CHAPTER 4 GENERAL MOTION OF A PARTICLE IN THREE DIMENSIONS --------------------------------------------------------------------------------------------------------- Note to instructors there is a tpo in equation 4..4. The rane of the projectile is v sin R NOT v sin... --------------------------------------------------------------------------------------------------------- 4. (a) ˆ V ˆ V F V i j kˆ V z F c( iz ˆ + ˆjz + k ˆ ) (b) F V iˆ ˆjβ kˆ γz + β+ γz (c) F V ce iˆ + ˆjβ + kˆ γ (d) V V V F V eˆ ˆ ˆ r eθ eφ r r θ rsinθ φ n F eˆ cnr r 4. (a) (b) (c) (d) iˆ ˆj kˆ F z z iˆ ˆj kˆ F z kˆ ( ) z iˆ ˆj kˆ F kˆ ( ) z z conservative conservative conservative
eˆ ˆ ˆ r eθr eφrsinθ θ θ φ F r sin r kr n conservative 4. (a) (b) iˆ ˆj kˆ F k z c c c z ( c ) iˆ ˆj kˆ F z z cz ˆ c ˆ ˆ cz z i j k + + + c c also cz z + implies that c as it must 4.4 (a) constant V,, z + mv at the oriin E + m v E at (,,) E + β + γ + mv mv
(b) v ( + ) (c) v m β + γ m β γ + + V m F m V m β V mz γ z z v v + β + γ m v v m + β + γ 4.5 (a) F i ˆ + ˆj on the path : dr id ˆ + ˆjd (,) F dr F d + F d d + d, on the path alon the -ais: and on the line : (,) F dr F (,) d+ F d F is conservative. dr id ˆ dr ˆjd (b) F i ˆ ˆj on the path : (,) F dr F d + F d d d, and, with F dr d d,, on the -ais: and, with on the line : on the -ais,, (,) F dr F d d (,) F dr F dr F d d,,
and, with, (,) on this path F dr + (,) F is not conservative. F dr d, 4.6 From Eample.., V ( z) V z m re r z ( + ) z mre + re From Appendi D, V z V z e + + + z z mre + + re re mz mre + mz + r With mr e an additive constant, z V ( z ) mz re F V kˆ V( z ) z ˆ z km + z re re ˆ z F km re m F, m F dz z mz m dz re h z zdz dz v z re h v h z re rv e z h r eh+ e z mz m re 4
re z h r rev e r r v h r e e z From Appendi D, ( ) r r v v h + + 4r + e ( hz r), e + + + 8 4 e e z z v z v z h + re From Eample.., h And with ( ) +, e v v r e v v h + r e 4.7 For a point on the rim measured from the center of the wheel: r ib ˆ cosθ ˆjbsin θ vt θ ωt, so r îv sinθ ˆ jvcosθ b v iˆv sinθ ˆjv cosθ Relative to the round, For a particle of mud leavin the rim: bsinθ and v v cosθ So v v t v cosθ t and b sinθ vt cosθ t At maimum heiht, v : v cosθ t vcosθ vcosθ hbsinθ v cosθ v h bsinθ + Maimum h occurs for cos θ dh b cosθ dθ v cosθ sin θ 5
b sinθ v 4 v b cos θ sin θ 4 v b v b b v hma + + v v v Measured from the round, b v h ma b+ + v 4 b The mud leaves the wheel at θ sin v 4.8 Rcosφ and ( cos ) R cosφ so t v cos Rsinφ v t v t vt t v t t and ( sin ) Rcosφ Rcosφ Rsinφ ( v sin) vcos vcos R cos φ sinφ tancosφ v cos v cos v cos φ ( tan cosφ sinφ) ( sincosφ cossin ) R cos φ cos φ From Appendi B, sin ( θ + φ) sinθ cosφ + cosθsinφ v cos R sin cos φ ( φ ) dr v φ φ) d cos φ + R is a maimum for sin sin cos cos( Implies that cos ( φ) ( φ) From appendi B, so cos( φ) R cos sin sin cos θ + φ cosθ cosφ sinθ sinφ π π φ φ + 4 v π φ π φ cos + sin cos φ 4 4 ma φ 6
4.9 π φ π π φ π φ Now sin cos cos 4 + 4 4 v π φ Rma cos + cos φ 4 Aain usin Appendi B, cos θ cos θ sin θ cos θ v π v Rma cos φ π + cos φ + cos + φ + cos φ Usin cos π + θ sin θ, v Rma sinφ sin φ R ma v ( + sinφ ) (a) Here we note that the projectile is launched downhill towards the taret, which is located a distance h below the cannon alon a line at an anle φ below the horizon. is the anle of projection that ields maimum rane, R ma. We can use the results from problem 4.8 for this problem. We simpl have to replace the anle φ in the above φ result with the anle -φ, to account for the downhill h slope. Thus, we et for the downhill rane R ma v cos sin ( + ϕ) R cos ϕ The maimum rane and the anle is are obtained from the problem above aain b v ( + sinϕ ) π replacin φ with the anle -φ Rma and ϕ. cos ϕ h v ( + sinϕ ) v We can now calculate Rma sinϕ cos ϕ sinϕ h h Solvin for sinϕ sin ϕ + v v π But, from the above sinϕ sin cos sin h h Thus sin + v v 7
h h + h + v sin csc v v Finall h csc + v (b) h h h Solvin for R ma Rma sinϕ sin csc Substitutin for csc and solvin v h Rma + v 4. We can aain use the results of problem 4.8. The maimum slope rane from problem 4.8 is iven b v h Rma + sinϕ sinϕ Solvin for sinϕ h h sinϕ v v Thus cosϕ ma Rma cosϕ h sinϕ We can calculate cosϕ from the above relation for sinϕ h h v v cosϕ ( sin ϕ) Insertin the results for sinϕ and cosϕ into the above cosϕ v h ma h sinϕ v 4. We can simplif this problems somewhat b notin that the trajector is smmetric about a vertical line that passes throuh the hihest point of the trajector. Thus we have the followin picture 8
z z ma v h h δ R We have reversed the trajector so that h ( 9.8 ft), and, the heiht and rane within which Micke can catch the ball represent the startin point of the trajector. h (.8 ft) is the heiht of the ball when Micke strikes it at home plate. δ is the distance behind home plate where the ball would be hpotheticall launched at some anle to achieve the total rane R. (8 ft) is the distance the ball actuall would travel from home plate if not cauht b Micke. (Note, because of the smmetr, v is the speed of the ball when it strikes the round also at the same anle at which was launched. We will calculate the value of assumin a time-reversed trajector!) v sin v sincos () The rane of the ball R () The maimum heiht R ma tan R z v cos () The heiht at h tan v cos tan From () and insertin this into () ives v cos R R R R zma tan tan tan 4 4 4zma Thus, R and insertin this epression and the first previousl derived into () tan (4) h tan ( tan ) 4z ma Let u tan and we obtain the followin quadratic 4 + 4 u zmau zmah and solvin for u u z ma ± h zma and lettin h ε z ma, we et 9
u z ε h ma ( ε ) or u z z.475.9z.9zm a Thus, tan.8 9.4 Now solve for usin a relation identical to (4) h ma ma ma tan tan 4z ma. This result is the correct one Aain we obtain a quadratic epression for u tan which we solve as before. This time, thouh, the first result for u is the correct one to use u z ε h and we obtain ma h.9 ft tan 4. The and z positions of the ball vs. time are vt cos θ z vtcos θ sinθ t Since v v cos θ v The ho rizontal rane is R dr The maimum rane occurs @ dθ cos θ sin θ dr v cos θcos θ cos θ sin θ sin θ dθ Thus, cos θ cos θ cos θ sin θ sin θ Usin the identities: cos θ + cos θ and sin θ sinθ cosθ We et : + cosθ cos θ sinθ sinθ cosθ cos θ cos ( ) ( ) + θ )( cos θ cosθ ) or ( cos Thus cosθ, cosθ ( ± ) 6 π Onl the positive root applies for the θ -rane: θ cosθ ( + ).7676 θ 9 5 6 v 5ms Thus (b) for Rma 55.4 m @ θ 9 5 θ
dz (a) The maimum heiht occurs at dt vcos θsinθ T or at vcos θsinθ T v or H cos θ sin θ maimum at fied θ dh The maimum possible heiht occurs @ d dh v cos θsinθcosθ cos θsin θsin θ d Usin the above trionometric identities, we et ( + cosθ ) sin cos sin sin sin ( cos θ θ θ θ θ θ) sinθ + cosθ cosθ or There are -roots: sinθ, cosθ, cosθ The first two roots ive minimum heihts; the last ives the maimum Thus, Hma 8.9m @ θ cos 7 4. The trajector of the shell is iven b Eq. 4.. with r replacin z z r r where r s vco θ z vsinθ r r Thus, r z rtanθ sec θ v sec θ + tan θ Since We have: r r tan θ tan r θ + z+ v v (r,z) are taret coordinates. The above equation ields two possible roots: 4 tanθ v v zv r ± r The roots are onl real if 4 v zv r The critical surface is therefore: 4 v zv r 4.4 If the velocit vector, of manitude s, makes an anle θ with the z-ais, and its
projection on the -plane make an anle φ with the -ais: s sinθ cosφ, and F F sinθ cosφ m s sinθ sinφ, and F F sinθ sinφ m z scosθ, and Fz m+ Fr cosθ mz Since F ( r cs c + + z ), the differential equations of motion are not separable. m cs z sinθ cosφ cs d d ds d m m ms cs dt ds dt ds s θ d c c ds γds, where γ m m ln ln ln γ s e γ s φ Similarl e γ s r r z γma ma 4.5 From eqn 4..6, ln γ + + γ γ γ u u From Appendi D: ln ( u) u for u < γ ma γma γ ma γ ma 4 ln + terms in γ z ma ma m a ma ma + γ γ γ + terms in γ z ma + ma γ γ 9 z ma ± + 4γ 6γ γ 6 ma z γ ± + 4γ 4γ Since ma >, the + sin is used. From Appendi D: 6γz 8γz 6γz + + + γ 8 terms in z 8γ z ma + + + terms in γ 4γ 4γ
For z 8z ma γ + z v sin and z v sin : v sin 4v sin sin ma γ + 4.6 Acos( ωt+ ), Aω sin ( ωt+ ) from, from A, Acosωt Bcos( ωt+ β ), ωbsin ( ωt+ β) kb k + m with k 4A, ω A and ω : m B 6 + ( A 9ω A ) 5A ω B 5A Then 4A 5Acos β and ω A 5ωAsin β 4 β 5 5 5Acos ωt6.9 cos sin 6.9 Since maimum and displacements are ± A and ± 5A, respectivel, the motion takes place entirel with in a rectanle of dime nsion A and A. β 6.9 6.9 AB cos From eqn 4.4.5, tan ψ A B 4 ( A)( 5A) cos( 6.9 ) tan ψ 5 A 5A 4 ψ tan 9. 4.7 V m F k π m k A cos t A cos( t + ) m + π
V m 4π m ( π β ) Bcos t+ V mz 9π mz z z Ccos π t+ γ π Since z at t, β γ π Acos πt Asinπt Aπ cosπt Since + + and z v z, v Aπ v A π v sinπt π Bsin π t, Bπ cos π t v π B v B π v sin π t π z Csin π t, z Cπ cosπ t v z Cπ v C π v z sin π t π Since ω π, t min ω π, and ω π the ball does retrace its path. z π π n n π n ω ω ω The minimum time occurs at n, n, n. z 4
t min π π AB cos 4.8 Equation 4.4.5 is tan ψ A B Transformin the coordinate aes z to the new aes z b a rotation about the z-ais throuh an anle ψ iven, from Section.8: cosψ + sinψ, sinψ + cosψ or, cosψ sinψ, and sinψ + cosψ cos From eqn. 4.4.: + sin A AB B Substitutin: cos ψ cosψ sinψ + sin ψ A cos cosψ sinψ + ( cos ψ sin ψ) cosψ sin AB ψ + ( sin ψ + cosψ sinψ + cos ψ) sin B For to be a major or minor ais of the ellipse, the coefficient of must vanish. cos ψ sin ψ cos cos sin ( cos sin ) ψ ψ ψ ψ + A AB B From Appendi B, cosψ sinψ sin ψ and cos ψ sin ψ cos ψ sinψ cos cosψ sinψ + A AB B tan ψ cos B A AB AB cos tan ψ A B 4.9 Shown below is a face-centered cubic lattice. Each atom in the lattice is centered within a cube on whose 6 faces lies another adjacent atom. Thus each atom is surrounded b 6 nearest neihbors at a distance d. We nelect the influence of atoms that lie at further distances. Thus, the potential ener of the central atom can be approimated as 6 V cr i i d 5
r d + + z z + + ( + + + ) + r d d z d d d n From Appendi D, ( + ) + n+ n( n ) + 4. z r d + + + + d d + + z + + z + + terms in d d d d 4 + ( + + ) + + r d z d d 4 d r d + ( + + z ) + + d d d ( ) + + + + + + r d z d d z + + z d + + r d d r d ( + + z ) + + d d d r + r d ( + + z ) + ( + ) d d Similarl: r + r4 d + + z + + d d r 5 r6 d ( z ) z + + + + + d d 6 V cd + + z + + + + z d d d ( ) 6cd + cd + + z V A+ B + + z d + terms in d 6
îv z ˆkB ĵe qb m F qe qb qe qb + m F q( E+ v B) v B ( i ˆ+ ˆj + kz ˆ ) kb ˆ ib ˆ ˆjB F iqb ˆ + ˆjq EB m F qb qb m qe qb qb ee eb eb + m m m m m m ee eb + ω + ω, ω m m ee Acos + ω + ( ωt+ θ ) ω m Aω sin ( ωt+ θ ), so θ ee, so A ω m ω + ee a( cosωt), a ω m ω + qb + ω ωa( cosωt) m ωa + ωa ωt ( ) cos ( ω ) sin a t+ a ωt asinωt+ bt, b ωa mz F z z zt + z 4. b mv + mqh m v b h b mv h Fr mcosθ + R b 7
h cosθ b h mv m m R m h b h h b b b b b the particle leaves the side of the sphere when R b h, i.e., b above the central plane 4. mv mh + at the bottom of the loop, h b h b v so mv mb, v b mv Fr m+ R b mv R m + m + m m b 4. From the equation for the ener as a function of s in Eample 4.6., m E ms + 4A s, s is underoin harmonic motion with: " k" ω m 4A A Since s 4Asinφ, φ increases b π radians durin the time interval: π A T π ω For ccloidal motion, and z are functions of φ so the undero a complete ccle ever time φ chanes b π. Therefore, the period for the ccloidal motion is one-half the period for s. A T T π ------------------------------------------------------------------------------------------------------ 8