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Pm#&IG K92$).a4.2h#*h'$*+ ."#>Jq2)C'$^4 I 23 AB u U c t = I =R4 $j''()."#>J$).a4*y 7,$$"m 1h#& $>p#.X#$I Cõu 7: Cho một bán cầu đặc đồng chất, khối lợng m, bán kính R, tâm O. 1. Chứng minh rằng khối tâm G của bán cầu cách tâm O của nó một đoạn là d = 3R/8. 2. Đặt bán cầu trên mặt phẳng nằm ngang. Đẩy bán cầu sao cho trục đối xứng của nó nghiêng một góc nhỏ so với phơng thẳng đứng rồi buông nhẹ cho dao động (Hình 1). Cho rằng bán cầu không trợt trên mặt phẳng này và ma sát lăn không đáng kể. Hãy tìm chu kì dao động của bán cầu. Cõu 8$2'g$"&/*'E*+ Ke #G#&'E,$-8P_#&*.P_'=4+'K923_ !;C,$H#&!~#Ko c4)1:#&1d'*+ .a4!;C=4+''-.X#$K92.R*\I @:#&1d'.P_' 1f2K92*+ 8:q2'E.+'j#&,I][c4)$-8P_#& '()8:q21:#&1d'K9'()!;C#-Iu*+ $Q.R*#92.EKe #G#& =o .a4'$X4 >'!7#&'()*+ 8i' r ,$H#&.p#$P$6#$KO )I6*c4~#&.PQ#&*9Ke * P_'K9,$2Z#& $Q&)#,R u8t' Ke =o .a4'$X4 >'!7#&'()8i' r .#8t'Ke !u#&8h8a# $j#$0 =I4!;C,$H#&'-.X#$m\*9#-KL*+ Ke ,$-8P_#&A}*B ~Cq>'.X#$.+8L#'()8i'.R3)4.EKe !)2.+#&.^4$:) Cõu 9E*h'$."##$P$6#$I 7."# .P_' ?'$."#.#$"4."# $ 7."# .P_' ?'$.`#.#$"4."# $ A } BI4+# !;C $4a#'Z*'E$"3- i'Z*I6*=R4 $j''PQ#&.+!:#&."# 12#&*h'$3)4,$.E#&,$2>nI Cõu 10 : $4>#$3>#&.g#3o''E=PL'3E#& FMà*K92') H '()*+ =92c4)#&."#In$.G K92)#H K9') H '() =92c4)#& ."##9C*+ $"4."# $ \n F $6!:#&c4)#&."#=o .a4 1" `4I$2$D#&3-8z#&$FvvNI YM 3 -'.+>#$3>#& 12#&'$;#,$H#&'FYI *l3,$-8P_#&f8f' 12#* f F I Y ,&.+8L#."# ?'$'()f8f' 12#fFvI I I ?#$'H#& $2> '(),*82h!S#&89*') - I I4 $)C=j'qh =D#&=j'qh Fà*./#& $Q&V#&4C`#$"4."# $&V))#H K9') H 1`# $6 -' .+8L##$0 '()f8f' 12#c4)#&."#,$ L)#H 'E&> 1X=D#&=)2#$`4x Cõu 11: 12#& $?#&$"*'();#&K^&)2 $2)>#$3>#&,$2Z#&'>'$&V)$),$f$sJ 89)F**,$2Z#& '>'$ u*G J$%#&$),$f.#*9#89UF*I I&4/#J$> 1)>#$3>#&.g#3o'= ,$2Z#&'>'$&V)K;#3>#&8`# J89'*I?#$=PL'3E#&>#$ 3>#&.g#3o'!2#&4/#J$> 1)I I&4/#J$> 1)>#$3>#& 1o#&'E=PL'3E#&#D* 12#&,$2Z#& uY à * ữ v à *I )IW>'.X#$KX 1?&a#K;# 14#& ;*#$0 *9 h.E#$V#&=j'qh.g#3o''()>#$3>#& 1o#&'$2K;#3>#& 1S#& #$)4I =IhKX 1? 1`#*9#'>'$K;# 14#& ;*'*'E#$V#&=j'qh.g#3o'#92'$2K;#3>#& 1S#&#$)4I < I m k m k r r M A …E' L84H#89MN #`#&E',$t'qh84H#891FY …n$ LKX 1? ),$t'qh L*G 17m• KL&E' L=D#& &$ In$.E )8E Jqt'KL*G 17IeC,$m #&29,$2Z#&< $6,$H#&'E )8E1),$[*G 17I †2!7#&.X#$8?$9*3-3#'$2 )*&>'< • )'E 3# 3# gh i OI J OI OJ = 12#&.E<• F@w &$ FMN w OI J Fƒ ‡1Fv I eC< F@ Y Pg#& i< F@ Y ĐÁP ÁN …L ) LF< ),$t'qh<•'$?#$89=>#,?#$'().PQ#& 1:##`# $%#&&E'KL*G 'a4 h•IU2.E ) <• 14C^# $%#&c4)*G 17 u.X#$84e ,$t'qh>#$3>#&# 3#F# 3#1 4C1)3#1FN 1FY …E'8E h•1),$[*G 'a4=D#&#`#&E'8"'$'() )8E32KL ) L<89 UF‡1FMN ‡Y FN 4.R*•mn 14#&.R*'4#& 1:#\] ),$t'qh L*G 17KL&E'1FY # 3#1F# 3#ˆ 3#ˆF ˆFFMN I …n$.E )8E32#&32#&KL ) L#`#&E'8"'$ 1" `4IR*mKX 1? I)'E < F<n )#1F@ )#Y F@ Y Y I Y…4&E' L*G 178L#$g#&E' L&L$h# $6>#$3>#&3OJ$Z#qh 29#J$a#,$H#&'E )3>#&8E1),$[*G 17I )'E3# &$ F 34C1) &$ FMN …n 84e#n$ )3>#& L*G J$%#&'(),$-KL&E' LMN '$‰'E )3>#&8E,$m*G 17#4.R* Lm 1`# .2h# I …)4,$Ku)K)'$h*Ke 'EKe# -'K $)#$'EKe# -'&E' ω I …]Z2 29#*H*f#.+#&8P_#& *K l F* l K… ω lm ⇒ K FK… ω l v AB …]Z2 29##z#&8P_#& *K F ω lm … *K ⇒ K F ω l …K AB uABK9AB l v Y =⇒ ω AYB †J!7#&.X#$8Š.+#&#z#& ω F\ *3 *)q Y Y A B M v v l ml mg l gl ⇔ =µ ϕ ⇒ ϕ = µ €$Pg#& 16#$!)2.+#& I 23A Bx A c t ω ϕ = + 12#&.E A l B K rad s m ω = = A B 23 A B 3# A B x cm Ac cm t v A cm ϕ ϕ π ϕ = − = − = = → → = = = eC I 23A BA Bx c t cm π = + …) $0C8:q2#r#Ncm '>'8a#'$‹#8`# J'>'$#$)4*+ '$4,6!2.E8:q2#r# 8a# $j h $Q.R* I t t T − = + KLt 2 89 $Q.R*8:q2#r#Ncm 8a# $jI …)q>'.X#$ $Q.R*8:q2#r#Ncm8a# $j$)35!7#&JJKf' gc4)C )'E,R u $Q.R*=)#.a4.#8t'8:q2#r#Ncm8a# $j $6Kf' gc4)C*+ &E' Œ I l Y N l YM OM t ω π π π = = − = N A B v t s π → = …U2.E $Q.R*8:q2#r#Ncm8a# $j89 N vƒ MI A B v v t s π π π = + = + t''E*)3> h ]'()Ke 8: q2=#!h#&*+ .2h# NA B mg l m K µ ∆ = = …) $0C'E$)]'()Ke J$7 $4+'K92'$^4'$4CR#.+#&'()Ke #4Ke . …n$'2#=d=:.P_',$2Z#&q*2*f#c4># ?#$'() $)#$K9'2#=dc4)#$'$- c4)C\89 A B Y Y Y I ml mx m l x= + = + …€$Pg#& 16#$'$4CR#.+#&'()'2#8o'89 A •B 3# 3# Y θ = θ θ d l I mg mgx dt )C A B •• • • 3# Y Y Y + θ + θ = − θ + ÷ l x m l x mxx mg …L'>'!)2.+#&#$[#E 1m $9#$ Y A B • • •• + θ θ θ + + = + + g x l xx l x l x …4'2#=d=:10 '$e* $63i $)C.pq 12#&*+ '$4,6!)2.+#&89,$H#&.>#&,R )=[c4)3-$h#& $j'() J$Pg#& 16#$K9K 8h A Y B •• A B + θ θ + = + g x l l x …U2.E a#3-&E''()!)2.+#&89 A Y B A B + ω = + g x l l x N Ž Ž Ž O C 1 C 2 x 3)#&J$Z8t'8:q2#r#N** $6]89=`# 1><AKX 1? B8t'Ke .3)#& 1>*9 8:q2&~#N** $6]89=`#J$Z<AKX 1? B …†J!7#&.#$84e =Z2 29##z#&8P_#& ) ?#$.P_'.+&Z* 2h.+'i'.h3)4 *y8a#c4)<89$D#&3-K9=D#& )q NA B m mg x m K µ ∆ = = …) -''()Ke .p'$^48a# $jMj#&KLKe .c4)] $f2'$^43)#& 1>8a# $j>J!7#&.X#$84e =Z2 29##z#&8P_#& ).P_' [ ] M )q )q )q )q A B A B A B A B A Y B A Y B m m m m mvKA K l mg A A x A x A x A x l µ ∆ − + = = + −∆ + − ∆ + − ∆ + − ∆ − ∆ M vNA l Bv m s→ = …+8"'$J$)'()$)3E#& h*+ .R*'>'$\]#$V#&.2h#! K9! 89 A B d d π π ϕ λ ∆ = − + KL Y YA B v cm f λ = = = …h89'i'.h&)2 $2)#4 A B A B M d d k d d k π π ϕ π λ λ ∆ = − + = → − = − $4+'\]#`# A B vwIIIwv M AB d d k AB k λ − < − = − < → = − 1`#.2h#\]'EY.R*'i'.h …h89'i' R4&)2 $2) A B A B A B M d d k d d k π π ϕ π λ λ ∆ = − + = + → − = + $4+'.2h#\] A B vwIIIwv M AB d d k AB k λ − < − = + < → = − 1`#.2h#\]'EY.R*'i' R4 …h.R* $4+'.2)#\]'>'$ 14#&.R**+ .2h#x'E$"4.PQ#&.'() $)3E#&89 d d x− = …R* $4+'.2h#\].j#&C`# $2Z*~# A B A BI M M d d x k x k λ λ − = = + → = + ABKL vwIIIwvk = − …U2.E )q *# Y Av BI ƒY„NA B M Y A BI Y„NA B M m x cm x cm = + = = + = …€$Pg#& 16#$!)2.+#& p#&$_J h'>'$\]#$V#&.2h#! K9! 89 I 23 A B I 23 A B A B M M M u c d d c t d d mm π π π π ω λ λ = − + + + + …).R* K9 .^4 $4+'*+ f8J#$e#\]89* `4.R*#`# AM BM AM BM b+ = + = 4C1)J !)2.+#&'() K9 89 I I 23 IY I 23 Y M M I I 23 IMN I 23 Y M M M M M M b u c c t u u b u c c t π π π π ω λ π π π π ω λ = + + + → = − = + + + h $Q.R* A B A B M M u mm u mm= → = − •n$ a#3- Nf Hz= ) $0C AM AB MB U U U= + '$j#& [U AB K4H#&J$)KLU MB #`#.2h#\],$H#& $R'$j) …@K9K6,$.EU AM K4H#&J$)U MB IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII …@K9'4+# $4a#'Z*K6,$.EU AM K4H#&J$)U MB IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII …'4+# $4a#'Z*K9 7."#K6,$.EU AM #&P_'J$)U MB IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII …'4+#'Z*'E."# 1m $4a#K9."# 1m $4a#@K6,$.E&E'8"'$J$)&V)U AB K9 U MB 89&E'#$d#IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII U2.E.2h#\]'E $R'$j)'4+#'Z*'E."# 1m $4a#1.+ i'Z*K9 7."#I •n$Z#z#&$+JW'$j) 7."#'$j)'4+#'Z*A1BI n$ Nf Hz= ) $0C w A YB C MB r L L C L C U V U U U U U Z Z= = + = → < → < !b $0C,$ z#& a#3-8`#c4>NHz $6Z L z#&Z C &Z*.#8t'Z L = Z C $6!:#&."# $"4!7#&*L.h 'i'.h#&$•)89 z#& a#3-8`#c4>NHz $6I z#& 1>& I U2.E,$Z#z#	C=X82hI •n$Z#z#&$+JW'$j)'4+#'Z*A1BK9$+J'$j) 7I …n$ Nf Hz= )'E$" Y Y Y A B C C AM r L L r AB r L C U V U V U U U U V U V U U U U = = = + = → = = = + − = Y N Y lN Y A B N Y N Y l A B NA B N C L Z C F Z L H r r π π − = Ω = → = Ω → = = Ω = Ω …Ub $0C8t' Nf Hz= $6qZC1)'+#&$Pm#& *)q Fl@#`##4 z#&{8`#c4>NHz $6&Z* $2Z*~#& I eC$+JW'$j)'4+#'Z*'E NA Bw N Y l A Br L H π = Ω = K9$+J'$j) 7 Y lN YA BC F − = …h F DUu AB →= *m':#D 2 .E#& w UCqUuuuu MMBAM =→====→ …L MB uTt Ml << &Z* u →U #`#D 1 *m 7 J$E#&."#c4) K9#&4/##$P#&,$H#&J$E#&."# c4)D 1 .P_' )'E UCqq =+− A„B …h FlM =+→= MBAMAB uuu A‚Bw, $_JABK9AB $6 h FlM ).P_' > + = < + −= CC UC u CC UC u Mb AM AƒB#`#$).H .^4=X'0* …)4 FlMm'$.+p#.X#$$).H .^4=X'0* )'E!:#&c4)$) 789./#&#$0 #`# B3#AB3#ABAB3#A B'23AB'23A l l tUCCtICCtUCCqCqC tUCCuCCuCCtUuu MBAMMBAM ωωϕωωω ωω −=++−⇔−=+→ =+→=+ 23 3# 23 C C U I q q c t a C C U C C i t q q c t a C C = = + + = = + + = I'23 AB I'23 AM MB C U q a u t C C C C C U q a u t C C C C = = + + = = + + h FlMAB $[)*~#AB#`# ).P_' = + = + C a CC UC C a CC UC $)CK92AB'$2 ) ( ) + + + = + = '23 '23I CC UC t CC UC u t CC UC u Mb AM A ) $0C w AM MB u u t #`#,$p#.X#$$) .H .^4=X'0*B 1. Do đối xứng, G nằm trên trục đối xứng Ox. Chia bán cầu thành nhiều lớp mỏng dày dx nhỏ. Một lớp ở điểm có toạ độ x= R sin , dày dx= Rcos.d có khối lợng dm = (Rcos ) 2 dx với Y @ Y * = nên: * !3#'23@ * q!* q l YM * == d = @Y *M @ '23 *M @ q M l M M = = = (đpcm) 2. Xét chuyển động quay quanh tiếp điểm M: gọi là góc hợp bởi OG và đờng thẳng đứng - mgd = I M . (1) biến thiên điều hoà với = *&! I O , I G , I M là các mômen quán tính đối với các trục quay song song qua O,G,M. Mô men quán tính đối với bán cầu là: I O = *@ N ; I O = I G + md 2 I M = I G + m( MG) 2 . Vì nhỏ nên ta coi MG = R-d I M = *@ N +m(R 2 2Rd) = *@ Y = @v &N *&! = T = &N @v e ';#=D#&,$'$P) >'!7#&8i'*&F, o l $d# 17'<q $%#&.j#& u 1`#q4-#&I< 1S#&KL]*L,$'E8i' >'!7#&I A B C 1 C 2 M D 1 D 2 H.2 Hình 2 < I < < q q Hình 1 !q h]*L•…€ 2 2 8 q , ∆ + FAKLq 2 89,$2Z#&'>'$&V)]*L32KL]'’B n$Ke 'E8.+q8:q2&~# 2 2 8 q∆ + …q •…€ 2 2 8 q q , ∆ + + F*qˆˆ ⇒ qˆˆ… , M* qF eCKe UKLJ$Pg#& 16#$qF\'23A ω + ϕ B 12#&.E , M* ω = $PKeC'$4,6!)2.+#&'()Ke F M* , π I$Q&)# u8t' >'!7#&8i'.#,$Ke !u#&8h8a# $j#$0 89 M* I , = = π n$ FqF\'23A ϕ BFq 2 F M• , F\ 3#ω ϕ F ⇒ \F M• , ϕ = π F\F ‚• , i' >'!7#&8`##$P$6#$KO R*!)2.+#&.^4$293)4,$ >'!7#&8i'• $6J$Z.j#&C`# ⇔ ≥ 12#&c4> 16#$*'$4CR#.+#& ⇔ FP B ®h max (F 2 ≥ ⇔ & 2 2 8 q \ , ∆ + + F&, \ M ≥ ⇒ • ≤ & $d#c K9c 89."# ?'$=Z# 1`#'() 7I I l l l =++ =++ =−= C q C q iL uuu qqi CABCAB 0C.h2$9* $f2 $Q&)# I =+ ′′ ii ω w KL II CCL CC + = ω K9 ( ) ϕω += tAi I'23I A…B n$ F 3#3#IIII 3#II '23I 〈⇒−==−= ′ −= ′ == ϕϕω ϕω ϕ UUUALiL Ai Ai AB 4C1) π ϕ −= K9 ω I L UU A − = eC − − = II I π ω ω tCos L UU i KL II CCL CC + = ω …†J!7#&J$Pg#& 16#$\#$q )#$ AK UeA hc I += λ F}\F„v‚I ƒ •Ff …†J!7#&J$Pg#& 16#$\#$q )#$ \W M mvA hc += λ F} W MAAK mvUe hchc +−= λλ …>J!7#&.X#$8Š.+#&#z#& AKMM Uemvmv += \W \W F} B A W λλ −= m hc v MA $)C3- smv MA lIMN v W = …n$2Z#&K;#FY**F} D ai = λ $)C3- m µλ v = )BX 1?&a#K;# 14#& ;*#$0 *9 h.E#$V#&=j'qh'()>#$3>#& 1o#&'$2K;#3>#& 1S#&#$)489K;#.[ =e' 1S#&K;# ?*=e' … ) U dtd xx λ == $)C3-qFY‚** =B$V#&=j'qh'()>#$3>#& 1o#&'$2K;#3>#& h qF„'* $2Z*~# BA MNI m ka D kx µλ λ =⇒= …)'E BA„vBAY‚ mm µλµ ≤≤ M„ ≤≤⇒ k w ,#&4C`#F},F‚ƒIIM eC'E„=j'qh'$2K;#3>#& hKX 1?qF„'*I …u.E ) ?#$.P_'=PL'3E#&'>'=j'qh = λ v„NwvwNMwMƒwMNwMNwY‚vA m µ B . đờng thẳng đứng - mgd = I M . (1) biến thi n điều hoà với = *&! I O , I G , I M là các mômen quán tính đối với các trục quay song song qua O,G,M. Mô men quán tính đối với bán cầu là: . `4I$2$D#& 3-8 z#&$FvvNI YM 3 -& apos;.+>#$3>#& 12# &'$;#,$H#&'FYI *l3, $-8 P_#&f8f' 12# * f F I Y ,&.+8L#."# ?'$'()f8f' 12# fFvI I I. một bán cầu đặc đồng chất, khối lợng m, bán kính R, tâm O. 1. Chứng minh rằng khối tâm G của bán cầu cách tâm O của nó một đoạn là d = 3R/8. 2. Đặt bán cầu trên mặt phẳng nằm ngang. Đẩy bán