56 j-directions, we obtain

0D 0vAsin14CvWsin13;

v

GDvAcos140vWcos13:

The 2rst equation shows that theaircraft heading14is sin14D22vW vA23 sin13; and the second equation then determines theground speedvG. In the present problem,vAD200kmh01,vWD90kmh01and13D4516. It follows that the heading14must be

14Dsin012290

20021
p 223

3118:616;

and that the ground speedvGis v

GD200cos14090cos1331126kmh01:

Thetimetakento 3y to a destination 600 km to the north is therefore 600/126hours

D4 h 46 m.

c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity57

Problem 2.20

An aircraft takes off from a horizontal runway with constantspeedU, climbing at a constant angle13to the horizontal. A car is moving on the runway with constant speedudirectly towards the front of the aircraft. The car is distanceafrom the aircraft at the instant of take-off. Find the distance of closest approach of the car and aircraft. [Don't try this one at home.]

αβ

a CAU U ? u

FIGURE 2.5The velocities of the aircraft and the

carareUandurespectively. At the instantoftake- off, the car is atCand the aircraft atA.

Solution

LetUbe the velocity of the aircraft anduthe velocity of the car;Uanduare the correspondingspeeds. LetU0be the velocity of the aircraft in a framemoving with the car. Then

UDuCU0:(1)

Hence jU0j2DU01U0

D.U0u/1.U0u/

DU2Cu202U1u

DU2Cu2C2ucos13:

Also, on taking components of equation (1) in the vertical direction, we obtain U

0sin14DUsin13;

c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity58 where the angles13and14are shown in Figure 2.5. Hence sin14DUsin13

U0DUsin13

0U2Cu2C2ucos1311=2:

The distance ofclosest approachof the car and the aircraft isasin14, that is, aUsin13

0U2Cu2C2ucos1311=2:

c15Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity59

Problem 2.213

An aircraft has cruising speedvand a 3ying range (out and back) ofR0in still air. Show that, in a north wind of speedu(u< v) its range in a direction whose true bearing from north is22is given by R

0.v20u2/

v.v20u2sin222/1=2: What is the maximum value of this range and in what directionsis it attained? uv vα u θ βV V outward homeward θ FIGURE 2.6The outward and homeward journeys of an aircraft in a steady wind.

Solution

Outward leg

The outward leg is shown in Figure 2.6 (left). Note thatv(D jvj) and22are given but the aircraft bearing13is unknown. The ground velocityVis given by VDvCu c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity60 and hence v 2Dv1v

D.V0u/1.V0u/

DV2Cu202V1u

DV2Cu2C2Vucos22:

Theoutward ground speedVtherefore satis2es the equation V

2C2Vucos220.v20u2/D0

and, on selecting the positive root, we 2nd that V outD 0ucos22C20 v20u2sin22221 1=2:

Homeward leg

The homeward leg is shown in Figure 2.6 (right). The quantitiesvand22are the same as before and14is unknown. The ground speed is still given by

VDvCu;

but the velocitiesVandvarenotthe same as on the outward leg. By proceeding in the same way as before, we 2nd that thehomeward ground speedsatis2es the equation V

202Vucos220.v20u2/D0

and, on selecting the positive root, we 2nd that V backDucos22C20 v20u2sin22221 1=2:

The range

The rangeRis restricted by the 3ying time which must not exceed2R0=v. Since the times taken to 3y out and back areR=VoutandR=Vbackrespectively,Ris determined by R

VoutCR

VbackD2R0

v; c

15Cambridge University Press, 2006

Chapter 2 Velocity, acceleration and scalar angular velocity61 that is

RD2R0VoutVback

v .VoutCVback/: On substituting in the values that we have obtained forVoutandVback, we 2nd that the3ying rangein the direction whose true bearing from north is22is given by

RDR0.v20u2/

v.v20u2sin222/1=2:

This range takes its maximum value when22D 61

231(that is, in directions at

right angles to the wind). In these cases, the range is R 022
10u2 v223 1=2 ; which is still less than the range in still air. c15Cambridge University Press, 2006

Chapter Three

Newton's laws of motion

and the law of gravitation c

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Chapter 3 Newton's laws of motion and the law of gravitation63

Problem 3.1

Four particles, each of massm, are situated at the vertices of a regular tetrahedron of sidea. Find the gravitational force exerted on any one of the particles by the other three. Three uniform rigid spheres of massMand radiusaare placed on a horizontal table and are pressed together so that their centres are at the verticesof an equilateral triangle. A fourth uniform rigid sphere of massMand radiusais placed on top of the other three so that all four spheres are in contact with each other. Find the gravitational force exerted on the upper sphere by the threelower ones.

FIGURE 3.1The particlesA,B CandD

eachhavemassmandarelocatedatthever- tices of a regular tetrahedron. A B CD M Nα

Solution

By the law of gravitation, each of the particlesB,CandDattracts the particleA with a force of magnitude m 2G a2: By symmetry, the resultant forceFpoints in the directionANand its magnitudeF can befoundby summingthecomponentsof thecontributingforcesinthisdirection. Hence

FD322m2Gcos13

a223 ; where13is the angle shown in Figure 3.1. The angle13can be found by elementary c

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Chapter 3 Newton's laws of motion and the law of gravitation64 geometry. By an application of Pythagoras, BM

2DBC20CM2Da2020

1 2a21 2D3 4a2; and, sinceNis the centroid of the triangleBCD, BND2 3BMDa p 3:

A second application of Pythagoras then gives

AN

2DAB20BN2Da2022a

p 323
2 D 2 3a2: and so cos13DAN ABDq 2 3: Hence theresultant forceexerted on particleAby the particlesB,CandD acts in the directionANand has magnitude FDp 6m2G a2: Since the four balls are spherically symmetric masses, their gravitational effect is the same as if each one were replaced by a particle of massMat its centre. These four `particles' form a regular tetrahedron of side2a. Hence, the gravitational force exerted on the upper ball by the three lower ones acts vertically downwards and has magnitude p 6m2G 4a2: c15Cambridge University Press, 2006 Chapter 3 Newton's laws of motion and the law of gravitation65

Problem 3.2

Eight particles, each of massm, are situated at the corners of a cube of sidea. Find the gravitational force exerted on any one of the particles by the other seven. Deduce the total gravitational force exerted on the four particles lying on one face of the cube by the four particles lying on the opposite face.

FIGURE 3.2The particlesA, ...,Heach

have massmand are located at the vertices of a cube. G A C D E F HH

ααβ

B

Solution

By the law of gravitation, each of the particlesB,C, ...,Hattracts the particleA with a force of magnitude m 2G R2; whereRis the distance between them. By symmetry, the resultant forceFpointsin the directionAGand so its magnitudeFcan be found by summing the components of the contributing forces in this direction. Hence

FD322m2G

a223 cos13C3 m2G 0p 2a12! cos14C m2G 0p 3a12! where13.14are the angles shown in Figure 3.2. By using elementary geometry, it is c

15Cambridge University Press, 2006

Chapter 3 Newton's laws of motion and the law of gravitation66 easily found that cos13Dr 1

3;and cos14Dr

2 3: Hence theresultant forceexerted on particleAby the particlesB,C, ...Hacts in the directionAGand has magnitude FDm2G a2 p 3Cr 3 2C1 3! : By symmetry, the resultant force exerted by the particlesE,F,G Hon the particlesA,B,C,Dpoints in the directionAEand its magnitudeF0can be found by summing the components of the contributing forces in thisdirection. Now the resultant force thatallthe other particles exert on particleAhas magnitudeFand the component ofthisforce in the directionAEisFcos13. Since the the forces that B,C,Dexert onAhave zero component in this direction,Fcos13is equal to the resultant force thatE,F,G,Hexert onA, resolved in the directionAE. It follows that the resultant force exerted by the particlesE,F,G Hon the particlesA,B,C,

Dpoints in the directionAEand has magnitude

F

0D4Fcos13D4m2G

a222 1C1 p 2C1 3p 323
: c15Cambridge University Press, 2006 Chapter 3 Newton's laws of motion and the law of gravitation67

Problem 3.3

A uniform rod of massMand length2alies along the intervalOE0a;a?of thex-axis and a particle of massmis situated at the pointxDx0. Find the gravitational force exerted by the rod on the particle. Two uniform rods, each of massMand length2a, lie along the intervalsOE0a;a? andOEb0a;bCa?of thex-axis, so that their centres are a distancebapart (b>2a). Find the gravitational forces that the rods exert upon each other. FP xdx {O x ?x

FIGURE 3.3The rod and the particle.

Solution

Consider the elementOEx;xCdx?of the rod which has massM dx=2aand exerts an attractive force of magnitude m.M dx=2a/G .x00x/2 onP(see Figure 3.3). We must now sum these contributions but, since the rod is a continuous distribution of mass, this sum becomes an integral. Theresultant force F

1exerted by the rod is therefore given by

F 1DmMG 2aZ a 0adx .x00x/2 D mMG 2a24 1 x00x25 xDa xD0a D mMG 2a22 1 x00a01 x0Ca23 D mMG x020a2: c

15Cambridge University Press, 2006

Chapter 3 Newton's laws of motion and the law of gravitation68 Considerthe elementOEx0;x0Cdx0?of thesecond rod which has massMdx0=2a. The force exerted by the 2rst rod on this element is towardsOand of magnitude .M dx=2a/MG .x00x/2: We must now sum these contributions but, since the second rodis also a continuous distribution of mass, this sum also becomes an integral. Theresultant forceF2that each rod exerts on the other is therefore given by F 2DM2G 2aZ bCa b0adx 0 x020a2 D M2G 4a2Z bCa b0a22 1 x00a01 x0Ca23 dx 0 D M2G 4a2h ln.x00a/0ln.x0Ca/i bCa b0a D M2G

4a2ln22b2

b204a223 : c15Cambridge University Press, 2006 Chapter 3 Newton's laws of motion and the law of gravitation69

Problem 3.4

A uniform rigid disk has massMand radiusa, and a uniform rigid rod has mass M

0and lengthb. The rod is placed along the axis of symmetry of the disk with one

end in contact with the disk. Find the forces necessary to pull the disk and rod apart. [Hint.Make use of the solution in the `disk' example.]

Solution

Let the axisOzbe perpendicular to the disk withOat the centre, and suppose that the rod occupies the interval024z24b. Consider the elementOEz;zCdz?of the rod which has massM0dz=b. The force exerted by the disk on thiselement acts towards

Oand has magnitude

2MM 0G a2b 10z

0z2Ca211=2!

dz; on using the result of Example 3.6. We must now sum these contributions but, since the rod is a continuous distribution of mass, this sum becomes an integral. The resultant forceFthat the disk exerts on the rod is therefore given by

FD2MM0G

a2bZ b 0 10z

0z2Ca211=2!

dz D 2MM0G a2b24 z020 z2Ca2211=225b0 D 2MM0G a2b20 aCb00a2Cb211=221 : This is theforce neededto pull the rod and the disk apart. c15Cambridge University Press, 2006 Chapter 3 Newton's laws of motion and the law of gravitation70

Problem 3.5

Show that the gravitational force exerted on a particleinsidea hollow symmetric sphere is zero. [Hint.The proof is the same as for a particleoutsidea symmetric sphere, except in one detail.]

FIGURE 3.4The particlePis inside a hol-

low gravitating sphere. O P brdv RR θ

Solution

The only difference that occurs when the particlePisinsidea hollow sphere is that the polar coordinateris now alwaysgreaterthan the distanceb(see Figure 3.4). The range of the variableRis thenr0b24R24rCband the integral overR(see the book , p. 66) is replaced by Z rCb r0b22

1Cb20r2

R223 dRDZ rCb r0b22

10r20b2

R223 dR D 24

RCr20b2

R25 RDrCb RDr0b D20 .rCb/C.r0b/21 020 .r0b/C.rCb/21 D0: (WhenPis outside the sphere, the corresponding value is4r.) It follows that the gravitational force exerted on a particle inside a hollow sphere is zero. c15Cambridge University Press, 2006 Chapter 3 Newton's laws of motion and the law of gravitation71

Problem 3.6

A narrow hole is drilled through the centre of auniformsphere of massMand radiusa. Find the gravitational force exerted on a particle of massmwhich is inside the hole at a distancerfrom the centre.

Solution

In this solution, we will neglect the material that was removed from the sphere to make the hole. When the particlePis a distancerfrom the centre, it is (i) exterior to a uniformsolidsphere of radiusrand mass.r=a/3M, and (ii) interior to a uniformhollowsphere with inner radiusrand outer radiusa. The solid sphere exerts the same force as that of a particle ofmass.r=a/3M located at the centre, while (from Problem 3.5) the hollow sphere exerts no resultant force. Theresultant forceexerted onPwhen it is inside the hole is therefore m

0.r=a/3M1G

r2C0D22mMG a323 r: This force acts towards the centre and is proportional tor. Hence, in the absence of any other forces (such as air resistance or boiling lava!),Pwill performsimple harmonic oscillationsin the hole. Note that this result applies only to auniform sphere. c15Cambridge University Press, 2006 Chapter 3 Newton's laws of motion and the law of gravitation72

Problem 3.7

A symmetric sphere, of radiusaand massM, has its centre a distanceb(b>a) froman in2niteplanecontaininga uniformdistributionof mass33per unit area. Find the gravitational force exerted on the sphere. b P R θ r dA

OFα

P FIGURE 3.5The particlePis distancebfrom a unform gravi- tating plane.

Solution

This is basically the same problem as that in Example 3.6 where the disk now has in2nite radius (and therefore in2nite mass). Despite this,the gravitational force it exerts onPis still 2nite. The force on the sphere is the same as that on a particlePof massMlocated at its centre. Consider the elementdAof the plane shown in Figure 3.5. This element has mass33dAand attracts the sphere with a force whose component perpendicular to the plane is

22M.33dA/G

R223 cos13: We must now sum these contributions but, since the plane is a continuous distrib- ution of mass, this sum becomes an integral. Theresultant forceFthat the plane exerts onPis therefore given by

FDM33GZ

Pcos13

R2dA; c

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Chapter 3 Newton's laws of motion and the law of gravitation73 wherePis the region occupied by the mass distribution. This integral is most easily evaluated using polar coordinates. In this casedAD .dr/.r d22/Dr dr d22, and the integrand becomes cos13

R2DRcos13

R3Db .r2Cb2/3=2; wherebis the distance ofPfrom the plane. The ranges of integration forr,22are

024r24 1and0242224231. We thus obtain

FDbM33GZ

rD1 rD0Z

22D231

22D022

1 .r2Cb2/3=223 r dr d22: Since the integrand is independent of22, the22-integration is trivial leaving

FD231bM33GZ

rD1 rD0r dr .r2Cb2/3=2

D231bM33Gh

0.r2Cb2/01=2irD1

rD0

D 0231bM33G22

001 b23

D231M33G:

This is thegravitationalforceexerted on the sphere. It seems strange that this force is independent of the distanceb, but this is because the attracting mass distribution is anin2nite plane. c15Cambridge University Press, 2006 Chapter 3 Newton's laws of motion and the law of gravitation74

Problem 3.83

Two uniform rigid hemispheres, each of massMand radiusaare placed in contact with each other so as to form a complete sphere. Find the forces necessary to pull the hemispheres apart.

FIGURE 3.6The solid hemispheresHCand

H

0attract each other.

z r dv+H H -O θ

Solution

Let the two hemispheresHCandH0be as shown in Figure 3.6. We wish to calculate the forceFthatHCexerts onH0. SinceHCexerts no resultant force uponitself,Fis equal to the force thatHCexerts on thewhole sphereof mass2M. It is tempting to say that this is equal to the force thatHCexerts on a particle of mass2Mlocated atO. However, this is not true since the mass ofHCliesinside the whole sphere. We must therefore proceed in the same manner as in Problem 3.6. Consider the volume elementdvofHC. This has mass32dv, where32is the constant density. This element attracts the whole sphere with a force which is the same as if the sphere were replaced by a particle of mass.r=a/3.2M/at its centre.

The component of this force in thez-direction is

222M.r=a/3.32dv/G

r223 cos22: We must now sum these contributions but, sinceHCis a continuous distribution of c

15Cambridge University Press, 2006

Chapter 3 Newton's laws of motion and the law of gravitation75 mass, this sum becomes an integral. Theresultant forceFis therefore given by

FD2M32G

a3Z H

Crcos22dv:

This integral is most easily evaluated using spherical polar coordinatesr,22,

36. In this casedvD.dr/.r d22/.rsin22d36/Dr2sin22dr d22d36, and the integral

becomes

FD2M32G

a3Z rDa rD0Z

22D31=2

22D0Z

36D231

36D0r3sin22cos22dr d22d36

D

431M32G

a322 ZrDa rD0r3dr23 Z22D31=2

22D0sin22cos22d22!

D

431M32G

a320 1

4a421h

0 1

4cos222i

31=2
0 D 1

231M32aG:

Finally, on using the relationMD2

331a332, we 2nd that theresultant forcethat

H

Cexerts onH0is

FD3M2G

4a2: This is theforce neededto pull the hemispheres apart. c15Cambridge University Press, 2006

Chapter Four

Problems in particle dynamics

c

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Chapter 4 Problems in particle dynamics77

Problem 4.1

Two identical blocks each of massMare connected by a light inextensible string and can move on the surface of aroughhorizontal table. The blocks are being towed at constant speed in a straight line by a rope attached to one of them. The tension in the tow rope isT0. What is the tension in the connecting string? The tension inthe tow rope is suddenly increased to4T0. What is the instantaneous acceleration of the blocks and what is the instantaneous tension in the connecting string? TT

FFMMT0

FIGURE 4.1The two blocks are linked together and towed by a forceT0.

Solution

Let the tension in the connecting string beTand the frictional force acting on each block beF(see Figure 4.1). The two frictional forces are equal because the blocks are physically identical and are travelling at the same speed. (i) Suppose 2rst that the whole system is moving atconstant speed. Then the blocks have zero acceleration and theirequations of motionare therefore T

00T0FD0;

T0FD0:

Hence TD1

2T0andFD1

2T0:

Thetensionin the connecting string is therefore1

2T0. (ii) Suppose that the tension in the tow rope is increased to4T0and that the system then has accelerationaat timet. Theequations of motionfor the two blocks then become 4T

00T0FDMa;

T0FDMa:

At any instant, the two blocks have the same speed and so the two frictional forces do remainequal. However, we have no right to suppose that, as the c

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Chapter 4 Problems in particle dynamics78

speed of the system increases, the frictional forces will remainconstant. But, at the instant that the tension in the tow rope is increased, the speed is (as yet) unchanged and it will still be true thatFD1

2T0. Thus,at this instant, we

have 7

2T00TDMa;

T01

2T0DMa:

Hence

TD2T0andaD3T0

2M: Hence theinstantaneousacceleration of the blocks is3T0=2Mand thein- stantaneoustension in the connecting string is2T0. (If it happens to be true thatFis independent of the speed of the blocks, these values will remain constant in the subsequent motion.) c15Cambridge University Press, 2006

Chapter 4 Problems in particle dynamics79

Problem 4.2

A body of massMis suspended from a 2xed pointOby an inextensible uniform rope of massmand lengthb. Find the tension in the rope at a distancezbelow O. The point of support now begins to rise with acceleration2g. What now

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