92 Mudah dan Aktif Belajar Fisika untuk Kelas XI A. Momentum Linear JWe[Sb TW`VS iS`Y TWcYWcS] VW`YS` ]WUWbSeS` eWceW`ef V[bSde[]S` _W_[^[][ _a_W`ef_( ;WdSc`iS _a_W`ef_ dS`YSe TWcYS`ef`Y bSVS _SddS VS` ]WUWbSeS` TW`VS eWcdWTfe( Fa_W`ef_ iS`Y V[_[^[][ dfSef TW`VS SVS^SZ ZSd[^ bWc]S^[S` S`eScS _SddS VS` ]WUWbSeS` TW`VS bSVS dSSe eWceW`ef( IWcdS_SS` _a_W`ef_ dfSef TW`VS iS`Y TWc_SddS B VS` TWcYWcS] VW`YS` ]WUWbSeS` P dWUScS _SeW_Se[d V[cf_fd]S` dWTSYS[ TWc[]fe( K 7 BP ".l+# DWeWcS`YS`4 K 7 _a_W`ef_ TW`VS "]Y_)d# B 7 _SddS TW`VS "]Y# P 7 ]WUWbSeS` TW`VS "_)d# Fa_W`ef_ SVS^SZ dWTfSZ TWdScS` gW]eac iS`Y ScSZ`iS dS_S VW`YS` ScSZ ]WUWbSeS` TW`VS( /G779 3;KHDC _W`YW_f]S]S` Zf]f_ YWcS] iS`Y ]WVfS V[]S[e]S` VW`YS` _a_W`ef_( BS _W`YSeS]S` _a_W`ef_ dWTSYS[ ]fS`e[eSd YWcS]( Af]f_ BB GWhea` _W`YSeS]S` TSZhS& bWcfTSZS` _a' _W`ef_ TW`VS e[Sb dSefS` hS]ef dWTS`V[`Y VW`YS` YSiS cWdf^eS` iS`Y TW]Wc¥S bSVS TW`VS VS` TWcScSZ dS_S VW`YS` ScSZ YSiS eWcdWTfe( JWUScS _SeW_Se[d& VSbSe V[ef^[d dWTSYS[ TWc[]fe( 4 7 : :H K ".l,# :ALM=H==I !*S(" dWeScS VW`YS` TW`ef] ! 7 B= f`ef] dWTfSZ bSce[]W^ ef`YYS^ VW`YS` _SddS B "]a`deS`#( IWcZSe[]S` ZfTf`YS``iS VS^S_ TW`ef] TWc[]fe [`[( ! 7 : :H K 7 : B" # :H P 7 B : :H P 7 B= ".l-# ISVS eWac[ cW^Se[g[eSd ":`VS S]S` _W_bW^S¥Sc[`iS V[ DW^Sd OBB VS^S_ TSZSdS` ?[d[]S FaVWc`#& Af]f_ BB GWhea` f`ef] bSce[]W^ ef`YYS^ TW`ef] 4 7 B= e[VS] TWc^S]f& eWeSb[ Af]f_ BB GWhea` TW`ef] 4 7 : :H K eWeSb TWc^S]f( 1. Hukum Kekekalan Momentum Linear C[]S bSce[]W^ _W`YS^S_[ eaeS^ YSiS ^fSc 4 `a^& :ALM=H==I !*S(" VSbSe V[ef^[d dWTSYS[ TWc[]fe( : :H K 7* SeSf K 7 eWeSb ".l.# Tugas Anda 4.1 Jika Anda bersepeda dengan kecepatan tinggi, manakah yang memiliki momentum lebih besar, sepeda atau Anda? Apakah hal tersebut dapat memberikan penjelasan, mengapa Anda akan terpelanting ke depan ketika sepeda berhenti mendadak? +( :bS]SZ dWe[Sb TW`VS iS`Y _W_[^[][ _a_W`ef_ ¥fYS _W_[^[][ W`WcY[9 ,( C[]S VfS TfSZ TW`VS _W_[^[][ W`WcY[ ][`We[] dS_S TWdSc& eWeSb[ _SddS`iS TWcTWVS& TW`VS _S`S iS`Y _W_[^[][ _a_W`ef_ ^WT[Z TWdSc9 -( :bS iS`Y V[_S]dfV VW`YS` ]aWX[d[W` cWde[efd[9 .( ;a^S : TWc_SddS , ]Y TWcYWcS] VW`YS` ]WUWbSeS` , _)d( ;a^S ; TWc_SddS + ]Y TWcYWcS] TWc^ShS`S` ScSZ eWcZSVSb TW`VS : VW`YS` ]WUWbSeS` 0 _)d( JWeW^SZ TWcef_Tf]S`& ]WUWbSeS` Ta^S ; _W`¥SV[ - _)d( ;WcSbS]SZ ]WUWbSeS` S]Z[c Ta^S :9 ;A>AGOH HAHKAG=E=LD FJIMAK 9JHAINOH$ 6HKOGM$ @=I <OH>OF=I$ FALE=F=IG=C MJ=G%MJ=G >ALDFON @=G=H >OFO G=NDC=I& Tes Kompetensi Awal Kecepatan yang berada pada kerangka satu-dimensi, penulisan notasi vektornya dapat diganti dengan notasi skalar. Ingatlah
Momentum, Impuls, dan Tumbukan 93 :ALM=H==I !*S*" _W`iSeS]S` TSZhS ¥[]S YSiS ^fSc eaeS^ bSVS dWTfSZ bSce[]W^ SVS^SZ `a^& _a_W`ef_ ^[`WSc eaeS^ $ VSc[ d[deW_ SVS^SZ eWeSb( IWc`iSeSS` eWcdWTfe V[]W`S^ VW`YS` .I@IB 0;@;@7A7C 2DB;CHIB 1?C;7F iS`Y V[`iSeS]S` VW`YS` KShS^ 7 KS]Z[c ".l/# :ALM=H==I !*S+" _W`YS`Vf`Y bW`YWce[S` TSZhS _a_W`ef_ ^[`WSc eaeS^ bSVS hS]ef ShS^ dS_S VW`YS` _a_W`ef_ ^[`WSc eaeS^ bSVS hS]ef S]Z[c( B`YSe& :ALM=H==I !*S+" SVS^SZ TW`ef] gW]eac( H^WZ ]ScW`S [ef& ScSZ _a_W`ef_ ^[`WSc`iS TWcYS`ef`Y bSVS ScSZ( 2. Momentum Linear dan Impuls IWcZSe[]S` 5=H>=L *&'( JWTfSZ _aT[^ TWc_SddS B eWc^WeS] bSVS T[VS`Y VSeSc TWcYWcS] VW`YS` ]WUWbSeS` P+ & ]W_fV[S` V[TWc[ YSiS eWeSb 4 dWZ[`YYS ]WUWbSeS``iS TWcfTSZ _W`¥SV[ P, dW^S_S !H( =Sc[ Af]f_ BB GWhea`& V[bWca^WZ bWcdS_SS` 4 ( B= ".l0# ISVS YWcS] ^fcfd V[bWcUWbSe TWc^S]f = 7 " ! , + H P P ".l1# =Sc[ :ALM=H==I !*S," VS` :ALM=H==I !*S-" V[VSbSe]S` 4 7 B " ! , + " # H P P ".l2# FW`fcfe Af]f_ BB GWhea` iS`Y V[]S[e]S` VW`YS` _a_W`ef_ ^[`WSc _W`iSeS]S` TSZhS YSiS "4# iS`Y V[TWc[]S` bSVS TW`VS TWdSc`iS dS_S VW`YS` bWcfTSZS` _a_W`ef_ "K# TW`VS bWc dSefS` hS]ef "H#( E[ZSe :ALM=H==I !*S("& iS[ef 4 7 : :H K ( ;WcSce[& :ALM=H==I !*S." dWdfS[ VW`YS` :ALM=H==I !*S("( H^WZ ]ScW`S [ef& YSiS cSeS'cSeS`iS SVS^SZ ! # !H K 4 " # ! , + H K K dWZ[`YYS ! # ' " + 4 PP HB B ".l3# ;WcVSdSc]S` :ALM=H==I !*S/"& :`VS _W`YWeSZf[ TSZhS ZSd[^ ]S^[ YSiS "4# VS` dW^S`Y hS]ef ^S_S`iS YSiS TW]Wc¥S "!H# dS_S VW`YS` bWcfTSZS` _a_W`ef_ SeSf dW^[d[Z S`eScS _a_W`ef_ S]Z[c "BP, # VS` _a_W`ef_ ShS^ "BP+ # TW`VS f`ef] YSiS 4 iS`Y eWeSb( G[^S[ 4!H V[]W`S^ dWTSYS[ [_bf^d iS`Y V[^S_TS`Y]S` VW`YS` 6( =W`YS` VW_[][S`& VSbSe V[d[_bf^]S` TSZhS [_bf^d iS`Y TW]Wc¥S bSVS dfSef TW`VS dS_S VW`YS` bWcfTSZS` _a_W`ef_ iS`Y eWc¥SV[ bSVS TW`VS eWcdWTfe( 6 7 4!H ".l+*# DWeWcS`YS`4 6 7 [_bf^d "Gd# 4 7 YSiS "G# !H ( dW^S`Y hS]ef "d# Gambar 4.1 Sebuah mobil yang mengalami percepatan sehingga kecepatannya berubah dari v1 menjadi v2 . v2 v1 Tokoh Werner Heisenberg (1901–1976) Ia dilahirkan di Duisberg, Jerman. Ia mempelajari fisika teoritis di Munich, di tempat ini pun ia menjadi penggemar ski dan pendaki gunung. Hasil pemikirannya yang terkenal ialah prinsip ketidakpastian Heisenberg yang didasarkan pada konsep momentum, yaitu foton yang digunakan untuk mengamati posisi elektron memiliki momentum yang relatif sama. Oleh karena itu, ketika terjadi tumbukan antara foton dan elektron akan mengubah posisi elektron. Sumber:Conceptual Physics,1998
94 Mudah dan Aktif Belajar Fisika untuk Kelas XI JWTfSZ Ta^S TWc_SddS /* Y ¥SefZ VSc[ ]We[`YY[S` 1&, _WeWc( JWeW^SZ _W`f_Tf] ^S`eS[& Ta^S _W_S`ef^ ]W_TS^[ "TWc^ShS`S` VW`YS` ScSZ dW_f^S# VW`YS` ]We[`YY[S` -&, _( A[ef`Y^SZ4 S( _a_W`ef_ Ta^S dWTW^f_ VS` dWdfVSZ _W`f_Tf] ^S`eS[5 T( YSiS iS`Y V[TWc[]S` ^S`eS[ bSVS Ta^S ]We[]S Ta^S _W`iW`efZ ^S`eS[ dW^S_S *&*, d( 7=Q=>0 =[]WeSZf[4 B 7 /* Y 7 / m +*l, ]Y >+ 7 1&, _ >, 7 -&, _ !H 7 *&*, d L`ef] YWcS] ¥SefZ TWTSd& TWc^S]f bWcdS_SS` dWTSYS[ TWc[]fe( J 1= + " ,=> J 2 = , ,=> = " ! ! , +* 1&,!= –+, m/s = " ! ! , "+*# -&, = 2 m/s ScSZ ]W TShSZ "l# ScSZ ]W SeSd"%# S( E + 7 BJ+ ( "/ m +*l, ]Y#"l+, _)d# 7 l *&0 ]Y_)d "ScSZ`iS ]W TShSZ# E , 7 BJ,( "/ m +*l, ]Y#"2 _)d# 7 *&. ]Y_)d "ScSZ`iS ]W SeSd# CSV[& _a_W`ef_ Ta^S dWTW^f_ ef_Tf]S` SVS^SZ l*&0 ]Y_)d VS` dWeW^SZ ef_Tf]S` SVS^SZ *&. ]Y_)d( T( B_bf^d iS`Y V[TWc[]S` Ta^S4 / ( E, l E + ( *&. l "l*&0# 7 + ]Y_)d @SiS iS`Y V[]Wc¥S]S` ^S`eS[ bSVS Ta^S dW^S_S , m +*l, VWe[] SVS^SZ , 7 ! / H 7 ! , + ]Y_)d , +* d " $ 7 /* G CSV[& YSiS iS`Y V[TWc[]S` ^S`eS[ /* G _W^ShS` ScSZ YWcS] ¥SefZ Ta^S( h1 h2 Tes Kompetensi Subbab A 8ALE=F=IG=C @=G=H >OFO G=NDC=I& +( JWTfSZ Ta^S dWbS] TWc_SddS *&. ]Y V[eW`VS`Y Zac[ja`' eS^ a^WZ bW`¥SYS YShS`Y VW`YS` ]WUWbSeS` +** _)d( KW`ef]S` YSiS iS`Y V[TWc[]S` a^WZ ]S][ bW`¥SYS YShS`Y eWcZSVSb Ta^S ¥[]S dW^S`Y hS]ef ]S][ _W`iW`efZ Ta^S4 S( . m +*l, d 5 T( , m +*l- d( ,( JWacS`Y d[dhS _W`YYW`YYS_ Ta^S TWc_SddS *&/ ]Y VS` _W^W_bSc]S``iS dWUScS Zac[ja`eS^ VW`YS` ]WUWbSeS` ShS^ ,* _)d( ;a^S eWcdWTfe _W_TW`efc e[S`Y dW^S_S , m +*l, d dWZ[`YYS TWcTS^[] ScSZ VW`YS` ]WUWbSeS` +/ _)d( KW`ef]S`4 S( [_bf^d iS`Y V[TWc[]S` e[S`Y bSVS Ta^S5 T( YSiS iS`Y V[TWc[]S` e[S`Y bSVS Ta^S5 VS` U( bWcUWbSeS` Ta^S "bW_f]f^# dW^S_S TWcdW`efZS`( B. Tumbukan IWc[de[hS ef_Tf]S` VSbSe :`VS eW_f[ bSVS VfS TW`VS iS`Y dS^[`Y TWceSTcS]S` dSef dS_S ^S[`( IWc_S[`S` iS`Y _W_S`XSSe]S` dWUScS ^S`Ydf`Y bWc[de[hS ef_Tf]S` SVS^SZ bWc_S[`S` T[^[Sc VS` Tah^[`Y( KSZf]SZ :`VS SefcS` bWc_S[`S`'bWc_S[`S` eWcdWTfe9 ISVS fcS[S` dfTTST dWTW^f_`iS& eW^SZ V[]WeSZf[ TSZhS [_bf^d dS_S VW`YS` bWcfTSZS` _a_W`ef_( ;Wc[]fe [`[ S]S` V[¥W^Sd]S` bWcfTSZS` _a_W`ef_ iS`Y eWc¥SV[ bSVS dWe[Sb TW`VS iS`Y dS^[`Y TWcef_Tf]S`& dWceS ]S[eS` S`eScS dWe[Sb bWcfTSZS` _a_W`ef_ eWcdWTfe( Impuls sebanding dengan perubahan momentum. Ingatlah Kata Kunci • impuls • momentum linear Jika berlaku Hukum Kekekalan Momentum Linear, momentum linear awal sama dengan momentum linear akhir. Ingatlah Contoh 4.1
Momentum, Impuls, dan Tumbukan 95 sebelum tumbukan ketika tumbukan setelah tumbukan FA vA vB vA vB vA vB Gambar 4.2 Hukum Kekekalan Momentum Linear berlaku pada dua benda yang bertumbukan. DWcWeS : VS` ]WcWeS ; dWTW^f_ TWcef_Tf]S` _Sd[`Y'_Sd[`Y _W_[^[][ _SddS B: VS` B; & VS` ]WUWbSeS` %: VS` %; ( DWVfS ]WcWeS eWcdWTfe TWcSVS bSVS dSef T[VS`Y VSeSc VS` _W_[^[][ ScSZ YWcS] iS`Y dS_S( C[]S ]WUWbSeS` ]WcWeS : ^WT[Z TWdSc VSc[bSVS ]WUWbSeS` ]WcWeS ;& ]WcWeS : bSVS dSSe eWceW`ef S]S` _W`STcS] ]WcWeS ;( DWe[]S ]WcWeS : _W`STcS] ]WcWeS ;& dWdfS[ VW`YS` Af]f_ BBB GWhea`& ]WcWeS : S]S` _W_TWc[]S` YSiS S]d[ dWTWdSc ,: VS` ]WcWeS ; S]S` _W_TWc[]S` YSiS cWS]d[ dWTWdSc ,; ( ;SYS[_S`S TWdSc YSiS cWS]d[ eWcdWTfe bSVS ]WcWeS ;9 C[]S Af]f_ DW]W]S^S` Fa_W`ef_ E[`WSc TWc^S]f bSVS bWc[de[hS ef_Tf]S` S`eScS ]WcWeS : VS` ;& TWdSc ]WVfS YSiS eWcdWTfe S]S` dS_S TWdSc( :]S` eWeSb[& ScSZ ]WVfS YSiS TWc^ShS`S` ScSZ dWZ[`YYS dWUScS _SeW_Se[d VSbSe V[ef^[d]S` dWTSYS[ TWc[]fe( KShS^ ( KS]Z[c K: ShS^ % K; ShS^ 7 K: S]Z[c% K; S]Z[c B: P: % B; P; ( B: P: !% B; P; ! ".l++# DWeWcS`YS`4 B: P:% B; P; 7 ¥f_^SZ _a_W`ef_ dWTW^f_ ef_Tf]S` B: P: !% B; P; ! 7 ¥f_^SZ _a_W`ef_ dWdfVSZ ef_Tf]S` <a`eaZ ^S[` VSc[ bWc[de[hS ef_Tf]S` iS`Y VSbSe V[^S]f]S` bW`VW]SeS` VW`YS` Af]f_ DW]W]S^S` Fa_W`ef_ E[`WSc SVS^SZ bSVS bWc_S[`S` 8DB& 8DB 97F( =[ TWTWcSbS ]aeS TWdSc& bWc_S[`S` [`[ V[dWTfe ¥fYS _aT[^ dW`YYa^( FaT[^ eWcdWTfe V[YWcS]]S` a^WZ W`WcY[ ^[dec[] iS`Y V[bWca^WZ VSc[ ¥Sc[`Y' ¥Sc[`Y ^[dec[] V[ SeSd`iS( DWUWbSeS` _S]d[_f_ dWe[Sb _aT[^ SVS^SZ dS_S ]ScW`S W`WcY[ ^[dec[] iS`Y V[dfb^S[ a^WZ ¥Sc[`Y'¥Sc[`Y ^[dec[] V[ SeSd`iS ¥fYS dS_S( :]S` eWeSb[& TWcSe bW`f_bS`Y`iS TWcTWVS& ]WUWbSeS` _aT[^ ¥fYS TWcTWVS( DWe[]S eWc¥SV[ TW`efcS` S`eScS VfS _aT[^ SeSf ^WT[Z& WXW] Yf`US`YS` iS`Y bS^[`Y ]WcSd S]S` V[eWc[_S a^WZ _aT[^ VW`YS` TWcSe bW`f_bS`Y bS^[`Y ]WU[^( Gambar 4.3 Hukum kekekalan momentum terjadi pada bom-bom car yang bertumbukan. Sumber:Dokumentasi Penerbit F[dS^`iS& SVS VfS TfSZ TW`VS iS`Y dS^[`Y TWcef_Tf]S`& dWbWce[ bSVS 5=H>=L *&( TWc[]fe( ,( =fS TfSZ ]W^WcW`Y : VS` ; _Sd[`Y'_Sd[`Y TWc_SddS *&2 ]Y VS` *&0 ]Y& TWcYWcS] dWScSZ VW`YS` ]WUWbSeS` -* _)d VS` /* _)d( Iad[d[ ]W^WcW`Y V[TfSe dWVW_[][S` dWZ[`YYS ]W^WcW`Y ; _W`f_Tf] ]W^WcW`Y : VSc[ TW^S]S`Y( JWeW^SZ ef_Tf]S`& ]WVfS`iS TWcYWcS] TWcdS_SS`( KW`ef]S` ]WUWbSeS` ]WVfS ]W^WcW`Y eWcdWTfe( Tes Kompetensi Subbab B 8ALE=F=IG=C @=G=H >OFO G=NDC=I& +( JWacS`Y `W^SiS` TWc_SddS .* ]Y TWcSVS V[ SeSd bWcSZf 0* ]Y( IWcSZf TWcYWcS] VW`YS` ]W^S¥fS` +* _)d( KW`ef]S` ]WUWbSeS` bWcSZf ¥[]S `W^SiS` _W^a_bSe VW`YS` ]WUWbSeS` +/ _)d VW`YS` ScSZ4 S( ]W VWbS`& dWScSZ VW`YS` ScSZ YWcS] bWcSZf5 T( ]W TW^S]S`Y& TWc^ShS`S` VW`YS` ScSZ YWcS] bWcSZf5 VS` U( ]W dS_b[`Y& eWYS] ^fcfd VW`YS` ScSZ YWcS] bWcSZf( AB A B vA vB vAB
96 Mudah dan Aktif Belajar Fisika untuk Kelas XI C. Jenis Tumbukan :VS e[YS _SUS_ ef_Tf]S` iS`Y V[bW^S¥Sc[ bSVS ba]a] TSZSdS` ef_Tf]S`& iS[ef ef_Tf]S` ^W`e[`Y dW_bfc`S& ef_Tf]S` ^W`e[`Y dWTSY[S`& VS` ef_Tf]S` e[VS] ^W`e[`Y dS_S dW]S^[( JfTTST [`[ ZS`iS _W_TSZSd ef_Tf]S` S`eScS VfS TW`VS iS`Y ^[`eSdS` ]WVfS`iS TWcSVS bSVS dSef YSc[d ^fcfd( 1. Tumbukan Lenting Sempurna =S^S_ ]WZ[VfbS` `iSeS& ef_Tf]S` ^W`e[`Y dW_bfc`S e[VS] bWc`SZ eWc¥SV[ ]ScW`S dWeW^SZ ef_Tf]S` SVS dWTSY[S` W`WcY[ iS`Y TWcfTSZ TW`ef] _W`¥SV[ W`WcY[ bS`Sd& W`WcY[ Tf`i[& VS` W`WcY[ ^S[``iS( :]S` eWeSb[& VS^S_ bW_TSZSdS` [`[& ef_Tf]S` ^W`e[`Y dW_bfc`S V[S`YYSb VSbSe eWc¥SV[( :`YYSbS` VW_[][S` dWc[`Y V[Yf`S]S` a^WZ bScS X[d[]ShS` f`ef] _W`iWVWcZS`S]S` dfSef _SdS^SZ& iS`Y V[¥SV[]S` dWTSYS[ dfSef ^S`VSdS` eWac[ iS`Y _W`VSdSc[ bWc[de[hS ef_Tf]S` ^S[``iS( ISVS ef_Tf]S` ^W`e[`Y dW_bfc`S& TWc^S]f Af]f_ DW]W]S^S` Fa_W`' ef_ E[`WSc( >`WcY[ ][`We[] eaeS^ iS`Y V[_[^[][ TW`VS dWTW^f_ VS` dWdfVSZ ef_Tf]S` SVS^SZ eWeSb( >`WcY[ baeW`d[S^ TW`VS e[VS] V[bWcZ[ef`Y]S` ]ScW`S ]WVfS TW`VS TWcYWcS] VS^S_ dSef T[VS`Y VSeSc VW`YS` ]We[`YY[S` iS`Y dS_S( Tugas Anda 4.2 Anggota polisi sering menggunakan rompi anti peluru dalam melaksanakan tugasnya. Dalam beberapa kejadian, rompi tersebut bisa tembus oleh peluru. Mengapa hal tersebut bisa terjadi? v1 v2 v1 ' v2 ' m1 m2 m1 m2 m1 m2 sebelum tumbukan saat tumbukan setelah tumbukan IWcZSe[]S` 5=H>=L *&*( =fS TfSZ TW`VS bSVS T[VS`Y VSeSc TWcYWcS] TWc^ShS`S`( DW_fV[S`& dWeW^SZ eWc¥SV[ ef_Tf]S` ]WVfS TW`VS eWcdWTfe TWcYWcS] TWc^ShS`S` ScSZ VSc[ ScSZ dW_f^S( DWUWbSeS` dWe[Sb TW`VS SVS^SZ P+ ! VS` P, !( JWdfS[ VW`YS` Af]f_ DW]W]S^S` Fa_W`ef_ E[`WSc& V[VSbSe]S` bWcdS_SS` dWTSYS[ TWc[]fe( B+ P+ % B, P, 7 B+ P+ ! % B, P, ! B+ "P+ N P+ !# 7 B, "P, ! N P, # ".l+,# FW`fcfe Af]f_ DW]W]S^S` >`WcY[ D[`We[]& V[VSbSe]S` bWcdS_SS` TWc[]fe( +@+ % +@, 7 +@+ ! % +@, ! + , B+ J + , % + , B, J , , 7 + , B+ J + ! , % + , B, J , ! , ".l+-# =Sc[ :ALM=H==I !*S')" S]S` V[bWca^WZ bWcdS_SS` dWTSYS[ TWc[]fe( B+ J + , % B, J , , 7 B+ J + ! , % B, J , ! , B+ "J + , N J+ ! , # 7 B, "J , ! , N %, , # B+ "J + % J+ !#"J + N J+ !# 7 B, "J , ! % J , #"J , ! N J, # ".l+.# C[]S cfSd ]S`S` :ALM=H==I !*S'*" V[TSY[ VW`YS` cfSd ]S`S` :ALM=H==I !*S'("& VS` cfSd ][c[ :ALM=H==I !*S'*# V[TSY[ VW`YS` cfSd ][c[ :ALM=H==I !*S'("& S]S` V[ZSd[^]S` "J + % J+ !# 7 "J , % J, !# "J + !l J, !# 7 l "J + N J, # SeSf + , + , ! ! + J J J J + - " " # / 1 , . " ".l+/# Gambar 4.4 Tumbukan lenting sempurna.
Momentum, Impuls, dan Tumbukan 97 =fS TW`VS iS`Y TWc_SddS , ]Y VS` . ]Y TWcYWcS] TWc^ShS`S` ScSZ VW`YS` ]WUWbSeS` 0 _)d VS` . _)d( KW`ef]S` ]WUWbSeS` ]WVfS TW`VS dWeW^SZ ef_Tf]S` ¥[]S ef_Tf]S` ^W`e[`Y dW_bfc`S VS` ef_Tf]S` ^W`e[`Y dWTSY[S`& ; 7 *&.( 7=Q=>0 =[]WeSZf[4 B) 7 , ]Y& B* 7 . ]Y& J ) 7 0 _)d& J * 7 l . _)d( IWcZSe[]S`( H^WZ ]ScW`S ]WVfS TW`VS TWcYWcS] TWc^ShS`S` ScSZ _S]S ]WVfS ]WUWbSeS` ZScfd TWc^ShS`S` eS`VS( ISVS Ua`eaZ daS^ [`[& J : V[S`YYSb bad[e[X VS` J ; V[S`YYSb `WYSe[X( =Sc[ Af]f_ DW]W]S^S` Fa_W`ef_ V[VSbSe B: J : % B; J ; 7 B: J : ! % B; J ; ! ", ]Y#"0 _)d# % ". ]Y#"l. _)d# 7 ", ]Y# J : ! % ". ]Y# J ; ! +, _)d l +0 _)d 7 , J : !% . J; ! l, _)d 7 "J : !% , J ; !# _)d "$# ISVS ef_Tf]S` ^W`e[`Y dW_bfc`S& ; 7 + _S]S J J J J + - " "/ 1 " , . A B A B ’ ’ 7 + & l 0 . J J " " " A B ’ ’ ( ) 7 + N J : ! % J; ! ( +* _)d "$$# =Sc[ W^[_[`Sd[ bWcdS_SS` "$# VS` "$$#& V[bWca^WZ l, 7 J : ! % , J ; ! & J : ! 7 l, l , J ; _S]S lJ : ! % J ; ! 7 +* _)d J; ! 7 +* _)d %J : ! J; ! 7 +* _)d l , _)d l ,J ; ! J; ! ( 2 - _)d • Pada tumbukan lenting sempurna, koefisien restitusinya adalah e = 1. • Pada tumbukan lenting sebagian, koefisien restitusinya 0 < e < 1, sedangkan pada tumbukan tidak lenting sama sekali, koefisien restitusinya adalah e = 0. Ingatlah Contoh 4.2 AScYS + bSVS :ALM=H==I !*S'." _W`iSeS]S` @D;<?G?;C F;GH?HIG? f`ef] ef_Tf]S` ^W`e[`Y dW_bfc`S( JWUScS f_f_& :ALM=H==I !*S'." V[ef^[d + , + , J J ! ! ; J J + - " " # / 1 , . " ".l+3# =S^S_ ZS^ [`[& ; SVS^SZ ]aWX[d[W` cWde[efd[( :ALM=H==I !*S'/" TWc^S]f f`ef] dW_fS ¥W`[d ef_Tf]S`( 2. Tumbukan Lenting Sebagian dan Tumbukan Tidak Lenting Sama Sekali Af]f_ ]W]W]S^S` _a_W`ef_ TWc^S]f f`ef] dW_fS ¥W`[d ef_Tf]S`( :]S` eWeSb[& Zf]f_ ]W]W]S^S` W`WcY[ ][`We[] ZS`iS TWc^S]f bSVS ef_Tf]S` ^W`e[`Y dW_bfc`S( ISVS ef_Tf]S` ^W`e[`Y dWTSY[S`& dWTSY[S` W`WcY[ ][`We[] dWeW^SZ ef_Tf]S` S]S` TWcfTSZ TW`ef] _W`¥SV[ W`WcY[ ^S[` dWZ[`YYS ¥f_^SZ W`WcY[ ][`We[] dWeW^SZ ef_Tf]S` S]S` ^WT[Z ]WU[^ VSc[bSVS dWTW^f_ ef_Tf]S`( AScYS ]aWX[d[W` cWde[efd[ ";# bSVS :ALM=H==I !*S'," TWcSVS bSVS [`eWc' gS^ * 6 ; 6 +( ISVS ef_Tf]S` e[VS] ^W`e[`Y dS_S dW]S^[& dWTSY[S` TWdSc W`WcY[ ][`We[]`iS TWcfTSZ TW`ef] dWZ[`YYS ¥f_^SZ W`WcY[ ][`We[] dWTW^f_ ef_Tf]S` ^WT[Z TWdSc VSc[bSVS dWeW^SZ ef_Tf]S`& VS` ; 7 *( ISVS ef_Tf]S` e[VS] ^W`e[`Y dS_S dW]S^[& ]WVfS TW`VS TWcdSef dWeW^SZ ef_Tf]S` VS` TWcYWcS] TWcdS_S'dS_S VW`YS` ]WUWbSeS` dS_S( :`VS VSbSe _W_bWcZSe[]S` S][TSe ^S[` VSc[ ef_Tf]S` ¥W`[d [`[(
98 Mudah dan Aktif Belajar Fisika untuk Kelas XI Tugas Anda 4.3 Jika Anda sedang menggunakan sepatu, cobalah tendang sebuah bola sepak. Kemudian, bukalah sepatu dan tendang kembali bola tersebut. Jika diasumsikan gaya yang Anda berikan selama menyentuh sama besar, selidikilah tendangan mana yang membuat bola terlempar jauh? Apa hubungan kejadian ini dengan konsep impuls? =fS TfSZ TW`VS eWc^WeS] bSVS dSef T[VS`Y VS` dWYSc[d( ;W`VS bWceS_S TWc_SddS B+ 7 2 ]Y VS` TW`VS ]WVfS B, 7 +, ]Y( DWVfS TW`VS TWcYWcS] TWc^ShS`S` ScSZ VW`YS` ]WUWbSeS` J + 7 ,* _)d VS` J , 7l +* _)d( KW`ef]S` ]WUWbSeS` dWe[Sb TW`VS dWeW^SZ eWc¥SV[ ef_Tf]S`( ;WcSbS W`WcY[ ][`We[] iS`Y TWcfTSZ _W`¥SV[ W`WcY[ bS`Sd& ¥[]S eWc¥SV[4 S( ef_Tf]S` ^W`e[`Y dW_bfc`S5 T( ef_Tf]S` ^W`e[`Y dWTSY[S`& ; 7 *&,/5 VS` U( ef_Tf]S` e[VS] ^W`e[`Y dS_S dW]S^[( 7=Q=>0 =[]WeSZf[4 B+ 7 2 ]Y5 B, 7 +, ]Y J + 7 ,* _)d5 J , 7 l +* _)d S( L`ef] ef_Tf]S` ^W`e[`Y dW_bfc`S4 =Sc[ Af]f_ DW]W]S^S` Fa_W`ef_& V[VSbSe]S` B+ J + % B, J ' 7 B+ J + ! % B, J , ! "2 ]Y#",* _)d# % "+, ]Y# "l+* _)d# 7 2 ]Y J + !% +, ]Y J, ! +* ]Y_)d 7 , ]Y J + ! % - ]Y J , ! "$# L`ef] ef_Tf]S` ^W`e[`Y dW_bfc`S& ; 7 +& :ALM=H==I !*S'," _W`¥SV[ l ,* _)d +* _)d ! + - J J " / 1 " " , . 1 2 ’ ’ 7 + -* 7 l J+ !%J , ! "$$# FW`dfTde[efd[]S` :ALM=H==I !#" ]W :ALM=H==I !##" _W`YZSd[^]S` J + ! 7 l +0 _)d( J , ! 7 +. _)d( IWcfTSZS` W`WcY[ ][`We[] >`WcY[ ][`We[] dWTW^f_ ef_Tf]S`4 +@+ % +@, ( + , B+ J + , % + , B, J , , 7 + , "2 ]Y#",* _)d#, % + , "+, ]Y#"+* _)d#, 7 ,(,** C( >`WcY[ ][`We[] dWeW^SZ ef_Tf]S`4 +@+ !% +@'! 7 + , B+ J + ! , % + , B, J , ! , 7 + , "2 ]Y#"l+0 _)d#, % + , "+,]Y#"+. _)d#, 7 ,(,** C( m1 m2 m1 m2 v1 v2 v1 v2 m1 = 8 kg m2 = 12 kg sebelum tumbukan setelah tumbukan Contoh 4.3 Tantangan untuk Anda Apakah Anda pernah menonton balapan? Mengapa di beberapa bagian sirkuit diberi pagar dari ban (tyre wall)? =Sc[ bWcdS_SS` "$$#& V[bWca^WZ NJ) ! % J* ! ( +* _)d NJ) ! % 2 - _)d 7 +* _)d J ) ! 7 l ,, - _)d 7 l + 1 - _)d CSV[& ]WUWbSeS` ]WVfS TW`VS dWeW^SZ ef_Tf]S` ^W`e[`Y dW_bfc`S SVS^SZ J : ! 7 l + 1 - _)d VS` %; ! 7 2 - _)d( DWVfS ]WUWbSeS` _W`YS^S_[ bWcfTSZS` eS`VS( :ce[`iS& ]WVfS TW`VS eWcdWTfe _W`YS^S_[ bWcfTSZS` ScSZ YWcS]( C[]S :`VS _W`VSbSe]S` `[^S[ ]WUWbSeS` e[VS] TWcfTSZ ScSZ dWeW^SZ ef_Tf]S`& TWcSce[ bWcZ[ef`YS` :`VS TW^f_ TW`Sc( ISVS ef_Tf]S` ^W`e[`Y dWTSY[S`& ; 7 *&. dWZ[`YYS ! ! ! ! 0 _)d . _)d ) * J J " " " " 7 *&. l J : ! % J ; ! 7 . _)d "$$$# =Sc[ W^[_[`Sd[ bWcdS_SS` "$# VS` "$$$#& V[VSbSe J ; ! 7 + - _)d5 J : ! 7 +* - _)d( CSV[& f`ef] ef_Tf]S` ^W`e[`Y dWTSY[S`& ]WUWbSeS` TW`VS : dWeW^SZ ef_Tf]S` SVS^SZ +* - _)d VS` ]WUWbSeS` TW`VS ; SVS^SZ + - _)d(
Momentum, Impuls, dan Tumbukan 99 CSV[& bSVS ef_Tf]S` ^W`e[`Y dW_bfc`S& W`WcY[ ][`We[] dWTW^f_ VS` dWdfVSZ ef_Tf]S` SVS^SZ eWeSb( T( L`ef] ef_Tf]S` ^W`e[`Y dWTSY[S`& ; 7 *&,/4 =Sc[ Af]f_ DW]W]S^S` Fa_W`ef_& V[VSbSe ZSd[^ dWbWce[ :ALM=H==I "$#( ISVS ef_Tf]S` ^W`e[`Y dWTSY[S`& ; 7 *&,/& :ALM=H==I !*S/" _W`¥SV[ 1&/ _)d 7 l %+ ! % J , ! "$$$# JfTde[efd[]S` :ALM=H==I !#" ]W VS^S_ :ALM=H==I !###" dWZ[`YYS V[bWca^WZ ZSd[^ J + ! 7 l,&/ _)d J , ! 7 / _)d >`WcY[ ][`We[] dWeW^SZ ef_Tf]S` +@+ ! % +@, ! 7 + , B+ J + ! , % + , B, J , ! , 7 + , "2 ]Y#"l,&/ _)d#, % + , "+, ]Y#"/ _)d#, 7 ,/ % +/* 7 +1/ ¥af^W CSV[& W`WcY[ ][`We[] iS`Y TWcfTSZ _W`¥SV[ bS`Sd SVS^SZ ;C;F=? @?C;H?@ G;8;AIB HIB8I@7C N ;C;F=? @?C;H?@ G;H;A7> HIB8I@7C 7 ,(,** ¥af^W l +1/ ¥af^W 7 ,(*,/ ¥af^W( U( L`ef] ef_Tf]S` e[VS] ^W`e[`Y dS_S dW]S^[& ; 7 *4 H^WZ ]ScW`S ; ( * _S]S J+ ! 7 J , ! 7 J! dWZ[`YYS :ALM=H==I !#" S]S` _W`¥SV[ +* 7 / J + ! 7 / J , ! 7 / J! J! 7 , _)d( >`WcY[ ][`We[] dWeW^SZ ef_Tf]S` +@+ ! % +@, ! 7 " + , B+ % + , B, # J! , 7 "+* ]Y# ", _)d#, 7 .* ¥af^W( CSV[& TS`iS]`iS W`WcY[ ][`We[] iS`Y TWcfTSZ _W`¥SV[ bS`Sd SVS^SZ ",(,** l .*# ¥af^W 7 ,(+0* ¥af^W( m1 m2 v1 v2 m1 = 8 kg m2 = 12 kg sebelum tumbukan m1 m2 setelah tumbukan v' m1 m2 v1 v2 m1 = 8 kg m2 = 12 kg sebelum tumbukan m1 m2 setelah tumbukan v1 ' v2 ' Kata Kunci • tumbukan lenting sebagian • tumbukan lenting sempurna • tumbukan tidak lenting sama sekali +( JWTfSZ Ta^S iS`Y _W_[^[][ _a_W`ef_ & _W`f_Tf] V[`V[`Y VS` _W_S`ef^( Kf_Tf]S` eWcdWTfe TWcd[XSe ^W`e[`Y dW_bfc`S VS` ScSZ`iS eWYS] ^fcfd( KW`ef]S`^SZ TWdSc`iS bWcfTSZS` _a_W`ef_ Ta^S( ,( JWTfSZ TW`VS : TWc_SddS . ]Y eWcYS`ef`Y bSVS dWfeSd eS^[ iS`Y bS`¥S`Y`iS - _( ;W`VS ; iS`Y TWc_SddS . ]Y _W`f_Tf] TW`VS : VW`YS` ]WUWbSeS` _W`VSeSc dWTWdSc 0 _)d( C[]S ef_Tf]S` iS`Y eWc¥SV[ W^Sde[d dW_bfc`S& Z[ef`Y^SZ dfVfe d[_bS`YS` _S]d[_f_( -( =fS TfSZ TW`VS _Sd[`Y'_Sd[`Y TWc_SddS +, ]Y VS` +2 ]Y TWcYWcS] TWc^ShS`S` ScSZ( DWUWbSeS` dWe[Sb TW`VS -* _)d( ISVS dfSef dSSe& ]WVfS TW`VS eWcdWTfe dS^[`Y TWcef_Tf]S` V[ SeSd ^S`eS[ iS`Y ^[U[`( Tes Kompetensi Subbab C 8ALE=F=IG=C @=G=H >OFO G=NDC=I& KW`ef]S` ]WUWbSeS` dWe[Sb TW`VS dWeW^SZ ef_Tf]S` VS` W`WcY[ ][`We[] iS`Y TWcfTSZ _W`¥SV[ W`WcY[ bS`Sd& ¥[]S eWc¥SV[4 S( ef_Tf]S` ^W`e[`Y dW_bfc`S5 T( ef_Tf]S` ^W`e[`Y dWTSY[S`& ; 7 *&.5 VS` U( ef_Tf]S` e[VS] ^W`e[`Y dS_S dW]S^[( .( JWTfSZ TS^a] TWc_SddS *&/ ]Y eWc^WeS] V[ SeSd ^S`eS[( DaWX[d[W` YWdW]S` TS^a] VW`YS` ^S`eS[ SVS^SZ *&,( IW^fcf iS`Y TWc_SddS +* YcS_ V[eW_TS]]S` ]W ScSZ TS^a] Z[`YYS TWcdScS`Y V[ VS^S_`iS VS` TS^a] TWcYWdWc dW¥SfZ dSef _WeWc( ;WcSbS]SZ ]WUWbSeS` bW^fcf dSSe S]S` _W`YW`S[ TS^a] "= 7 +* _)d, #9 "@f`S]S` 4 < ] G 7 + , BJ, # Tantangan untuk Anda Berikut ini ukuran diameter, massa jenis, dan periode rotasi dari planet-planet. Dari data tersebut, berapakah momentum yang dimiliki setiap planet? Apakah ada hubungan antara momentum dan periode rotasi planet tersebut? :G=IAN /&/, -&3/ -&-- *&03 #!FH" +,(01/ 0(13* +.,(32* +,*(/.* " ,- ¥S_ /0! ,. ¥S_ -1! 3 ¥S_ -/! +* ¥S_ .*! ;f_[ FScd Pfb[eWc JSefc`fd
100 Mudah dan Aktif Belajar Fisika untuk Kelas XI Gambar 4.5 Tumbukan lenting sebagian pada benda jatuh bebas. D. Tumbukan Lenting Sebagian pada Benda Jatuh Bebas Kf_Tf]S` ^W`e[`Y dWTSY[S` TWc^S]f ¥fYS bSVS dWTfSZ TW`VS iS`Y TWcYWcS] ¥SefZ TWTSd( =S^S_ ]Sdfd [`[& iS`Y TWce[`VS] dWTSYS[ TW`VS ]WVfS SVS^SZ ^S`eS[( IWcZSe[]S` 5=H>=L *&+( =W`YS` VW_[][S`& ]WUWbSeS` TW`VS ]WVfS dWTW^f_ VS` dWdfVSZ ef_Tf]S` SVS^SZ `a^( JWTfSZ Ta^S ¥SefZ TWTSd VSc[ ]We[`YY[S` > eWcZSVSb ^S`eS[( DWUWbSeS` Ta^S dWdSSe dWTW^f_ VS` dWdfVSZ _W`f_Tf] ^S`eS[ _W_W`fZ[ bWcdS_SS` J 7 ,=> ".l+1# DWUWbSeS` Ta^S dWTW^f_ VS` dWdfVSZ _W`f_Tf] ^S`eS[ _W_W`fZ[ bWcdS_SS` J + 7 + " ,=> VS` J + ! 7 % + , ! => ".l+2# KS`VS `WYSe[X bSVS J + _W`S`VS]S` ScSZ Ta^S ]W TShSZ VS` eS`VS bad[e[X bSVS J + ! _W`S`VS]S` ScSZ Ta^S ]W SeSd( DaWX[d[W` cWde[efd[ S`eScS Ta^S VS` ^S`eS[ VSbSe V[bWca^WZ VSc[ bWcdS_SS` ! ! " # " + , + , J J ! ! ; J J + + , !* , * => ; => + - " # "/ 1 " " , . # >! ; > ".l+3# :ALM=H==I !*S'/" VSbSe V[Yf`S]S` bSVS TW`VS ¥SefZ ]W ^S`eS[ VS` ]W_fV[S` _W_S`ef^ TWTWcSbS ]S^[( JWTSYS[ Ua`eaZ& bWcZSe[]S` ]Sdfd bSVS 5=H>=L *&,( ISVS ]Sdfd eWcdWTfe& bWcdS_SS` ]aWX[d[W` cWde[efd[ _W`¥SV[ v1 v h 2 1 h2 Gambar 4.6 Tinggi pantulan bola yang mengalami tumbukan lenting sebagian. h1 h2 h3 h4 m m m m /( JWTfSZ TW`VS : TWc_SddS , ]Y iS`Y TWcYWcS] VW`YS` ]W^S¥fS` 0 _)d TWcef_Tf]S` VW`YS` TW`VS ; TWc_SddS . ]Y iS`Y V[S_( JWeW^SZ ef_Tf]S`& TW`VS : TWcYWcS] _f`Vfc VW`YS` ]W^S¥fS` + _)d( KW`ef]S`4 S( ]WUWbSeS` TW`VS ; dWeW^SZ ef_Tf]S`5 T( W`WcY[ iS`Y TWcfTSZ S][TSe ef_Tf]S`5 VS` U( ]aWX[d[W` cWde[efd[ f`ef] ef_Tf]S` eWcdWTfe( 0( JWTfSZ ecf] TWc[d[ _fSeS` VW`YS` _SddS -* ea` iS`Y TWcYWcS] VW`YS` ]WUWbSeS` +* _)d _W`STcS] dWTfSZ _aT[^ TWc_SddS ,* ea` iS`Y dWVS`Y V[S_( JWeW^SZ ef_Tf]S`& ]WVfS`iS TWcYWcS] VW`YS` ]WUWbSeS` J( A[ef`Y `[^S[ J( 1( IW^fcf TWc_SddS -* Y V[eW_TS]]S` VS` TWcdScS`Y bSVS dWTfSZ TS^a] ]Sif( DW_fV[S`& d[deW_ "TS^a] VS` bW^fcf# TWcYWcS] VW`YS` ]WUWbSeS` + _)d& TWcSbS]SZ ]WUWbSeS` bW^fcf ]We[]S _W`YW`S[ TS^a]9 "_SddS TS^a] 0 ]Y# 2( Ka`a iS`Y TWc_SddS-/ ]Y _W^a`USe VW`YS` ]WUWbSeS` - _)d VSc[ dWTfSZ bWcSZf ]WU[^ iS`Y V[S_( FSddS bWcSZf SVS^SZ -* ]Y( ;WcSbS]SZ ]WUWbSeS` bWcSZf dWeW^SZ Ka`a _W^a_bSe9 3( =fS _aT[^ _S[`S` _Sd[`Y'_Sd[`Y TWcYWcS] VW`YS` ]WUWbSeS` -0 ]_)¥S_ VS^S_ ScSZ TWc^ShS`S` VS` TWcef_Tf]S` dWUScS e[VS] W^Sde[d dS_S dW]S^[( FSddS ]WVfS _aT[^ _Sd[`Y'_Sd[`Y *&/ ]Y VS` *&,/ ]Y( A[ef`Y TWcSbS ]WUWbSeS` ]WVfS _aT[^ eWcdWTfe dWeW^SZ ef_Tf]S`( ;WcSbS]SZ W`WcY[ ][`We[] iS`Y Z[^S`Y9 +*( @WcTa`Y bW`YS`Y]fe TScS`Y TWc_SddS +,(*** ]Y TWcYWcS] bSVS cW^`iS VW`YS` ]WUWbSeS` . _)d( K[TS' e[TS TScS`Y dWTWcSe - ea` V[eS_TSZ]S` VS^S_ YWcTa`Y eWcdWTfe eWYS] ^fcfd ScSZ YWcS] YWcTa`Y( A[ef`Y ]WUWbSeS` YWcTa`Y dW]ScS`Y(
Momentum, Impuls, dan Tumbukan 101 Perhatikan gambar berikut. Sebuah ayunan yang bandulnya bermassa M dinaikkan pada ketinggian H dan dilepaskan. Pada bagian terendah dari lintasannya, bandul membentur suatu balok bermassa m yang mula-mula diam di atas permukaan dataran yang licin. Jika setelah tumbukan kedua benda saling menempel, ketinggian h yang dapat dicapai keduanya adalah .... a. ' ) 0 2 ( * m m+M 2 H b. ' ) 0 2 ( * m m+M H c. ' ) 0 2 ( * M m+M 2 H d. ' ) 0 2 ( * M m+M H 2 e. ' ) 0 2 ( * M m+M 2 H 2 Soal UMPTN Tahun 2001 Pembahasan: Mv = (M + m)v' M 2gH = (M + m) 2gH M2 2gH = (M + m) 2 2gh h = ! 2 M H m+M 2 = ' ) 0 2 ( * M m+M 2 H Jawaban: c JWTfSZ Ta^S : V[¥SefZ]S` "¥SefZ TWTSd# VSc[ dfSef e[e[] I( Ea]Sd[ e[e[] I V[ef`¥f]]S` bSVS YS_TSc( ;a^S TWcYWcS] TWTSd dWbS`¥S`Y eS^S`Y ^[U[` "¥Sc['¥Sc[ eS^S`Y 5#( =[ f¥f`Y eS^S`Y eWcVSbSe Ta^S ; iS`Y _SddS`iS , ]S^[ _SddS :( DWVfS`iS TWcef_Tf]S` dWUScS W^Sde[d dW_bfc`S( S( A[ef`Y ]WUWbSeS` Ta^S : dWdSSe dWTW^f_ _W`f_Tf] Ta^S ;( T( A[ef`Y ]WUWbSeS` Ta^S ; dWdSSe dWeW^SZ TWcef_Tf]S` VW`YS` Ta^S :( U( ;WcSbS e[`YY[ _S]d[_f_ iS`Y V[USbS[ a^WZ TW`VS ;9 "JaS^ FW`f¥f H^[_b[SVW JS[`d B`Va`Wd[S# 7=Q=>0 IWcZSe[]S` YS_TSc( K[`YY[ Ta^S : VSc[ bWc_f]SS` eS^S`Y dWTW^f_ ¥SefZ SVS^SZ L 5 7 eS` $ #L 7 5 eS` $ _SddS Ta^S :& B: 7 B _SddS Ta^S ;& B; 7 , B S( DWUWbSeS` Ta^S : dWdSSe dWTW^f_ _W`f_Tf] Ta^S ; dS_S VW`YS` ]WUWbSeS` bSVS bad[d[ ,& _S]S 1 2 BJ: , 7 B=L 7 B=5 eS`$ J : 7 , eS` =5 $ T( DWUWbSeS` Ta^S ; dWdSSe dWeW^SZ TWcef_Tf]S` VW`YS` Ta^S : _W_W`fZ[ Af]f_ DW]W]S^S` Fa_W`ef_4 B: J : % B; J ; 7 B: J : !% B; J ; ! BJ: % * 7 BJ: !% , BJ; ! J : 7 J : ! % J ; ! , eS` =5 $ 7 J : !% J ; !(((((((((("$# Af]f_ DW]W]S^S` >`WcY[ FW]S`[]4 + , B: J : , % + , B; J ; ,7 + , B: J : ! , % + , B; J ; ! , + , BJ: ,7 + , BJ: ! , % BJ; ! , B=5 eS`$ 7 + , BJ: ! , % BJ; ! , =5 eS` $ 7 + , J : ! , % J ; ! , (((((((("$$# =Sc[ bWcdS_SS` "$# VS` "$$#& S]S` V[bWca^WZ =5eS` $ 7 + , " , eS` =5 $ l , J ; !#, % J ; ! , =5eS` $ 7 + , ", =5 eS`$ l . J ; ! , eS` =5 $ % . J ; ! , # % J ; ! , =5eS`$ 7 =5eS`$ l ,J ; ! , eS` =5 $ % - J ; ! , Contoh 4.4 ; 7 , + > > 7 - , > > 7 . - > > ".l,*# A P B R $ x R P $ B R R 2 1 X 3 4 5 A M H m Pembahasan Soal h
102 Mudah dan Aktif Belajar Fisika untuk Kelas XI Tes Kompetensi Subbab D 8ALE=F=IG=C @=G=H >OFO G=NDC=I& +( JWTfSZ Ta^S V[^WbSd]S` VSc[ ]We[`YY[S` +* _WeWc V[ SeSd ^S`eS[( ;a^S _W`f_Tf] ^S`eS[ VS` V[bS`ef^]S` ]W_TS^[( C[]S ]aWX[d[W` cWde[efd[ ef_Tf]S``iS *&0& TWcSbS]SZ e[`YY[ Ta^S iS`Y V[bS`ef^]S`( ,( JWTfSZ TW`VS V[^WbSd]S` VSc[ dfSef eW_bSe dW¥SfZ +,/ _ V[ SeSd ^S`eS[( JWeW^SZ _W`YW`S[ ^S`eS[& Ta^S _W_S`ef^ ]W_TS^[ VS` TW`VS `S[] ^SY[ dWe[`YY[ ./ _( A[ef`Y ]aWX[d[W` cWde[efd[ ef_Tf]S` TW`VS eWcdWTfe( -( JWTfSZ Ta^S V[^WbSd]S` VSc[ YWVf`Y iS`Y e[`YY[`iS /* _( JWeW^SZ _W`YW`S[ ^S`eS[& Ta^S eWcdWTfe V[bS`' ef^]S` ]W_TS^[( ISVS bS`ef^S` bWceS_S& Ta^S _W`USbS[ ]We[`YY[S` > _WeWc V[ SeSd ^S`eS[ VS` bSVS bS`ef^S` ]WVfS _W`USbS[ ]We[`YY[S` / _WeWc( A[ef`Y^SZ >( E. Ayunan Balistik Kf_Tf]S` e[VS] ^W`e[`Y TWc^S]f ¥fYS bSVS bWc[de[hS Sif`S` TS^[de[]& iS[ef TW`VS iS`Y eWcV[c[ SeSd dWbaea`Y TS^a] ]Sif iS`Y V[[]Se]S` bSVS dWfeSd eS^[( DW_fV[S`& V[YS`ef`Y]S` dWVW_[][S` cfbS SYSc TS^a] VSbSe TWcYWcS] VW`YS` TWTSd( :if`S` dWbWce[ [`[ VSbSe V[Yf`S]S` dWTSYS[ dS^SZ dSef UScS f`ef] _W`Yf]fc ]W^S¥fS` bW^fcf( IW^fcf iS`Y S]S` V[f]fc ]W^S¥f' S``iS V[eW_TS]]S` VSc[ dWTfSZ dW`¥SeS ]W ScSZ TS^a] eWcdWTfe( IWcZSe[]S` 5=H>=L *&-( F[dS^]S` TS^a] iS`Y eWcYS`ef`Y _W_[^[][ _SddS B; & dWVS`Y]S` bW^fcf eWcdWTfe _W_[^[][ _SddS BI ( IW^fcf iS`Y V[eW_TS]]S` eWcdWTfe TWcdScS`Y VS^S_ TS^a] VS` TWcYWcS] TWcdS_S'dS_S VW`YS` ]WUWbSeS` J! dWZ[`YYS TS^a] TWcSif` VS` _W`USbS[ ]We[`YY[S` >( FW`fcfe Af]f_ DW]W]S^S` Fa_W`ef_ E[`WSc& V[VSbSe bWcdS_SS` BI J b 7 "BI % B; #J! J b 7 ! " ! 4 * 4 B BJ B ".l,+# Gambar 4.7 Ayunan balistik mencapai tinggi maksimum hmaks = (1 – cos % ). Kata Kunci • koefisien restitusi J ; ! 7 , - , eS` =5 $ DWUWbSeS` Ta^S : dWdfVSZ ef_Tf]S` SVS^SZ =5 eS` $ 7 + , J : ! , % " , - , eS` =5 $ # , 7 + , J : ! , % 2 3 =5 eS`$ J : !7 $ , eS` 3 =5 _)d( CSV[& ]WUWbSeS` Ta^S : dWeW^SZ ef_Tf]S` SVS^SZ J : !7 $ , eS` 3 =5 _)d( U( K[`YY[ _S]d[_f_ iS`Y V[USbS[ Ta^S ; dWeW^SZ ef_Tf]S` V[bWca^WZ _W^S^f[ bWcdS_SS` W`WcY[ _W]S`[] V[ bad[d[ . VS` /( +I;. % +D;. 7 +I;/ % +D;/ * % + , B; J ; ! , 7 B; =M % * " + , #", B#" , - , eS` =5 $ # , 7 , B=M B " . 3 ,=5 eS`$ # 7 , B=M M 7 5 " . 3 eS`$ #SeSf M 7 5 "+ % . 3 eS`$ # VSc[ VSdSc eS^S`Y( CSV[& e[`YY[ _S]d[_f_ Ta^S ; dWeW^SZ ef_Tf]S` SVS^SZ M 7 5 "+ % . 3 eS`$ # VSc[ VSdSc eS^S`Y( vp mp mp + mb % vmp hmaks
Momentum, Impuls, dan Tumbukan 103 JWbaea`Y TS^a] ]Sif TWc_SddS .&3/ ]Y eWcYS`ef`Y bSVS dWfeSd eS^[( JWTfSZ bW^fcf iS`Y _SddS`iS /* Y V[eW_TS]]S` VSc[ dWTfSZ dW`SbS` ]W ScSZ TS^a] dWZ[`YYS bW^fcf eWcdWTfe TWcdScS`Y V[ VS^S_`iS( ;S^a] eWcdWTfe TWcSif` VS` `S[] dWe[`YY[ 2* U_ VSc[ ]WVfVf]S` dW_f^S( C[]S bWcUWbSeS` YcSg[eSd[ V[ eW_bSe eWcdWTfe +* _)d, & eW`ef]S` ]WUWbSeS` bW^fcf dWTW^f_ _W`YW`S[ TS^a]( 7=Q=>0 =[]WeSZf[4 BI 7 /* Y 7 *&*/ ]Y5 = 7 +* _)d, B; 7 .&3/ ]Y5 > 7 2* U_ 7 *&2 _ J 4 7 " # 4 * " 4 B B B ,=> 7 ! ! ! ! ! , *&*/ .&3/ ]Y , +* _)d *&2 _ *&*/ ]Y 7 .** _)d CSV[& ]WUWbSeS` bW^fcf dWTW^f_ _W`YW`S[ TS^a] SVS^SZ .** _)d( Contoh 4.5 FW`fcfe Af]f_ DW]W]S^S` >`WcY[ FW]S`[]& W`WcY[ ][`We[] iS`Y V[_[^[][ bW^fcf VS` TS^a] bSVS bad[d[ 4 dS_S VW`YS` W`WcY[ baeW`d[S^ iS`Y V[_[^[][ bW^fcf VS` TS^a] ]We[]S _W`USbS[ ]We[`YY[S` _S]d[_f_( JWUScS _SeW_Se[d& ]WSVSS` eWcdWTfe VSbSe V[ef^[d]S` dWTSYS[ TWc[]fe( + , "BI % B; #J! , 7 "BI % B; #=> J! 7 ,=> ".l,,# =Sc[ :ALM=H==I !*S('" VS` :ALM=H==I !*S(("& V[bWca^WZ ! " # " # , 4 * 4 4 B B J => B ".l,-# DWeWcS`YS`4 BI 7 _SddS bW^fcf "]Y# B; 7 _SddS TS^a] "]Y# J I 7 ]WUWbSeS` bW^fcf "_)d# J! 7 ]WUWbSeS` dWeW^SZ bW^fcf VS` TS^a] TWcef_Tf]S` "_)d# Gaya dorong pada roket dihasilkan akibat semburan gas panas dengan kecepatan tinggi. Gas panas tersebut dihasilkan dari proses pembakaran hidrogen cair dan oksigen cair. Untuk mengatasi pengaruh gravitasi Bumi, roket memiliki percepatan 3 kali percepatan gravitasi Bumi. Oleh karena itu, gaya dorong yang harus diberikan oleh roket harus lebih dari 3 × 107 N. Rocket propulsion is produced by hot gas expelled from the rear of the rocket at high velocity. The hot gas made from burning of liquid hydrogen and liquid oxygen. To abandon the earth’s gravitational force, rocket have to get acceleration three times faster than earth’s gravitational acceleration. So, rocket propulsion should be more 3 × 107 N. Informasi untuk Anda Information for You V[^WbSd bSVS dSSe iS`Y TWcdS_SS` VS` ]WVfS`iS TWcef_Tf]S` dWUScS W^Sde[d dW_bfc`S( A[ef`Y^SZ ]W`S[]S` ]WVfS Ta^S eWcdWTfe dWeW^SZ ef_Tf]S`( -( JWTfSZ TW`VS eWcYS`ef`Y bSVS dWfeSd eS^[ iS`Y bS`¥S`Y`iS ,* U_( ;W`VS eWcdWTfe V[ef_Tf] a^WZ TW`VS ^S[` iS`Y _SddS`iS . ]S^[ _SddS TW`VS dW_f^S( C[]S ef_Tf]S``iS W^Sde[d dWTSY[S`& VW`YS` `[^S[ ; 7 *&1/& TWcSbS]SZ ]WUWbSeS` TW`VS iS`Y _W`f_Tf] SYSc TW`VS iS`Y eWcYS`ef`Y eWcdWTfe _W`USbS[ e[e[] eWce[`YY[ VSc[ ^[`eSdS``iS9 +( JWTfSZ TS^a] TS¥S TWc_SddS , ]Y eWcYS`ef`Y bSVS dWfeSd eS^[ bS`¥S`Y( ;S^a] eWcdWTfe V[eW_TS] a^WZ dWTfe[c bW^fcf TWc_SddS ,* Y VW`YS` ]WUWbSeS` _W`VSeSc /** _)d( C[]S ef_Tf]S``iS W^Sde[d dWTSY[S` "; 7 *&0+0*#& Z[ef`Y ]We[`YY[S` _S]d[_f_ iS`Y VSbSe V[USbS[ TS^a] eWcdWTfe( ,( =fS Ta^S ]WU[^ iS`Y _Sd[`Y'_Sd[`Y TWc_SddS + ]Y VS` , ]Y V[YS`ef`Y]S` bSVS dWfeSd eS^[ *&, _( JWe[Sb Ta^S _f^S' _f^S V[TWc[ Sif`S` dWTWdSc 3*k eWcZSVSb YSc[d gWce[]S^ dWbWce[ bSVS YS_TSc( DW_fV[S`& ]WVfS Ta^S Tes Kompetensi Subbab E 8ALE=F=IG=C @=G=H >OFO G=NDC=I&
104 Mudah dan Aktif Belajar Fisika untuk Kelas XI F. Gaya Dorong Roket :_T[^^SZ dWTfSZ TS^a` ]ScWe& ^S^f e[fb^SZ TS^a` eWcdWTfe Z[`YYS _W`YW_TS`Y& `S_f` ¥S`YS` dS_bS[ _W^Wefd( CWb[e^SZ f¥f`Y ^fTS`Y eW_bSe fVScS _Sdf] VW`YS` eS`YS`& ]W_fV[S` ^WbSd]S`( :`VS S]S` _W^[ZSe TS^a` ]ScWe eWcdWTfe eWcVaca`Y a^WZ fVScS iS`Y ]W^fSc VSc[ VS^S_ TS^a` ]SdWe( Ic[`d[b dWbWce[ [`[ _WcfbS]S` bc[`d[b ]Wc¥S ca]We( Fa_W`ef_ TS^a` dWTW^f_ V[^WbSd]S` SVS^SZ dS_S VW`YS` `a^& ]ScW`S J 7 *( DWe[]S TS^a` V[^WbSd]S`& _a_W`ef_ TS^a` _W`¥SV[ E ( BTS^a`J TS^a`( IWcfTSZS` _a_W`ef_`iS bWc dSefS` hS]ef SVS^SZ dS_S VW`YS` YSiS Vaca`Y iS`Y V[^S]f]S` W_TfdS` fVScS eWcZSVSb TS^a`& iS`Y ScSZ`iS TWc^ShS`S` VW`YS` ScSZ ]W^S¥fS` TS^a`( H^WZ ]ScW`S bc[`d[b [`[ dS_S VW`YS` bc[`d[b YSiS Vaca`Y ca]We& bWcfTSZS` _a_W`ef_ bSVS ca]We VSbSe V[ef^[d]S` dWTSYS[ TWc[]fe( , 7 ! BJ H ! ! 7 E H ! ! ".l,.# DWeWcS`YS`4 , 7 YSiS Vaca`Y ca]We "G# J 7 ]WUWbSeS` ca]We "_)d# @SiS Vaca`Y iS`Y V[^S]f]S` YSd bSVS TS^a` VSbSe V[S`S^aY[]S` dWTSYS[ YSiS Vaca`Y TSZS` TS]Sc bSVS ca]We( Gambar 4.8 Gaya dorong pada balon. Tes Kompetensi Subbab F 8ALE=F=IG=C @=G=H >OFO G=NDC=I& +( =[]WeSZf[ YSd bS`Sd iS`Y ]W^fSc VSc[ ca]We _W_[^[][ ]W^S¥fS` ,** _)d( KW`ef]S` TWdSc YSiS Vaca`Y dWTfSZ ca]We iS`Y _Wd[``iS VSbSe _W`iW_Tfc]S` YSd bS`Sd ZSd[^ VSc[ bW_TS]ScS` VW`YS` ]W^S¥fS` bS`UScS` YSd /* ]Y)d( ,( FSddS eaeS^ dWTfSZ ca]We eWc_Sdf] TSZS` TS]Sc`iS SVS^SZ +(,** ]Y( JW_TfcS` YSiS bS`Sd iS`Y ]W^fSc VSc[ ca]We _W_[^[^][ ]W^S¥fS` ,/* _)d( C[]S V[ZScSb]S` ca]We VSbSe _W^S¥f VW`YS` bWcUWbSeS` ,&/ _)d, & eW`ef]S`^SZ ]W^S¥fS` bWcfTSZS` _SddS YSd V[ VS^S_ ca]We eWcdWTfe( .( ;a^S : iS`Y _SddS`iS - ]Y TWcSVS bSVS bad[d[ dWbWce[ eWc^[ZSe bSVS YS_TSc( A B ;a^S eWcdWTfe S]S` _W`f_Tf] Ta^S ; iS`Y TWc_SddS , ]Y( C[]S ]aWX[d[W` cWde[efd[ ef_Tf]S` *&/& TWcSbS _WeWc]SZ Ta^S ; S]S` `S[]9 "bS`¥S`Y eS^[ *&, _#(
Momentum, Impuls, dan Tumbukan 105 Rangkuman +( Fa_W`ef_ iS`Y V[_[^[][ dfSef TW`VS SVS^SZ ZSd[^ bWc]S^[S` S`eScS _SddS VS` ]WUWbSeS` bSVS dSSe eWceW`ef( Fa_W`ef_ _WcfbS]S` TWdScS` gW]eac iS`Y ScSZ`iS dS_S VW`YS` ScSZ ]WUWbSeS``iS( K ( B P B_bf^d SVS^SZ ZSd[^ ]S^[ YSiS VS` dW^S`Y hS]ef dW^S_S YSiS TW]Wc¥S( 6 ( 4 !H B_bf^d iS`Y TW]Wc¥S bSVS dfSef TW`VS TWdSc`iS dS_S VW`YS` bWcfTSZS` _a_W`ef_ iS`Y V[S^S_[ TW`VS( 6 ( B !P 7 BP, l BP+ ,( Af]f_ DW]W]S^S` Fa_W`ef_ _W`iSeS]S` TSZhS ¥f_^SZ _a_W`ef_ dWTW^f_ VS` dWdfVSZ ef_Tf]S` SVS^SZ dS_S& iS[ef B) P) % B* P* 7B) P) !% B* P* ! -( Af]f_ BB GWhea` _W`iSeS]S` TSZhS YSiS iS`Y V[TWc[]S` bSVS dfSef TW`VS& TWdSc`iS dS_S VW`YS` bWcfTSZS` _a_W`ef_ TW`VS dWe[Sb dSef dSefS` hS]ef( ! " # # ! ! B B , + H H K P P 4 .( DaWX[d[W` cWde[efd[ SVS^SZ f]fcS` ]W^W`e[`YS` "W^Sde[d[eSd# dWTfSZ ef_Tf]S`( + - " # "/ 1 , . " + , + , J J ! ! ; J J /( Af]f_ DW]W]S^S` Fa_W`ef_ TWc^S]f bSVS dW_fS ¥W`[d ef_Tf]S`( ISVS ef_Tf]S` ^W`e[`Y dW_bfc`S& TWc^S]f B+ J + % B, J ,7 B+ J + ! % B, J , !& J , ! l J + ! 7 l"J , l J + #! SeSf ]aWX[d[W` cWde[efd[ ; 7 +( ISVS ef_Tf]S` e[VS] ^W`e[`Y dS_S dW]S^[& ]WVfS TW`VS iS`Y TWcef_Tf]S` S]S` TWcdSef VS` TWcYWcS] TWcdS_S dWeW^SZ ef_Tf]S`( ISVS ¥W`[d ef_Tf]S` [`[& Af]f_ DW]W]S^S` >`WcY[ D[`We[] e[VS] TWc^S]f& W`WcY[ dWTSY[S` Z[^S`Y _W`¥SV[ W`WcY[ ]S^ac& W`WcY[ Tf`i[& VS` _W`iWTST]S` bWcfTSZS` hf¥fV bSVS TW`VS( B+ J + % B, J ,7 "B+ % B, # J! ISVS ef_Tf]S` ^W`e[`Y dWTSY[S`& TWc^S]f bWcdS_SS` B+ J + % B, J ,7 B+ J + ! % B, J , ! + - " # "/ 1 , . " + , + , J J ! ! ; J J & * 6 ; 6 + 0( ;W`VS iS`Y TWcYWcS] ¥SefZ TWTSd _WcfbS]S` ¥W`[d ef_Tf]S` ^W`e[`Y dWTSY[S`( DaWX[d[W` cWde[efd[`iS SVS^SZ # > ! ; > SeSf ; 7 , + > > 7 - , > > 7 . - > > 1( DWUWbSeS` bW^fcf VSbSe V[f]fc VW`YS` Sif`S` TS^[de[] VW`YS` _W_S`XSSe]S` bc[`d[b ef_Tf]S` e[VS] ^W`e[`Y dS_S dW]S^[( ! " # " # , 4 * 4 4 B B J => B
106 Mudah dan Aktif Belajar Fisika untuk Kelas XI Peta Konsep Kf_Tf]S` 9JHAINOH$ 6HKOGM$ @=I <OH>OF=I Fa_W`ef_ B_bf^d Af]f_ DW]W]S^S` Fa_W`ef_ Kf_Tf]S` K[VS] EW`e[`Y JS_S JW]S^[ Kf_Tf]S` EW`e[`Y JWTSY[S` Kf_Tf]S` EW`e[`Y JW_bfc`S Setelah mempelajari bab ini, tentu Anda dapat memahami konsep momentum, impuls, dan tumbukan. Anda juga tentu dapat menjelaskan jenis-jenis tumbukan yang mungkin terjadi dan hukum apa saja yang berlaku di dalamnya. Dari keseluruhan materi yang telah Anda pelajari, bagian manakah yang menurut Anda sulit? Coba Anda diskusikan dengan teman atau guru Fisika Anda. Refleksi Dengan mempelajari bab ini, Anda dapat mengetahui bahwa untuk mengukur kecepatan peluru tidak harus memasang alat ukur pada peluru. Anda cukup menghitung jejak yang ditinggalkannya, seperti pada bandul balistik. Nah, coba sebutkan manfaat lain mempelajari bab ini. _W_TSZSd V[Yf`S]S` f`ef] _W`YS`S^[d[d TWc^S]f bSVS ¥W`[d`iS TWc^S]f
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111 1& :DGDCG=C M=G=C M=NO E=Q=>=I R=IB K=GDIB NAK=N @=I FALE=F=IG=C K=@= >OFO G=NDC=I& 2 kg A B 1 kg 4 kg A B 1 kg m F 30° Tes Kompetensi Fisika Semester 1 +( IWcUWbSeS` dWTfSZ bSce[]W^ bSVS dSSe H SVS^SZ 0H D l .E( Ff^S'_f^S bSce[]W^ TWcYWcS] VW`YS` ]WUWbSeS` ,D( MW]eac ]WUWbSeS` bSce[]W^ bSVS dSSe H SVS^SZ (((( S( ", % -H#D l .H E T( ", l -H , #D % .H E U( ", % -H , #D l .H E V( "- l ,H , #D % .H E W( "- % ,H , #D l .H E ,( JWTfSZ bSce[]W^ TWcYWcS] dWdfS[ VW`YS` bWcdS_SS` bad[d[ F 7 "H , l .H#[ % ",H , l ,H#¥& F VS^S_ _WeWc VS` H VS^S_ VWe[]( ;WdSc ]WUWbSeS` TW`VS bSVS dSSe H 7 + VWe[] SVS^SZ (((( S( , _)d T( 2 _)d U( , , _)d V( , / _)d W( - - _)d -( =fS TfSZ TW`VS : VS` ; _Sd[`Y'_Sd[`Y TWc_SddS . ]Y VS` - ]Y V[ZfTf`Y]S` VW`YS` eS^[ "_SddS eS^[ V[STS[]S`# _W^S^f[ dWTfSZ ]Seca^ TWc_SddS , ]Y VS` TWc¥Sc['¥Sc[ ,* U_( C[]S TW`VS TWcbfeSc VW`YS` eS^[& bWcUWbSeS` dWe[Sb TW`VS SVS^SZ (((( "= 7 +* _)d, # S( *&1/ _)d, T( +&,/ _)d, U( +&/* _)d, V( +&1/ _)d, W( ,&** _)d, .( JWTfSZ bSce[]W^ _f^S'_f^S V[S_& ]W_fV[S` V[bWcUWbSe VS^S_ ^[`eSdS` _W^[`Y]Sc VW`YS` ¥Sc['¥Sc[ , _ _W`fcfe bWcdS_SS` bWcUWbSeS` 7 7 0H , l .H( IWcdS_SS` ]WUW' bSeS` dW`ec[bWeS^`iS SVS^SZ ((((( S( .H- l .H , T( +,H , l 2H U( 3H 0 l +,H / % .H . V( .H 0 l 0H / % .H . W( -0H . l ,.H - l +0H , /( JWTfSZ _aT[^ TWc_SddS +(*** ]Y _W^f`Ufc VW`YS` ]WUWbSeS` 1, ]_)¥S_& ]W_fV[S` V[cW_ VW`YS` YSiS ,( C[]S ]aWX[d[W` YWdW] ][`We[] S`eScS caVS VS` ¥S^S` *&, ]W_fV[S` TWcZW`e[ dWeW^SZ . VWe[]& TWdSc YSiS bW`YWcW_S` , SVS^SZ (((( S( /(*** G T( .(/** G U( -(/** G V( -(*** G W( .(*** G 0( IWcZSe[]S` YS_TSc TWc[]fe& ¥[]S TWTS` ; V[^WbSd]S`& TS^a] : eWbSe S]S` TWcYWcS]( C[]S bWcUWbSeS` YcSg[eSd[ +* _)d, & ]aWX[d[W` YWdW] deSe[d S`eScS TS^a] VS` _W¥S SVS^SZ (((( S( *&/ V( *&, T( *&. W( *&+ U( *&- 1( IWcZSe[]S` dfdf`S` TW`VS'TW`VS bSVS YS_TSc TWc[]fe( DaWX[d[W` YWdW] ][`We[] S`eScS TS^a] : VS` _W¥S SVS^SZ *&+,/( C[]S TWTS` ; V[^WbSd]S`& bWcUWbSeS` TW`VS : SVS^SZ (((( "= 7 +* _)d, # S( ,&,/ _)d, V( +&1/ _)d, T( ,&** _)d, W( +&/* _)d, U( +&** _)d, 2( IWcZSe[]S` YS_TSc TWc[]fe [`[( DaWX[d[W` YWdW] ][`We[] S`eScS TW`VS VW`YS` T[VS`Y _[c[`Y SVS^SZ *&/ VS` bWcUWbSeS` YcSg[eSd[ ;f_[ 7 =( ;WdSc`iS YSiS iS`Y V[TfefZ]S` f`ef] _W`Vaca`Y TW`VS VW`YS` _SddS B dfbSiS VSbSe TWcYWcS] ]W SeSd VW`YS` ]WUWbSeS` eWeSb SVS^SZ (((( S( *&/ B= T( *&2 B= U( +&- B= V( +&/ B= W( +&+ B=
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Tes Kompetensi Fisika Semester 1 113 ,*( FSddS ¥Sc['¥Sc[ b^S`We O dS_S VW`YS` 7 ]S^[ _SddS ;f_[ VS` TWc¥Sc['¥Sc[ 8 ]S^[ ¥Sc['¥Sc[ ;f_[( ;WcSe TW`VS V[ b^S`We O V[TS`V[`Y]S` VW`YS` TWcSe`iS V[ ;f_[ _W`¥SV[ (((( S( 78 ]S^[ T( 78' ]S^[ U( a b ]S^[ V( , 7 8 ]S^[ W( + 78 ]S^[ ,+( JWTfSZ dSeW^[e TWcSVS bSVS ]We[`YY[S` > V[ SeSd bWc_f]SS` ;f_[( C[]S V[]WeSZf[ ¥Sc['¥Sc[ ;f_[ SVS^SZ 5& bWc[aVW dSeW^[e eWcdWTfe _W`YW^[^[`Y[ ;f_[ SVS^SZ (((( "acT[e dSeW^[e V[S`YYSb TWcfbS ^[`Y]ScS`# S( " # , 5 > = " T( , 5 > " U( 5 > " V( , " # . 5 > = " W( , " # . , 5 > = " ,,( JWTfSZ TW`VS iS`Y TWcSVS V[ ;f^S` TWcYWcS] VW`YS` bWcUWbSeS` 7 ]ScW`S _W`VSbSe YSiS , ( C[]S TW`VS eWcdWTfe V[TShS ]W ;f_[ iS`Y YSiS YcSg[eSd[`iS W`S_ ]S^[ _SddS ;f^S` VS` TW`VS eWcdWTfe V[TWc[ YSiS , ¥fYS& bWcUWbSeS` SVS^SZ (((( S( 07 T( 7 0 U( 7 V( 0 7 W( 0 7 ,-( FW`fcfe Af]f_ BBB DWb^Wc& ]fSVcSe bWc[aVW dfSef b^S`We _W`YW^[^[`Y[ FSeSZSc[ dWTS`V[`Y VW`YS` (((( S( _SddS b^S`We T( ¥ScS] b^S`We ]W FSeSZSc[ U( bS`Y]Se e[YS _SddS b^S`We V( bS`Y]Se e[YS ¥ScS] b^S`We ]W FSeSZSc[ W( ]fSVcSe ¥ScS] FSeSZSc[ ]W b^S`We ,.( C[]S bWc[aVW dSeW^[e iS`Y _W`YacT[e ;f_[ VW`YS` ¥Sc[' ¥Sc[ 7 +,&2 m +*0 _ dWTWdSc -&3 ¥S_& bWc[aVW dSeW^[e iS`Y ¥Sc['¥Sc[ acT[e`iS 7 +3&, m +*0 _ SVS^SZ (((( S( 1&, ¥S_ V( .&2 ¥S_ T( 0&. ¥S_ W( ,&+ ¥S_ U( /&3 ¥S_ ,/( JWacS`Y `W^SiS` _W`Sc[] bWcSZf`iS VW`YS` dfVfe ./k eWcZSVSb ScSZ Zac[ja`eS^ VS` bWcSZf TWcb[`VSZ dW¥SfZ +* _( C[]S `W^SiS` eWcdWTfe _W`Sc[] bWcSZf VW`YS` YSiS +2* G& TWdSc fdSZS iS`Y V[^S]f]S` `W^SiS` eWcdWTfe SVS^SZ (((( S( 3** C V( 2/* C T( 3** 2 C W( 21/ C U( 3** 3 C ,0( JWTfSZ Ta^S TWc_SddS .** Y V[^W_bSc gWce[]S^ ]W SeSd Z[`YYS _W`USbS[ ]We[`YY[S` 2 _( IWcfTSZS` W`WcY[ baeW`d[S^ ]We[]S Ta^S TWcSVS bSVS ]We[`YY[S` . _ SVS^SZ (((( S( +, C V( l+0 C T( +0 C W( -, C U( l+, C ,1( JWTfSZ TW`VS ¥SefZ TWTSd VSc[ eW_bSe iS`Y e[`YY[`iS 2* _( C[]S W`WcY[ baeW`d[S^`iS _f^S'_f^S dWTWdSc .(*** C VS` = 7 +* _)d, & ]WUWbSeS` TW`VS eWbSe dWTW^f_ dS_bS[ V[ eS`SZ SVS^SZ (((( S( +* _)d V( -* _)d T( ,* _)d W( .* _)d U( ,/ _)d ,2( JWTfSZ _Wd[` ]W`VScSS` TWc_aeac _W_[^[][ VSiS dWTWdSc +** Zb( NS]ef iS`Y V[bWc^f]S` _Wd[` eWcdWTfe f`ef] _W`YZSd[^]S` YSiS -(1-* G VS` _W`YYWcS]]S` ]W`VScSS` dW¥SfZ 0* _ SVS^SZ (((( S( , dW]a` V( / dW]a` T( - dW]a` W( 0 dW]a` U( . dW]a` ,3( JWTfSZ YcS`Se TWc_SddS *&/ ]Y e[TS'e[TS _W^WVS] VS` bWUSZ _W`¥SV[ VfS TSY[S` VW`YS` bWcTS`V[`YS` _SddS bWUSZS` , 4 - VS` TWcYWcS] TWc^ShS`S` ScSZ( C[]S bWUSZS` iS`Y ]WU[^ _W_[^[][ ]WUWbSeS` ,.* _)d& bWcTS`V[`YS` W`WcY[ ][`We[] bWUSZS` bWceS_S VS` bWUSZS` ]WVfS dWdSSe dWeW^SZ YcS`Se eWcdWTfe _W^WVS] SVS^SZ (((( S( + 4 + V( 3 4 . T( - 4 , W( , 4 + U( , 4 - -*( @Sd bS`Sd iS`Y V[ZSd[^]S` a^WZ dWTfSZ ca]We _W_[^[][ ]W^S¥fS` .** _)d VS` eWc¥SV[ bW_TS]ScS` YSd dWTS`iS] 2* ]Y VS^S_ hS]ef dSef dW]a`( ;WdSc`iS YSiS Vaca`Y ca]We eWcdWTfe SVS^SZ (((( S( ,(*** G V( ,.(*** G T( +,(*** G W( -,(*** G U( +0(*** G
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115 A. Kinematika Gerak Rotasi B. Dinamika Gerak Rotasi C. Kesetimbangan Benda Tegar ER^[NUXNU 6[QN ZRYVUN` TR^NXN[ `aOaU _R¥^N[T N`YR` ]RY¥[PN` V[QNU5 BR[TN]N N`YR` `R^_ROa` _RYNYa ZR[RXaX `aOaU[dN XR`VXN ZaYNV OR^]a`N^5 @R`VXN XRQaN YR[TN[ QN[ XNXV[dN QV`RXaX$ VN ZR[TNYNZV TR^NX ^¥`N_V YROVU PR]N` QVON[QV[TXN[ ]NQN _NN` QV^R[`N[TXN[& =NY V[V QV]R^¥YRU NXVON` ]R[Ta^N[TN[ Z¥ZR[ V[R^_VN dN[T QVZVYVXV `aOaU[dN& @RZaQVN[$ VN XRZONYV ZR^R[`N[TXN[ `aOaU[dN XR`VXN UR[QNX ZR[QRXN`V NV^ NTN^ ZRZ]R^¥YRU Z¥ZR[ V[R^_VN dN[T YROVU OR_N^ _RUV[TTN XRPR]N`N[ _aQa`[dN ZR[WNQV YROVU XRPVY$ YNYa ZN_aX XR QNYNZ NV^ QR[TN[ UNYa_ `N[]N `R^QR[TN^ Oa[dV ]R^PVXN[ NV^& 9NYNZ ONO V[V$ 6[QN NXN[ ZRZ]RYNWN^V ]^V[_V]%]^V[_V] SV_VXN dN[T ZR[WNQV NPaN[ QN^V ]R^V_`VcN `R^_ROa`& 9NYNZ XRUVQa]N[ _RUN^V%UN^V$ 6[QN NXN[ ON[dNX ZR[RZaXN[ N]YVXN_V ]R[R^N]N[ X¥[_R] QV[NZVXN ^¥`N_V QN[ XR_R`VZON[TN[ OR[QN `RTN^& Gerak Rotasi dan Kesetimbangan Benda Tegar Peloncat indah menekuk tubuhnya ketika mulai berputar dan merentangkan tubuhnya ketika hendak mendekati air. Sumber: Contemporary College Physics, 1993 memformulasikan hubungan antara konsep torsi, momentum sudut, dan momen inersia berdasarkan Hukum II Newton, serta penerapannya dalam masalah benda tegar. Setelah mempelajari bab ini, Anda harus mampu: menerapkan konsep dan prinsip mekanika klasik sistem kontinu dalam menyelesaikan masalah. Hasil yang harus Anda capai: Bab 5
116 Mudah dan Aktif Belajar Fisika untuk Kelas XI )& 6]N dN[T QVZNX_aQ QR[TN[ Z¥ZR[ TNdN$ Z¥ZR[ X¥]RY$ QN[ Z¥ZR[ V[R^_VN5 *& 6]N UaOa[TN[ N[`N^N =aXaZ @RXRXNYN[ B¥ZR[% `aZ HaQa` QN[ =aXaZ >> @R]YR^5 <B?BHPI IBILBH>F>ME GKJNBL 4BM>G ;KO>NE A>J 7BNBOEI?>JC>J 2BJA> =BC>M# GBMF>G>JH>D NK>H$NK>H ?BMEGPO A>H>I ?PGP H>OED>J% +& HROa`XN[ WR[V_%WR[V_ XR_R`VZON[TN[& ?RYN_XN[ QR[TN[ ONUN_N 6[QN _R[QV^V& ,& 6]N ]R[TR^`VN[ QN^V `V`VX OR^N` OR[QN5 Tes Kompetensi Awal A. Kinematika Gerak Rotasi ?VXN 6[QN ZRZ]R^UN`VXN[ _ROaNU ]V^V[TN[ UV`NZ dN[T _RQN[T OR^]a`N^ N`Na ONYV[T%ONYV[T XV]N_ N[TV[ dN[T _RQN[T OR^]a`N^$ 6[QN NXN[ ZRYVUN` ONUcN ]V^V[TN[ UV`NZ QN[ ONYV[T%ONYV[T XV]N_ N[TV[ `R^_ROa` ZRYNXaXN[ TR^NX ^¥`N_V ]NQN _ROaNU _aZOa _`N_V¥[R^ !`R`N]" QNYNZ XR^N[TXN NPaN[ V[R^_VN `R^`R[`a& 8¥[`¥U YNV[ QN^V OR[QN `RTN^ dN[T _RQN[T OR^¥`N_V NQNYNU ON[ _R]RQN dN[T _RQN[T OR^]a`N^ QN[ X¥ZVQV ]a`N^& 1. Posisi Sudut dalam Gerak Rotasi ER^UN`VXN[ 4>I?>M +%( dN[T ZR[QR_X^V]_VXN[ _ROaNU OR[QN `RTN^ OR^OR[`aX YV[TXN^N[ dN[T OR^¥`N_V ]NQN _aZOa `R`N] 1 QN[ `RTNX Ya^a_ `R^UNQN] OVQN[T LM$ IV`VX E `R^YR`NX ]NQN OR[QN _RUV[TTN TN^V_ DE OR^¥`N_V OR^_NZN OR[QN `R^_ROa`& HaQa` * dN[T QVOR[`aX ¥YRU TN^V_ DE QN[ _aZOa-L ZR[QR_X^V]_VXN[ ]¥_V_V OR[QN _NN` OR[QN OR^¥`N_V& HaQa` * QVOR[`aX ¥YRU _ROaNU Oa_a^ G ]NQN YV[TXN^N[ OR^WN^V%WN^V F$ CVYNV _aQa` * !QNYNZ ^NQVN[" QV[dN`NXN[ QR[TN[ ]R^_NZNN[ G F * $ !-h)" @¥¥^QV[N` * ZR[R[`aXN[ ]¥_V_V ^¥`N_V QN^V OR[QN `RTN^ ]NQN _aN`a cNX`a$ * 4 * !H"& ?VXN ]NQN _NN` H ]¥_V_V _aQa` OR[QN * ZNXN ]NQN _NN` H" $ H$ ]¥_V_V _aQa`[dN * * " $ & 2. Kecepatan Sudut Rata-Rata dalam Gerak Rotasi HNZN UNY[dN QR[TN[ TR^NX YV[RN^$ QNYNZ TR^NX ^¥`N_V$ XRPR]N`N[ _aQa` ^N`N%^N`N QVQRSV[V_VXN[ _RONTNV ]R^ON[QV[TN[ N[`N^N ]R^aONUN[ _aQa` $* `R^UNQN] ]R^aONUN[ cNX`a ! $H "& ER^UN`VXN[ 4>I?>M +%)% <N^V_ DE ]NQN OR[QN `RTN^ OR^OR[`aX YV[TXN^N[ dN[T OR^¥`N_V ZRZOR[`aX _aQa` *) `R^UNQN] _aZOa-L ]NQN cNX`a H ) & @RZaQVN[$ ]NQN _NN` H * $ _aQa` dN[T QVOR[`aX[dN OR^aONU ZR[WNQV ** $ 9R[TN[ QRZVXVN[$ XRPR]N`N[ _aQa` ^N`N%^N`N[dN NQNYNU * ) ^N`N%^N`N * ) HH H * * * , % $ $ $ % $ !-h*" DYRU XN^R[N * ZRZVYVXV _N`aN[ QR^NWN`$ ^NQVN[$ N`Na ]a`N^N[$ XRPR]N`N[ _aQa` ^N`N%^N`N QN]N` QV[dN`NXN[ QR[TN[ _N`aN[ QR^NWN`' _RX¥[$ ^NQVN['_RX¥[$ N`Na ]a`N^N['_RX¥[& Gambar 5.1 Baling-baling kipas berotasi pada porosnya. Gambar 5.2 Sebuah benda tegar berbentuk lingkaran berotasi pada sumbu koordinat O. y x P r O * Gambar 5.3 Perpindahan sudut sebesar $* dari sebuah benda tegar yang berotasi pada saat t1 dan t2 . y x P,t2 P,t1 O *) ** $* s
Gerak Rotasi dan Kesetimbangan Benda Tegar 117 3. Kecepatan Sudut Sesaat dalam Gerak Rotasi 7NTNVZN[NXNU PN^N ZR[TaXa^ XRPR]N`N[ _aQa` _R_NN` QN^V _ROaNU OR[QN `RTN^ dN[T OR^TR^NX ^¥`N_V5 HNZN UNY[dN QR[TN[ TR^NX YV[RN^$ QNYNZ TR^NX ^¥`N_V$ XRPR]N`N[ _aQa` _R_NN` !, " QV`R[`aXN[ QNYNZ _RYN[T cNX`a ! $H " dN[T _N[TN` _V[TXN` ! $ &H ( " _RUV[TTN ]R^aONUN[ XRPR]N`N[[dN _N[TN` XRPVY& DYRU XN^R[N V`a$ XRPR]N`N[ _aQa` _R_NN` ]NQN TR^NX ^¥`N_V QV[dN`NXN[ QR[TN[ ]R^_NZNN[ ( ( YVZ YVZ H H H * , , $& $& $ $ $ $ N`Na : :H * , $ !-h+" 9R[TN[ QRZVXVN[$ XRPR]N`N[ _aQa` _R_NN` ZR^a]NXN[ `a^a[N[ ]R^`NZN QN^V ]R^_NZNN[ ]¥_V_V _aQa` _RONTNV Sa[T_V cNX`a& ER^UN`VXN[ T^NSVX * `R^UNQN] H ]NQN 4>I?>M +%*& ?VXN 2 1 $tt t $ % _N[TN` XRPVY N`Na ZR[QRXN`V [¥Y$ `V`VX F NXN[ _N[TN` QRXN` QR[TN[ E$ N`Na ONUXN[ OR^VZ]V` _RUV[TTN XRZV^V[TN[ !T^NQVR[" T^NSVX* `R^UNQN] H QV `V`VX E ZR^a]NXN[ XRPR]N`N[ _aQa` _R_NN`& DYRU XN^R[N V`a$ XRPR]N`N[ _aQa` _R_NN` QN]N` ]aYN QV[dN`NXN[ QR[TN[ ]R^_NZNN[ OR^VXa`& `N[ : :H * , ) $ $ !-h," 4. Percepatan Sudut dalam Gerak Rotasi HROaNU ]N^`VXRY dN[T OR^TR^NX ^¥`N_V NXN[ ZRZVYVXV ]R^PR]N`N[ _aQa` WVXN ]N^`VXRY `R^_ROa` ZR[TNYNZV ]R^aONUN[ XRPR]N`N[ _aQa`& 9R[TN[ QRZVXVN[$ ]R^PR]N`N[ _aQa` NQNYNU ]R^aONUN[ XRPR]N`N[ _aQa` _R`VN] _N`aN[ cNX`a _RUV[TTN _N`aN[[dN NQNYNU ^NQ'_* & a. Percepatan Sudut Rata-Rata ER^PR]N`N[ _aQa` ^N`N%^N`N QV[dN`NXN[ QR[TN[ YNZON[T ( & ?VXN _ROaNU ]N^`VXRY OR^TR^NX ZRYV[TXN^ QR[TN[ XRPR]N`N[ _aQa` ,1 ]NQN _NN` H ) $ XRZaQVN[ OR^aONU ZR[WNQV ,2 ]NQN _NN` H * $ ]R^PR]N`N[ _aQa` ^N`N%^N`N !( " ]N^`VXRY `R^_ROa` QNYNZ _RYN[T cNX`a DH QV[dN`NXN[ QR[TN[ ]R^_NZNN[ OR^VXa`& * ) * ) H H , , ( % $ % N`Na , ( $ $ $H !-h-" b. Percepatan Sudut Sesaat 6[QN `RYNU ZR[TR`NUaV ONUcN XRPR]N`N[ _aQa` _R_NN` !, " QV`R[`aXN[ QNYNZ _RYN[T cNX`a dN[T _N[TN` _V[TXN` ! $ &H ( " _RUV[TTN ]R^aONUN[ XRPR]N`N[[dN _N[TN` XRPVY& HNZN UNY[dN QR[TN[ XRPR]N`N[ _aQa` _R_NN` !, "$ ]R^PR]N`N[ _aQa` _R_NN` !( " WaTN ZR^a]NXN[ ]R^PR]N`N[ _aQa` ^N`N% ^N`N dN[T `R^WNQV QNYNZ _RYN[T cNX`a ! $H " dN[T _N[TN` _V[TXN`$ N`Na $ &H (& 9R[TN[ QRZVXVN[$ ]R^PR]N`N[ _aQa` _R_NN` QV[dN`NXN[ QR[TN[ ]R^_NZNN[ OR^VXa`& t1 t1 t(s) P Q 0 * ! ^NQ *) ** ) Gambar 5.4 Kecepatan sudut sesaat (, ) dapat ditentukan dari gradien grafik * terhadap t pada sebuah titik.
118 Mudah dan Aktif Belajar Fisika untuk Kelas XI ( YVZ H ( ( $ & $ ( YVZ H H , ( $ & $ $ $ : :H , ( $ !-h." 9R[TN[ ZR[_aO_`V`a_VXN[ :BMN>I>>J !+S)" XR QNYNZ :BMN>I>>J !+S,"$ QV]R^¥YRU ]R^_NZNN[ ]R^PR]N`N[ _aQa` _RONTNV OR^VXa`& * * : :: : :H :H :H :H , ** ( 2 4 $$ $ 6 8 3 5 !-h/" 9R[TN[ QRZVXVN[$ 6[QN `RYNU ZR[TR`NUaV ONUcN ]R^PR]N`N[ _aQa` _R_NN` ZR^a]NXN[ `a^a[N[ ]R^`NZN QN^V XRPR]N`N[ _aQa` _R_NN` N`Na `a^a[N[ XRQaN QN^V ]R^_NZNN[ ]¥_V_V _aQa` _RONTNV Sa[T_V cNX`a& HRYNV[ V`a$ ]R^PR]N`N[ _aQa` _R_NN` QN]N` ]aYN QV`R[`aXN[ OR^QN_N^XN[ T^NQVR[ TN^V_ _RUV[TTN T^NSVX , `R^UNQN] H QV _aN`a `V`VX& ER^UN`VXN[ T^NSVX , 4 ,!H" ]NQN 4>I?>M +%+& ER^PR]N`N[ _aQa` _R_NN` QV `V`VX 6 NQNYNU `N[(1 , _RQN[TXN[ ]R^PR]N`N[ _aQa` _R_NN` QV `V`VX 7 NQNYNU `N[ 2 ( . 5. Gerak Rotasi dengan Kecepatan Sudut Konstan ?VXN _ROaNU OR[QN `RTN^ OR^TR^NX ^¥`N_V QR[TN[ XRPR]N`N[ _aQa` !, " X¥[_`N[$ ]R^_NZNN[ TR^NX QN^V OR[QN `RTN^ `R^_ROa` QV[dN`NXN[ QR[TN[ ]R^_NZNN[ , : :H * $ :* $, :H * * # ( : ( H $ , :H # * * ! % ( H ( H $, :H # * * ! % ( H $,H ! ( * *, H H $ " !-h0" QR[TN[ 0 * NQNYNU _aQa` dN[T QV`RZ]aU ]NQN _NN` H 4 (& 6XN[ `R`N]V$ WVXN ]N^`VXRY `R^_ROa` ZR[TNYNZV ]R^aONUN[ XRPR]N`N[ _aQa` _RUV[TTN ZR[TUN_VYXN[ ]R^PR]N`N[ _aQa`$ ]R^_NZNN[ TR^NX ^¥`N_V QN^V OR[QN `RTN^ `R^_ROa` QN]N` QV`a^a[XN[ QN^V :BMN>I>>J !+S-"$ QR[TN[ [VYNV( `R`N]& ( : :H , $ : : , $( H , , , # ( : ( H $ (:H # , , ! % ( H ( H $( :H # , , ! % ( H $(H ! ( , ,( H H $ " !-h1" Kata Kunci • posisi sudut • kecepatan sudut • percepatan sudut Gambar 5.5 Percepatan sudut sesaat berdasarkan gradien garis singgung , = , (t) di titik t. , (rad/s1 ) , 2 , 1 t 1 t 1 t (s) B A ( 1 ( 2
Gerak Rotasi dan Kesetimbangan Benda Tegar 119 7R^QN_N^XN[ :BMN>I>>J !+S." QN[ :BMN>I>>J !+S/"$ QV]R^¥YRU ]R^_NZNN[ ]¥_V_V _aQa` a[`aX TR^NX ZRYV[TXN^ OR^N`a^N[ _RONTNV OR^VXa`& ! * * * , $ # # ( ( H : H :H ! ( ( H $ " , (H :H # ** , ( % $ " * ( ( ) * H H ! * ( ( ) * * *, ( H H $" " H !-h)(" Tugas Anda 5.1 Buatlah perbandingan antara persamaan posisi, kecepatan, dan percepatan untuk gerak lurus berubah beraturan dan gerak rotasi benda tegar. Tes Kompetensi Subbab A 7BMF>G>JH>D A>H>I ?PGP H>OED>J% )& HROaNU ^¥QN OR^WN^V%WN^V +( PZ OR^¥`N_V QR[TN[ _aQa` ^¥`N_V dN[T QV[dN`NXN[ QR[TN[ ]R^_NZNN[ * 4 !, ^NQ'_* "H * & IR[`aXN[YNU N& OR_N^ _aQa` * ]NQN cNX`a H ) 4 * _ QN[ H * 4 ,$(- QNYNZ _N`aN[ ^NQVN[ QN[ QR^NWN`3 O& WN^NX `RZ]aU ]N^`VXRY ]NQN `R]V ^¥QN _RYNZN V[`R^bNY cNX`a H ) QN[ H * 3 P& XRPR]N`N[ _aQa` ^N`N%^N`N QNYNZ _N`aN[ ^NQ'_ N[`N^N H ) QN[ H * 3 QN[ Q& XRPR]N`N[ _aQ` _R_NN` ]NQN ` 4 + _& *& HROaNU ^¥QN _R]RQN dN[T _RQN[T OR^]a`N^ ZRZVYVXV QVNZR`R^ +0 PZ& ?VXN _ROaNU `V`VX ]NQN `R]V ^¥QN ZR[RZ]RY -.$- XNYV ]NQN `N[NU$ OR^N]NXNU ]N[WN[T YV[`N_N[ dN[T QVYNYaV ¥YRU ^¥QN _R]RQN `R^_ROa`5 +& HROaNU ]V^V[TN[ 6)* OR^¥`N_V QR[TN[ XRPR]N`N[ _aQa` , dN[T QV[dN`NXN[ QR[TN[ ]R^_NZNN[ , 4 !0 ^NQ'_+ "H * & IR[`aXN[YNU ]R^PR]N`N[ _aQa` ^N`N%^N`N N[`N^N H ) 4 ) _ QN[ H * 4 * _ QN[ OR_N^ ]R^PR]N`N[ _aQa` _R_NN` ]NQN _NN` H 4 - _& B. Dinamika Gerak Rotasi 1. Momen Gaya (Torsi) /DB;C =7M7 N`Na `¥^_V ZR^a]NXN[ OR_N^N[ dN[T ZR[dRONOXN[ _ROaNU OR[QN `RTN^ !OR[QN dN[T `VQNX QN]N` OR^aONU OR[`aX" PR[QR^a[T a[`aX OR^¥`N_V `R^UNQN] ]¥^¥_[dN& B¥ZR[ TNdN `R^ZN_aX OR_N^N[ bRX`¥^ QN[ QVOR^V YNZON[T + !`Na"& 7NTNVZN[NXNU PN^N ZR[R[`aXN[ `¥^_V N`Na Z¥ZR[ TNdN dN[T ORXR^WN ]NQN OR[QN dN[T OR^¥`N_V5 ER^UN`VXN[ XRZONYV 4>I?>M +%, !>"& @a[PV Za^ dN[T ZRZVYVXV ]N[WN[T F QVTa[NXN[ a[`aX ZRZa`N^ Za^ ]NQN ^¥QN Z¥OVY& @R`VXN Za^ OR^]a`N^ NXVON` TNdN dN[T QVOR^VXN[ ]NQN Xa[PV Za^$ Za^ OR^]R^N[ _RONTNV R[T_RY N`Na ]a_N` _aZOa ]a`N^& B¥ZR[ TNdN QV^aZa_XN[ QR[TN[ ]R^_NZNN[ + 4 $ i ! !-h))" 7R^QN_N^XN[ :BMN>I>>J !+S''"$ QVXR`NUaV ONUcN `¥^_V N`Na Z¥ZR[ TNdN ZRZVYVXV _N`aN[ [Rc`¥[ ZR`R^ !CZ"& ER^UN`VXN[ 4>I?>M +%,!?" dN[T ZR[a[WaXXN[ QVNT^NZ Z¥ZR[ TNdN dN[T ORXR^WN ]NQN _NN` ZRZOaXN Za^ ^¥QN Z¥OVY& @¥Z]¥[R[ TNdN 3 dN[T ORXR^WN ]NQN ON`N[T $ ZRZOR[`aX _aQa` * $ N^NU ]R^]a`N^N[[dN _RN^NU QR[TN[ N^NU ]a`N^N[ WN^aZ WNZ& @¥Z]¥[R[ TNdN 3 dN[T ZR[dRONOXN[ Za^ OR^¥`N_V NQNYNU ; _V[ * $ dNV`a X¥Z]¥[R[ TNdN dN[T `RTNX Ya^a_ `R^UNQN] $& DYRU XN^R[N V`a$ OR_N^ Z¥ZR[ TNdN dN[T QVUN_VYXN[ NQNYNU + 4 F, _V[ * 4 ,: !-h)*" Gambar 5.6 (a) Gaya F dibutuhkan untuk membuka mur pada roda mobil. (b) Diagram momen gaya atau torsi yang bekerja saat membuka mur pada roda mobil. !N" r F y O r d x F F sin * + * !O"
120 Mudah dan Aktif Belajar Fisika untuk Kelas XI @R`R^N[TN[2 + 4 OR_N^ Z¥ZR[ TNdN !CZ" F 4 WN^NX _aZOa ^¥`N_V XR `V`VX `N[TXN] !Z" , 4 OR_N^ TNdN dN[T QVXR^WNXN[ !C" : 4 F _V[ * 4 YR[TN[ Z¥ZR[ !Z" ER^UN`VXN[ ]R^_NZNN[ Z¥ZR[ TNdN ]NQN :BMN>I>>J !+S''"& 7R_N^ Z¥ZR[ TNdN QVUN_VYXN[ QN^V ]R^XNYVN[ bRX`¥^& ?VXN _aQa` N[`N^N F QN[ TNdN , `RTNX Ya^a_ !* 4 1(f"$ NXN[ QV]R^¥YRU OR_N^ Z¥ZR[ TNdN + _ROR_N^ + 4 F, !-h)+" 6^NU QN^V Z¥ZR[ TNdN !+ " QN]N` QV`R[`aXN[ OR^QN_N^XN[ X¥[bR[_V N`Na XR_R]NXN`N[ OR^VXa` V[V& ?VXN N^NU ]a`N^N[ `¥^_V _RN^NU QR[TN[ N^NU ]a`N^N[ WN^aZ WNZ$ Z¥ZR[ TNdN !+ " OR^[VYNV ]¥_V`VS& ?VXN N^NU ]a`N^N[ `¥^_V OR^YNcN[N[ N^NU QR[TN[ N^NU ]a`N^N[ WN^aZ WNZ$ Z¥ZR[ TNdN !+ " OR^[VYNV [RTN`VS& 2. Momen Inersia 6[QN `R[`a ZR[TR`NUaV ONUcN _R`VN] OR[QN ZRZVYVXV XRPR[QR^a[TN[ a[`aX ZRZ]R^`NUN[XN[ XRNQNN[ TR^NX[dN& 9NYNZ TR^NX YV[RN^$ _R`VN] OR[QN dN[T QVNZ NXN[ `R`N] QVNZ& 6QN]a[ OR[QN dN[T _RQN[T OR^TR^NX Ya^a_ OR^N`a^N[ ZRZVYVXV XRPR[QR^a[TN[ a[`aX `R`N] OR^TR^NX Ya^a_ OR^N`a^N[$ XRPaNYV NQN ^R_aY`N[ TNdN dN[T ZRZR[TN^aUV[dN& =NY `R^_ROa` _R_aNV QR[TN[ =aXaZ > CRc`¥[& @RPR[QR^a[TN[ OR[QN a[`aX ZRZ]R^% `NUN[XN[ XRNQNN[[dN QV_ROa` ?C;FG?7 N`Na B7GG7& 9RZVXVN[ ]aYN UNY[dN QR[TN[ QV[NZVXN TR^NX ^¥`N_V& 7R[QN dN[T _RQN[T OR^¥`N_V ZRZVYVXV XRPR[QR^a[TN[ a[`aX `R`N] ZRZ]R^`NUN[XN[ TR^NX ^¥`N_V[dN& @RPR[QR^a[TN[ `R^_ROa` QV[NZNXN[ BDB;C ?C;FG?7& B¥ZR[ V[R^_VN dN[T OR^¥`N_V QV]R[TN^aUV ¥YRU ZN__N QN[ ]¥YN QV_`^VOa_V ZN__N `R^UNQN] _aZOa ]a`N^& a. Momen Inersia Benda Diskrit (Partikel) ?VXN _ROaNU ]N^`VXRY dN[T OR^ZN__N B OR^]a`N^ ZR[TRYVYV[TV _aZOa ]a`N^ dN[T OR^WN^NX F QN^V ]N^`VXRY `R^_ROa`$ QN]N`XNU 6[QN ZR[TUV`a[T OR_N^[dN R[R^TV XV[R`VX QN^V ]N^`VXRY `R^_ROa`5 ER^UN`VXN[ 4>I?>M +%-& 9NYNZ TR^NX ZRYV[TXN^$ XRPR]N`N[ YV[RN^ QV[dN`NXN[ QR[TN[ J 4 , ^$ QR[TN[ , XRPR]N`N[ _aQa`& DYRU XN^R[N V`a$ OR_N^ R[R^TV XV[R`VX ^¥`N_V QN^V ]N^`VXRY QN]N` QV[dN`NXN[ QR[TN[ * * ^¥`N_V ) * + B @ $ F J !-h)," 9N^V :BMN>I>>J !+S'*"$ QV]R^¥YRU [VYNV BF* dN[T ZR[dN`NXN[ Z¥ZR[ V[R^_VN QN^V ]N^`VXRY dN[T OR^TR^NX ZRYV[TXN^& B¥ZR[ V[R^_VN QVYNZON[TXN[ QR[TN[ -& - 4 BF* !-h)-" 9R[TN[ QRZVXVN[$ Z¥ZR[ V[R^_VN _ROaNU ]N^`VXRY _RON[QV[T QR[TN[ ZN__N ]N^`VXRY QN[ XaNQ^N` WN^NX N[`N^N ]N^`VXRY QN[ _aZOa ]a`N^[dN& B¥ZR[ V[R^_VN ZR^a]NXN[ OR_N^N[ _XNYN^ dN[T ZRZVYVXV _N`aN[ XTZ* & Gambar 5.7 Partikel bermassa m berputar mengelilingi sebuah sumbu putar yang berjarak r dari partikel tersebut. , r m v
Gerak Rotasi dan Kesetimbangan Benda Tegar 121 Gambar 5.8 Momen inersia benda pejal dihitung dengan metode integral terhadap r 2 dm. y x dm r o sumbu rotasi Gambar 5.9 Sebuah batang homogen memiliki massa m dengan panjang batang L. x L x dm y 0 x1 x2 b. Momen Inersia Benda Tegar HROaNU OR[QN `RTN^ `R^QV^V N`N_ _RWaZYNU ]N^`VXRY dN[T `R^]V_NU _N`a QR[TN[ dN[T YNV[[dN _R^`N WN^NX[dN `R`N]& B¥ZR[ V[R^_VN[dN ZR^a]NXN[ WaZYNU QN^V Z¥ZR[ V[R^_VN _RZaN ]N^`VXRY V`a$ dNX[V - 4 ' BF* & 6XN[ `R`N]V$ a[`aX _ROaNU OR[QN dN[T ZRZVYVXV QV_`^VOa_V ZN__N dN[T X¥[`V[a$ N`Na `VQNX QN]N` QV]V_NUXN[ _R]R^`V 4>I?>M +%.$ OR^YNXa ]R^_NZNN[ - 4 * F :B # !-h)." HROaNU OR[QN OR^ZN__N B dN[T OR^OR[`aX ON`N[T QR[TN[ ]N[WN[T . `NZ]NX _R]R^`V 4>I?>M +%/& 7R[QN `R^_ROa` ZRZVYVXV Z¥ZR[ V[R^_VN `R^UNQN] ]¥^¥_[dN QV D& B¥ZR[ V[R^_VN[dN QV`R[`aXN[ QR[TN[ PN^N V[`RT^N_V QN^V _RYa^aU ZN__N dN[T NQN$ QR[TN[ F NQNYNU WN^NX RYRZR[ ZN__N :B XR _aZOa ]a`N^& IV[WNa _ROaNU ON`N[T dN[T ZRZVYVXV ZN__N B _R^ON _NZN QN[ ]N[WN[T ON`N[T .& 9N^V :BMN>I>>J !+S',"$ QVXR`NUaV ONUcN F 4 L QN[ :B 4 / :/ . _R^`N ON`N_%ON`N_ V[`RT^NY[dN NQNYNU L ) QN[ L * $ NXN[ QV]R^¥YRU * * ) L L / -L :L . 2 4 $ 6 8 3 5 # !-h)/" J[`aX _aZOa ]a`N^ dN[T `R^YR`NX ]NQN aWa[T ON`N[T _R]R^`V 4>I?>M +%'& QR[TN[ L) 4 ( QN[ L * 4 . ZNXN Z¥ZR[ V[R^_VN ON`N[T ZR[WNQV $ # * ( / . - L :L . 4 + ( ) + . / L . . 0 7 9 / 1 4 ) + + / . . . 0 7 9 / 1 4 ) + /.* !-h)0" J[`aX _aZOa ]a`N^ dN[T `R^YR`NX ]NQN ]R^`R[TNUN[ ON`N[T$ _R]R^`V `NZ]NX ]NQN 4>I?>M +%''$ NXN[ QV]R^¥YRU L ) 4 h ) * . QN[ L * 4 ) * .& DYRU XN^R[N V`a$ Z¥ZR[ V[R^_VN ON`N[T NXN[ ZR[WNQV ! ! ) ) * * * + ) ) * * + + * ) + )) ) ) ) + * + * )* . . . . - L / / :L L . . / . ./. . % . 0 $ $ 7 9 / 1 . 0 $ % % $ 7 9 / 1 # !-h)1" B¥ZR[ V[R^_VN OR[QN dN[T _aZOa ]a`N^[dN `R^YR`NX ]NQN ]a_N` ZN__N QV_ROa` Z¥ZR[ V[R^_VN ]a_N` ZN__N !- ]Z "& ?VXN _aZOa ]a`N^[dN `VQNX `R^YR`NX ]NQN ]a_N` ZN__N$ a[`aX ZR[PN^V Z¥ZR[ V[R^_VN[dN$ QN]N` QVTa[NXN[ ]R^_NZNN[ OR^VXa` V[V dN[T QV_ROa` _RONTNV XNVQNU _aZOa _RWNWN^& - 4 - ]Z # B:* @R`R^N[TN[2 B 4 ZN__N OR[QN : 4 WN^NX QN^V ]a_N` ZN__N XR _aZOa ]a`N^ B¥ZR[ V[R^_VN QN^V OROR^N]N OR[`aX OR[QN QR[TN[ ]¥_V_V _aZOa `R^`R[`a QN]N` QVYVUN` ]NQN `NORY OR^VXa` V[V& Gambar 5.10 Momen inersia untuk sumbu putar yang terletak pada ujung batang adalah I = 1 3 ML2 . x L y 0 Gambar 5.11 Momen inersia untuk sumbu putar yang terletak pada pertengahan batang adalah I = 1 12 ML2 . x 1 2 L – 1 2 L 0 y
122 Mudah dan Aktif Belajar Fisika untuk Kelas XI =>?BH +%' B¥ZR[ >[R^_VN 7R^ONTNV 7R[QN L L R ON`N[T _VYV[QR^ ON`N[T _VYV[QR^ ZRYNYaV ]a_N` ZRYNYaV aWa[T ZRYNYaV _aZOa _VYV[QR^ - 4 ) * )* /. - 4 ) * + /. - 4 /4* 2BJA> 4>I?>M :KMKN 9KIBJ 5JBMNE> PV[PV[ `V]V_ Momentum sudut merupakan besaran vektor. Ingatlah R *$&$'"#) (#%!& ZRYNYaV _aZOa _VYV[QR^ ]RWNY - 4 ) * * /4 R L _VYV[QR^ ]RWNY _R]R^`V `NZ]NX ]NQN TNZON^ -4 ) ) * * , )* /4 /. O¥YN ]RWNY ZRYNYaV _NYNU _N`a TN^V_ _V[TTa[T[dN - 4 / * - /4 O¥YN ]RWNY ZRYNYaV QVNZR`R^ - 4 * * - /4 O¥YN OR^¥[TTN ZRYNYaV QVNZR`R^ - 4 * * + /4 R R R Sumber: Fundamental of Physics, 2001
Gerak Rotasi dan Kesetimbangan Benda Tegar 123 3. Hubungan Momen Gaya dan Percepatan Sudut ER^UN`VXN[ 4>I?>M +%'(& HROaNU OR[QN `RTN^ OR^ZN__N B OR^¥`N_V ]NQN `R^UNQN] `V`VX _aZOa 1& 6[QN `RYNU ZR[TR`NUaV ONUcN ]NQN OR[QN `RTN^ dN[T OR^¥`N_V TNdN dN[T ORXR^WN ]NQN OR[QN ZRYV]a`V _RYa^aU ONTVN[ OR[QN$ `R`N]V S¥Xa_ ]NQN TNZON^ `R^_ROa` QVcNXVYV ¥YRU `V`VX 2 dN[T OR^WN^NX F `R^UNQN] _aZOa ^¥`N_V$ <NdN `N[TR[_VNY ! ORXR^WN QV `V`VX 2 _RUV[TTN OR[QN OR^TR^NX ^¥`N_V QR[TN[ ]R^PR]N`N[ `N[TR[_VNY #` X¥[_`N[& BR[a^a` =aXaZ >> CRc`¥[$ OR_N^ TNdN `N[TR[_VNY `R^_ROa` NQNYNU ,` 4 B 7` !-h*(" ?VXN ^aN_ XV^V QN[ ^aN_ XN[N[ QN^V :BMN>I>>J !+S(&" QVXNYVXN[ QR[TN[ WN^V%WN^V F$ NXN[ QV]R^¥YRU ,` F 4 B 7` F =aOa[TN[ N[`N^N ]R^PR]N`N[ `N[TR[_VNY !7` " QN[ ]R^PR]N`N[ _aQa` !( " NQNYNU 7` 4 ( F& DYRU XN^R[N V`a$ QV]R^¥YRU ,` F 4 B( F * 4 B F*( !-h*)" IRYNU QVXR`NUaV ONUcN + 4 ,` F$ DYRU XN^R[N ,H `RTNX Ya^a_ F ZNXN QV]R^¥YRU ]R^_NZNN[ Z¥ZR[ TNdN _ROR_N^ + 4 B F *( 4 - ( !-h**" @R`R^N[TN[2 + 4 OR_N^ Z¥ZR[ TNdN !CZ" ( 4 OR_N^ ]R^PR]N`N[ _aQa` !^NQ'_" - 4 Z¥ZR[ V[R^_VN !XTZ* " Kata Kunci • benda diskrit • benda tegar • energi kinetik rotasi • energi kinetik translasi • momentum sudut • sumbu putar • torsi HROaNU ON`N[T U¥Z¥TR[ ZRZVYVXV ]N[WN[T . QN[ ZN__N B& 7N`N[T `R^_ROa` QVOR^V R[T_RY _RUV[TTN QN]N` OR^¥`N_V _R]R^`V `NZ]NX ]NQN TNZON^& 7N`N[T QNYNZ XRNQNN[ QVNZ ]NQN ]¥_V_V U¥^Ve¥[`NY& IR[`aXN[YNU OR_N^ ]R^PR]N`N[ _aQa` !(" QN[ OR_N^ ]R^PR]N`N[ `N[TR[_VNY ]NQN aWa[T ON`N[T !7` " XR`VXN ON`N[T QVYR]N_XN[& 6>Q>?0 <NdN OR^N` dN[T ORXR^WN ]NQN ON`N[T NQNYNU , 4 B =# QR[TN[ WN^NX : 4 ) * A QN^V ]a_N` ^¥`N_V !R[T_RY"& B¥ZR[ >[R^_VN ON`N[T dN[T QV]R^¥YRU QN^V =>?BH +%' NQNYNU - 4 ) + /.* & ER^_NZNN[ ]R^PR]N`N[ _aQa` ON`N[T NQNYNU * * ) ! "* + ) )* + + B= . ,: = - . /. B. + ( $$ $ $ ?NQV$ OR_N^ ]R^PR]N`N[ _aQa` ON`N[T NQNYNU + 4 * = . ( & 6QN]a[ OR_N^ ]R^PR]N`N[ `N[TR[_VNY ]NQN aWa[T ON`N[T NQNYNU 7` 4 ( F 4 + * = . . 4 + * =& mg 1 2 Contoh 5.1 Tantangan untuk Anda Seorang pemain akrobat akan membutuhkan tongkat ketika melakukan aksi berjalan di atas tali. Jika ada dua buah tongkat yang sama panjang tetapi beratnya berbeda, tongkat manakah yang harus dipilih? Gambar 5.12 Gaya F bekerja pada sebuah partikel P pada benda tegar menghasilkan torsi + . y x F r P O Lengan momen Garis kerja F * +
124 Mudah dan Aktif Belajar Fisika untuk Kelas XI 4. Momentum Sudut dan Hukum Kekekalan Momentum Sudut HR]R^`V UNY[dN ]NQN TR^NX `^N[_YN_V$ ]NQN TR^NX ^¥`N_V WaTN `R^QN]N` X¥[_R] Z¥ZR[`aZ dN[T QV_ROa` QR[TN[ Z¥ZR[`aZ _aQa`& ER[`V[T[dN XROR^NQNN[ Z¥ZR[`aZ _aQa` ]NQN TR^NX ^¥`N_V _NZN QR[TN[ ]R[`V[T[dN XROR^NQNN[ Z¥ZR[`aZ YV[RN^ ]NQN TR^NX `^N[_YN_V& ENQN 4>I?>M +%') `R^YVUN` _ROaNU ]N^`VXRY OR^ZN__N B QN[ ZRZVYVXV Z¥ZR[`aZ YV[RN^ L ]NQN ]¥_V_V M `R^UNQN] `V`VX ]a_N` 1& B¥ZR[`aZ _aQa` 8 `R^UNQN] `V`VX ]a_N` 1 NQNYNU 8 4 M i L !-h*+" 7R_N^[dN Z¥ZR[`aZ _aQa` QVOR^VXN[ ¥YRU ]R^_NZNN[ . 4 F E _V[ * !-h*," HaQa` * NQNYNU _aQa` dN[T QVOR[`aX ¥YRU M QN[ L& 6^NU Z¥ZR[`aZ _aQa` 8 `RTNX Ya^a_ `R^UNQN] OVQN[T dN[T QVOR[`aX ¥YRU M QN[ L& 6^NU 8 QN]N` QVOR[`aX ¥YRU XNVQNU `N[TN[ XN[N[$ dNV`a Nda[XN[ M XR N^NU L ZRYNYaV _aQa` `R^XRPVY QR[TN[ PN^N ZR[TR]NYXN[ WN^V%WRZN^V `N[TN[ XN[N[& >Oa WN^V dN[T `RTNX Ya^a_ QN[ ZR[a[WaX N^NU `R^`R[`a ZR[dN`NXN[ N^NU 8& B¥ZR[`aZ _aQa` ZR^a]NXN[ OR_N^N[ bRX`¥^ dN[T N^NU[dN ZR[TVXa`V XNVQNU `N[TN[ XN[N[& ?VXN QV`V[WNa _RPN^N _XNYN^$ :BMN>I>>J +%(* QN]N` QV`aYV_XN[ _RONTNV OR^VXa` !_aQa`* 4 1(f"& . 4 FE 4 FBJ 4 F B, F 4 BF*, DYRU XN^R[N BF* 4 -# :BMN>I>>J !+S(*" ZR[WNQV 8 4 - , !-h*-" DYRU XN^R[N OR_N^ + 4 -( QN[ OR_N^ . 4 -, ZNXN QV]R^¥YRU , $ $ ( dL d I I dt dt _RUV[TTN $( dL dt !-h*." :BMN>I>>J !+S(+" ZR[dN`NXN[ ONUcN WaZYNU `¥^_V RX_`R^[NY _NZN QR[TN[ YNWa ]R^aONUN[ Z¥ZR[`aZ _aQa`& :BMN>I>>J !+S(," _ROR[N^[dN QVQN]N` QN^V ]R[a^a[N[ ]R^_NZNN[ `¥^_V QN[ TNdN QNYNZ OR[`aX Z¥ZR[`aZ& 9N]N`XNU 6[QN ZR[a^a[XN[ ]R^_NZNN[ `R^_ROa`5 ?VXN ;@GH;FC7A + 4 ($ [VYNV " NXN[ X¥[_`N[ _RUV[TTN bRX`¥^ Z¥ZR[`aZ _aQa` `¥`NY _V_`RZ ]N^`VXRY `R`N] X¥[_`N[& ER^[dN`NN[ `R^_ROa` ZR^a]NXN[ =aXaZ @RXRXNYN[ B¥ZR[`aZ HaQa`$ dNV`a WVXN+ 4 ($ Z¥ZR[`aZ _aQa`[dN X¥[_`N[& 9R[TN[ ZR[TaONU _a_a[N[ ONTVN[%ONTVN[[dN$ XRYRZONZN[ ^¥`N_V - OR[QN `R^_ROa` NXN[ OR^aONU& ?VXN `VQNX NQN `¥^_V RX_`R^[NY$ ]R^aONUN[ XRYRZONZN[ ^¥`N_V - NXN[ QVVZON[TV ¥YRU XRPR]N`N[ _aQa`, OR[QN NTN^ _NYV[T ZR[VNQNXN[& DYRU XN^R[N V`a$ =aXaZ @RXRXNYN[ B¥ZR[`aZ HaQa` QN]N` QV`aYV_XN[ _RONTNV OR^VXa`& .) 4 .* N`Na - ) , ) 4 - * , ) !-h*/" ENQN 4>I?>M +%'*$ _R_R¥^N[T OR^QV^V QV N`N_ ]V^V[TN[ dN[T _RQN[T OR^]a`N^$ ]NQN XRQaN `N[TN[[dN ZN_V[T%ZN_V[T ZRZRTN[T _ROaNU ORON[& ?VXN _aZOa ]a`N^ QVNZOVY OR^VZ]V` QR[TN[ _aZOa `aOaU ¥^N[T V`a QN[ ZN__N ORON[ dN[T QV]RTN[T[dN `R^YR`NX WNaU QN^V _aZOa ]a`N^$ dNV`a _RWNaU ]N[WN[T ^R[`N[TN[ `N[TN[[dN$ Z¥ZR[ V[R^_VN ORON[ NXN[ OR^[VYNV OR_N^& =NY `R^_ROa` `R^WNQV XN^R[N ORON[ `R^YR`NX WNaU QN^V _aZOa ]a`N^& Gambar 5.14 Momentum sudut ketika kedua tangan direntangkan adalah I a, a . Gambar 5.13 Diagram momen gaya atau torsi yang bekerja pada waktu membuka mur sebuah roda mobil. * p p sin * r y O d x L
Gerak Rotasi dan Kesetimbangan Benda Tegar 125 Gambar 5.15 Momentum sudut ketika kedua tangan dirapatkan (I b, b ). Planet X C D A2 A1 A B Matahari Gambar 5.16 Lintasan sebuah planet mengelilingi Matahari berbentuk elips. HR¥^N[T ]R^R[N[T OR^ZN__N -( XT ZRY¥[PN` ZR[V[TTNYXN[ ]N]N[ `aZ]a QNYNZ X¥[SVTa^N_V Ya^a_ QR[TN[ XRPR]N`N[ _aQa` ($*- ]a`N^N[ ]R^ _RX¥[ `R^UNQN] ]a_N` ZN__N[dN& @RZaQVN[$ ]R^R[N[T ZR[TTaYa[TXN[ `aOaU[dN QN[ TaYa[TN[ `R^_ROa` OR^WN^V%WN^V XV^N%XV^N +( PZ& 9R[TN[ ZR[TN[TTN] ]R^R[N[T _RONTNV _ROaNU ON`N[T U¥Z¥TR[ QR[TN[ ]N[WN[T )$. Z _R_NN` _R`RYNU ZRY¥[PN`$ QN[ ZV^V] _ROaNU O¥YN U¥Z¥TR[ _NN` OR^TaYa[T$ ]R^XV^NXN[ XRPR]N`N[ _aQa`[dN _NN` OR^TaYa[T& 6>Q>?0 9VXR`NUaV2 / 4 -( XT3 ,) 4 ($*- ]a`N^N['_3 4 4 +( PZ3 . 4 )$. Z& ENQN _NN` NcNY$ ]R^R[N[T OR^OR[`aX ON`N[T QR[TN[ ]¥^¥_ QV `V`VX `R[TNU _RUV[TTN Z¥ZR[ V[R^_VN[dN NQNYNU - ) 4 ) )* /.* 4 ) )* !-( XT"!)$. Z"* 4 )($./ XTZ* ENQN _NN` ZR[TTaYa[T$ ]R^R[N[T OR^OR[`aX O¥YN QR[TN[ Z¥ZR[ V[R^_VN[dN NQNYNU - ) 4 * - /4* 4 * - !-( XT"!($+ Z"* 4 )$0 XTZ* Contoh 5.2 Hukum Kekekalan Momentum Sudut berlaku jika tidak ada momen gaya luar yang bekerja pada sistem. Ingatlah @R`VXN ]a`N^N[ _RQN[T OR^YN[T_a[T QN[ XRQaN `N[TN[ dN[T ZRZRTN[T ORON[ QV^N]N`XN[ ZR[QRXN`V `aOaU$ _R]R^`V ]NQN 4>I?>M +%'+$ ZNXN Z¥ZR[ V[R^_VN[dN NXN[ OR^Xa^N[T XN^R[N WN^NX ORON[ XR _aZOa ]a`N^[dN OR^Xa^N[T& BR[TV[TN` ]R^_NZNN[ - 4 B F*,$ XRPR]N`N[ _aQa` dN[T QVNYNZV NXN[ OR^`NZONU OR_N^& =NY V[V QV_RONOXN[ XN^R[N NQN[dN ]R[a^a[N[ Z¥ZR[ V[R^_VN& 9R[TN[ QRZVXVN[$ OR^YNXa UaOa[TN[ 77 88 - - , , $ 3 7 8 - - # QN[ , , 7 8 % !-h*0" @R`R^N[TN[2 - 7 4 Z¥ZR[ V[R^_VN XRNQNN[ 7 !XTZ* " ,a 4 OR_N^ XRPR]N`N[ _aQa` XRNQNN[ 7 !^NQ'_" - 8 4 OR_N^ Z¥ZR[ V[R^_VN XRNQNN[ 8 !XTZ* " ,b 4 XRPR]N`N[ _aQa` XRNQNN[ 8 !^NQ'_" 6]YVXN_V YNV[ QN^V =aXaZ @RXRXNYN[ B¥ZR[`aZ HaQa` NQNYNU TR^NX ]YN[R` QNYNZ ZR[TRYVYV[TV BN`NUN^V& ?¥UN[[R_ @R]YR^ ZR[dN`NXN[ ONUcN QNYNZ _RYN[T cNX`a $t dN[T _NZN$ _ROaNU ]YN[R` NXN[ ZR[dN]a YaN_ QNR^NU dN[T _NZN& ER^UN`VXN[ 4>I?>M +%',& HROaNU ]YN[R` OR^TR^NX QN^V ]¥_V_V 6 XR ]¥_V_V 7 _RUV[TTN YaN_ QNR^NU dN[T QVYNYaV[dN NQNYNU ') !TNZON^ dN[T QVN^_V^"& AaN_ QNR^NU ') NQNYNU ')4 ) * !NYN_"!`V[TTV"4 ) * !F"!F $* " 4 ) * F * $* ') $H 4 ) * F * $* $H 4 ) * F *, 4 * * BF B , 4 * . B ')4 * . B $H !-h*1" BR[a^a` =aXaZ >> @R]YR^$ YaN_ OVQN[T ') _NZN QR[TN[ YaN_ OVQN[T '* & ?VXN UNY V[V QVXNV`XN[ QR[TN[ =aXaZ @RXRXNYN[ B¥ZR[`aZ HaQa`$ Z¥ZR[`aZ _aQa` !"" ]YN[R` QNYNZ XRNQNN[ ) !YV[`N_N[ 67" QN[ ]NQN XRNQNN[ * !YV[`N_N[ )*" NQNYNU _NZN& @¥[_RXaR[_V Y¥TV_ QN^V UNY V[V NQNYNU XRYNWaN[ ]YN[R` QNYNZ ZR[TRYVYV[TV BN`NUN^V `VQNX X¥[_`N[& ENQN `V`VX ]R^VURYV¥[ !`V`VX ]NYV[T QRXN` QR[TN[ BN`NUN^V" XRYNWaN[[dN ]NYV[T OR_N^$ QN[ ]NQN `V`VX N]URYV¥[ !`V`VX ]NYV[T WNaU QR[TN[ BN`NUN^V" XRYNWaN[[dN ]NYV[T XRPVY&
126 Mudah dan Aktif Belajar Fisika untuk Kelas XI 5. Gerak Menggelinding 6[QN ]N_`V _R^V[T ZR[WaZ]NV OR[QN dN[T OR^TR^NX ZR[TTRYV[QV[T$ ZV_NY[dN _ROaNU O¥YN dN[T QVYRZ]N^XN[ QV N`N_ YN[`NV ZR[QN`N^$ N`Na ^¥QN%^¥QN _ROaNU XR[QN^NN[ dN[T ZRYNWa QV WNYN[ ^NdN& BR[TTRYV[QV[T NQNYNU ]R^V_`VcN OR^TR^NX[dN _ROaNU OR[QN _RPN^N `^N[_YN_V QN[ ^¥`N_V !4>I?>M +%'-!>" QN[ +%'-!?""& ENQN _NN` ZR[TTRYV[QV[T$ N]NOVYN `VQNX `R^WNQV _RYV] ZNXN XRQaN TR^NX OR^YN[T_a[T _RPN^N OR^_NZNN[ QN[ WN^NX dN[T QV`RZ]aU QNYNZ _N`a ]a`N^N[ _NZN QR[TN[ XRYVYV[T OR[QN& a. Menggelinding pada Bidang Datar HROaNU OR[QN !_VYV[QR^" `R^YR`NX QV N`N_ OVQN[T QN`N^ QN[ QVOR^V TNdN ! _R]R^`V ]NQN 4>I?>M +%'-!@"& 6TN^ _VYV[QR^ QN]N` ZR[TTRYV[QV[T ZNXN OVQN[T QN`N^ N`Na YN[`NV UN^a_ QVOaN` XN_N^& ?VXN YN[`NV QVOaN` YVPV[ ZNXN _VYV[QR^ UN[dN ZRYNXaXN[ TR^NX `^N[_YN_V QN[ `VQNX QV_R^`NV QR[TN[ TR^NX ^¥`N_V _RUV[TTN _VYV[QR^ QVXN`NXN[ `VQNX ZR[TTRYV[QV[T$ XN^R[N _VYV[QR^ QNYNZ XRNQNNN[ _RYV]& ?NQV$ ]NQN OVQN[T QN`N^ XN_N^$ OR[QN ZR[TNYNZV `^N[_YN_V QN[ TR^NX ^¥`N_V& 7R_N^[dN TR^NX `^N[_YN_V ZR[a^a` =aXaZ >> CRc`¥[ NQNYNU , O < = & B 7 J[`aX TR^NX ^¥`N_V QN^V _RZaN TNdN dN[T ORXR^WN ]NQN OR[QN$ UN[dN TNdN TR_RX dN[T ZR[TUN_VYXN[ Z¥ZR[ TNdN& ?NQV$ Z¥ZR[ TNdN NXN[ ZR[dRONOXN[ _VYV[QR^ OR^TR^NX ^¥`N_V `R^UNQN] ]¥^¥_[dN$ dNV`a QV `V`VX 1& ER^_NZNN[ Z¥ZR[ TNdN[dN QN]N` QV`aYV_XN[+ 4 -( QN[+ 4 < = 4$ 9N^V ]R^_NZNN[ `R^_ROa` NXN[ QV]R^¥YRU < = 4 - 4 ( ?VXN TNdN TR_RX < = dN[T ORXR^WN ]NQN _VYV[QR^ OR^[VYNV X¥[_`N[ QN[ QV_aO_`V`a_VXN[ XR QNYNZ ]R^_NZNN[ TR^NX `^N[_YN_V !, O < = 4 B7" ZNXN NXN[ QV]R^¥YRU 7 4 * ) , B 4 " !-h+(" J[`aX _VYV[QR^ ]RWNY dN[T ZRZVYVXV Z¥ZR[ V[R^_VN - 4 ) * B4* !QN^V =>?BH +%'"$ ]R^_NZNN[ ]R^PR]N`N[ TR^NX `^N[_YN_V NXN[ ZR[WNQV 7 4 * + , B !-h+)" @R`R^N[TN[2 , 4 OR_N^ TNdN dN[T ORXR^WN !C" 7 4 OR_N^ ]R^PR]N`N[ `^N[_YN_V !Z'_* " B 4 ZN__N OR[QN !XT" @RPR]N`N[ _aQa` _NN` ZR[TTaYa[T QVUV`a[T QR[TN[ ]R^_NZNN[ - ),) 4 - *,* N`Na ,* 4 ) * - - ,) $ _RUV[TTN ,* 4 * * )($./ XT Z )$ 0 XT Z i ($*- ]a`N^N['_ 4 )$,0 ]a`N^N['_& Gambar 5.17 (a) Benda bergerak translasi. (b) Benda bergerak rotasi. (c) Benda menggelinding pada bidang yang permukaannya kasar. !P" N O mg F f g , R F , 0 !O" R F v 0 , = 0 !N" R
Gerak Rotasi dan Kesetimbangan Benda Tegar 127 HROaNU _VYV[QR^ ]RWNY OR^ZN__N )- XT ZRZVYVXV WN^V%WN^V *( PZ& HVYV[QR^ `R^_ROa` QVQ¥^¥[T ¥YRU TNdN _ROR_N^ +( C _R]R^`V ]NQN TNZON^& IR[`aXN[YNU ]R^PR]N`N[ dN[T QVNYNZV _VYV[QR^ `R^_ROa`$ WVXN2 N& `VQNX NQN TNdN TR_RX N[`N^N _VYV[QR^ QN[ YN[`NV !_VYV[QR^ ZRYa[Pa^'_RYV]"3 O& `R^QN]N` TR_RXN[ _RUV[TTN _VYV[QR^ ZR[TTRYV[QV[T& 6>Q>?0 9VXR`NUaV2 B 4 )- XT3 , 4 +( C3 4 4 *( PZ 4 ($* Z& N& HVYV[QR^ ZRYa[Pa^ !_RYV]"3 +( C * *Z'_ & )- XT , 7 B $$ $ O& HNN` _VYV[QR^ ZR[TTRYV[QV[T$ Ta[NXN[ :BMN>I>>J !+S()" QR[TN[ - 4 ) * * B4 $ ZNXN ! ! * *+( C , * Z'_ & + + + )- XT , 7 B $$ $ HRXN^N[T$ P¥ON `R[`aXN[ ]R^PR]N`N[ _aQa`[dN !("& Contoh 5.3 F N mg f g Gaya gesek akan menyebabkan silinder menggelinding. Gaya gesek ini berperan dalam menghasilkan momen gaya untuk silinder. Ingatlah b. Menggelinding pada Bidang Miring 6[QN `RYNU ZR[TR`NUaV ONUcN _VYV[QR^ NXN[ ZR[TTRYV[QV[T WVXN `R^QN]N` TNdN TR_RX N[`N^N _VYV[QR^ QN[ OVQN[T _R[`aU& 9RZVXVN[ UNY[dN ]NQN OVQN[T ZV^V[T$ NTN^ _VYV[QR^ QN]N` ZR[TRYV[QV[T ZNXN UN^a_ NQN TNdN TR_RX N[`N^N _VYV[QR^ QN[ OVQN[T ZV^V[T& J[`aX TR^NX `^N[_YN_V$ OR^YNXa ]R^_NZNN[ B= _V[ * h < T 4 B7 QN[ 0 & B= P¥_ * $ HVYV[QR^ NXN[ OR^TR^NX ^¥`N_V WVXN _VYV[QR^ V[V ZRZVYVXV ]R^PR]N`N[ _ROR_N^ 7 & 7H _RUV[TTN NXN[ QV]R^¥YRU ]R^PR]N`N[ _aQa`( & ER^PR]N`N[ _aQa` QN^V TR^NX ^¥`N_V ]NQN _ROaNU _VYV[QR^ QV_RONOXN[ NQN[dN Z¥ZR[ TNdN QN^V TNdN TR_RX& ER^_NZNN[ TNdN TR_RX a[`aX OR[QN dN[T ZR[TTRYV[QV[T ]NQN OVQN[T ZV^V[T NQNYNU < = 4 - * 7 4 & <NdN YNV[ dN[T ORXR^WN ]NQN _VYV[QR^$ `R`N]V `VQNX ZR[VZOaYXN[ Z¥ZR[ TNdN$ _R]R^`V TNdN OR^N` `VQNX QV]R^UV`a[TXN[& =NY V[V QVXN^R[NXN[ TNdN `R^_ROa` ORXR^WN QN^V ]a_N` ^¥`N_V& 9R[TN[ ZRYNXaXN[ RYVZV[N_V QN^V QaN ]R^_NZNN[ QV N`N_$ QV]R^¥YRU ]R^_NZNN[ B= _V[ * O - * 7 4 4 B7 - * B=_V[ 7 - B 4 * $ " !-h+*" J[`aX _VYV[QR^ ]RWNY !- 4 ) * * B4 "$ ]R^_NZNN[ ]R^PR]N`N[[dN NQNYNU * * _V[ _V[ ) + * * B= B= 7 B4 B B 4 * * $ $ " * _V[ + 7 = $ * !-h++" HROaNU _VYV[QR^ ]RWNY U¥Z¥TR[ OR^ZN__N * XT ZRZVYVXV WN^V%WN^V + PZ& @RZaQVN[$ _VYV[QR^ `R^_ROa` QV`RZ]N`XN[ ]NQN OVQN[T ZV^V[T QR[TN[ _aQa` XRZV^V[TN[ -+f& IR[`aXN[YNU ]R^PR]N`N[ dN[T QVNYNZV _VYV[QR^ WVXN2 N" `VQNX NQN TR_RXN[ !YN[`NV YVPV["3 O" _VYV[QR^ ZR[TTRYV[QV[T& Contoh 5.4 Gambar 5.18 Sebuah benda yang menggelinding pada bidang miring. mg sin * mg mg cos * f g * N
128 Mudah dan Aktif Belajar Fisika untuk Kelas XI c. Beban Dihubungkan Melalui Katrol 9V @RYN_ K$ 6[QN `RYNU ZRZ]RYNWN^V `R[`N[T ORON[ N`Na OR[QN dN[T QVVXN` ¥YRU _Ra`N_ `NYV QN[ QVUaOa[TXN[ QR[TN[ XN`^¥Y& 6XN[ `R`N]V$ ZN__N XN`^¥Y QVNONVXN[ XN^R[N 6[QN ORYaZ ZRZ]RYNWN^V X¥[_R] Z¥ZR[ V[R^_VN& HRXN^N[T$ 6[QN `RYNU ZRZ]RYNWN^V Z¥ZR[ V[R^_VN& 9R[TN[ QRZVXVN[$ 6[QN QN]N` ZR[TTa[NXN[ X¥[_R] Z¥ZR[ V[R^_VN _RUV[TTN QN]N` ZRZ]R^UV`a[TXN[ ZN__N XN`^¥Y ]NQN _V_`RZ XN`^¥Y V[V& ER^UN`VXN[ 4>I?>M +%'/& HROaNU ORON[ OR^ZN__N B QVUaOa[TXN[ QR[TN[ _Ra`N_ `NYV `VQNX OR^ZN__N ZRYNYaV _ROaNU XN`^¥Y dN[T QN]N` OR^TR^NX ^¥`N_V& @N`^¥Y ZRZVYVXV ZN__N / QN[ OR^WN^V%WN^V 4$ _R^`N ZRZVYVXV Z¥ZR[ V[R^_VN -& INYV QV`N^VX ¥YRU TNdN 3 _RUV[TTN ORON[ OR^TR^NX XR N`N_ QN[ ZR[TNYNZV ]R^PR]N`N[ `^N[_YN_V _ROR_N^ 7& ER^_NZNN[ TR^NX `^N[_YN_V ]NQN OR[QN NQNYNU '3 > $ B & 5 h B= & B 7 -5 4 B= " B 7 !-h+," ER^_NZNN[ TR^NX ^¥`N_V ]NQN XN`^¥Y NQNYNU !+ 4 ( ( 3 $ 7 - 4 ,4 O 54 4 7 - 4 7 4 ! % * , 54 - !-h+-" ?VXN `RTN[TN[ `NYV ]NQN :BMN>I>>J !+S)*" QV_aO_`V`a_VXN[ XR :BMN>I>>J !+S)+" NXN[ QV]R^¥YRU ]R^_NZNN[ ! % % $ * , B= B7 4 7 - N`Na * , B= 7 - B 4 % $ " !-h+." ?VXN XN`^¥Y OR^a]N _VYV[QR^ ]RWNY QR[TN[ ZN__N /$ Z¥ZR[ V[R^_VN - 4 ) * /4* & :BMN>I>>J !+S)," ZR[WNQV * , B= 7 / B % $ " * ! * , B= 7 / B % $ " !-h+/" Gambar 5.19 Beban dihubungkan dengan katrol. !N" !O" T mg R R F T a 6>Q>?0 9VXR`NUaV2 B 4 * XT3 4 4 + PZ 4 + i)(h* Z3 * 4 -+f N& DYRU XN^R[N OVQN[T YVPV[$ _VYV[QR^ ]RWNY OR^TR^NX _R]N[WN[T OVQN[T QR[TN[ PN^N ZRYa[Pa^ !ZR[TTRYV[PV^"& ', B7 $ - B= _V[ * 4 B 7 7 4 = _V[ * 4 )( Z'_* _V[ -+f 4 )( Z'_* !($0" 4 0 Z'_* O& HNN` _VYV[QR^ ]RWNY ZR[TTRYV[QV[T$ Ta[NXN[ :BMN>I>>J !+S((" QR[TN[ - 4 ) * B4* $ _RUV[TTN QV]R^¥YRU 7 4 * + , B 4 * _V[ + B= B * 4 * !*"!)( Z'_ "!_V[ -+ " + & 4 - ) + Z_* & Gambar 5.20 (a) Gaya-gaya pada benda m. (b) Gaya-gaya pada katrol. , M F T T mg M R
Gerak Rotasi dan Kesetimbangan Benda Tegar 129 ER^UN`VXN[ TNZON^ QV _NZ]V[T& 9aN OaNU OR[QN dN[T ZN__N[dN B) 4 * XT QN[ B* 4 + XT QVUaOa[TXN[ QR[TN[ _Ra`N_ `NYV ZRYNYaV _ROaNU XN`^¥Y QR[TN[ WN^V%WN^V )( PZ QN[ ZN__N XN`^¥Y . XT& IR[`aXN[YNU2 N& ]R^PR]N`N[ YV[RN^ _V_`RZ3 O& `RTN[TN[ `NYV 5) QN[ 5* & 6>Q>?0 N& HV_`RZ ZaYN%ZaYN QNYNZ XRNQNN[ QVNZ& @R`VXN QVYR]N_$ _V_`RZ ZaYNV OR^TR^NX XR N^NU B* _RUV[TTN OR[QN B) [NVX QN[ B* `a^a[ 5) h B) = 4 B) 7 B* = h 5* 4 B* 7 5) 4 B) = # B) 7 5* 4 B* = h B* 7 ER^Ya QVXR`NUaV ONUcN `NYV QVN[TTN] U¥Z¥TR[& IV[WNa XN`^¥Y2 !+ 4 -( 3 - 4 ) * BX 4 * 3 -( 4 5* 4 h 5) 4 > 7 4 4 !5* h 5) "4 - ) * BX 47 4 !5* h 5) "4 ) * BX 7 4 !5* h 5) " HaO_`V`a_VXN[ UN^TN 5) QN[ 5* QN^V TR^NX `^N[_YN_V XR QNYNZ ]R^_NZNN[ TR^NX ^¥`N_V ZNXN QV]R^¥YRU ) * BX 7 4 L!B* = h B* 7" h !B) = # B) 7"M * ) ) * BBB @ 2 4 6 8 " " 3 5 7 4 !B* h B) " = 7 4 * ) * ) ! " ) * @ B B= BBB % " " % $ $ $ 2 4 " " 3 5 * * !+ XT * XT")( Z'_ )( XT Z'_ * )$*- Z'_ ) 0 XT . XT + XT * XT * ?NQV$ ]R^PR]N`N[ YV[RN^ _V_`RZ NQNYNU )$*- Z'_* & O& IRTN[TN[ `NYV 5) QN[ 5* 2 5) 4 B) = # B) 7 4 !* XT" !)( Z'_* " # !* XT" !)$*- Z'_* " 4 **$- C& 5* 4 B* = h B* 7 4 !+ XT" !)( Z'_* " h !+ XT" !)$*- Z'_* " 4 *.$*- C& ?NQV$ `RTN[TN[ `NYV 5) QN[ 5* OR^`a^a`%`a^a` NQNYNU **$- C QN[ *.$*- C& Contoh 5.5 Tantangan untuk Anda Dengan kecepatan yang sama, mobil manakah di antara mobil yang bergerak di atas jalan kasar dan mobil yang bergerak di atas jalan licin, yang terlebih dahulu sampai? Mengapa demikian? @R`R^N[TN[2 7 4 OR_N^ ]R^PR]N`N[ YV[RN^ !Z'_* " , 4 OR_N^ TNdN `N^VX!C" B 4 ZN__N OR[QN !XT" / 4 ZN__N XN`^¥Y !XT" m1 m2 T2 T1 a 6. Energi dalam Gerak Rotasi a. Energi Kinetik Rotasi HR`VN] OR[QN dN[T OR^TR^NX ZRZVYVXV R[R^TV XV[R`VX& HNZN UNY[dN QR[TN[ OR[QN dN[T ZRYNXaXN[ TR^NX ^¥`N_V$ OR[QN `R^_ROa` NXN[ ZRZVYVXV R[R^TV XV[R`VX ^¥`N_V& 9VXR`NUaV R[R^TV XV[R`VX `^N[_YN_V NQNYNU +@ 4 ) * BJ* !-h+0" BR[TV[TN` J 4 F, ZNXN +@ 4 ) * B !F, " * 4 ) * BF*, * & Gambar 5.21 Benda bergerak translasi dengan kecepatan v sambil berotasi dengan kecepatan sudut , . v , R T2 m1 . g T1 m2 . g a
130 Mudah dan Aktif Belajar Fisika untuk Kelas XI Gambar 5.22 Silinder yang mula-mula diam bergerak menggelinding. A B v h 9VXR`NUaV ONUcN BF* NQNYNU Z¥ZR[ V[R^_VN ZNXN a[`aX TR^NX ^¥`N_V ZR[WNQV +@ ^¥`N_V 4 ) * -, * !-h+1" @R`R^N[TN[2 +@ 4 R[R^TV XV[R`VX `^N[_YN_V !W¥aYR" +@ ^¥`N_V 4 R[R^TV XV[R`VX ^¥`N_V !W¥aYR" - 4 Z¥ZR[ V[R^_VN OR[QN !XTZ* " , 4 XRYNWaN[ _aQa` !^NQ'_" b. Energi Kinetik Translasi dan Rotasi IRYNU 6[QN XR`NUaV ONUcN _ROaNU OR[QN dN[T OR^TR^NX ZR[TTRYV[QV[T NXN[ ZRZVYVXV QaN TR^NXN[$ dNV`a TR^NX YV[RN^ QN[ TR^NX ^¥`N_V& <NOa[TN[ TR^NX YV[RN^ QN[ WN^NX ^¥`N_V QV_ROa` TR^NX `^N[_YN_V& <R^NX `^N[_YN_V ZRZVYVXV XRPR]N`N[ YV[RN^ J$ _RQN[TXN[ TR^NX ^¥`N_V[dN ZRZVYVXV XRPR]N`N[ _aQa` ,& 7R[QN dN[T ZR[TTRYV[QV[T ZRZVYVXV R[R^TV XV[R`VX `^N[_YN_V QN[ R[R^TV XV[R`VX ^¥`N_V& HRPN^N ZN`RZN`V_$ ZRZR[aUV ]R^_NZNN[ +@ `¥`NY 4 ) * BJ* # ) * -, * +@ `¥`NY 4 +@ # +@ ^¥`N_V !-h,(" @R`R^N[TN[2 +@ `¥`NY 4 R[R^TV XV[R`VX `¥`NY c. Hukum Kekekalan Energi Mekanik pada Gerak Rotasi ER^UN`VXN[ 4>I?>M +%((& HROaNU _VYV[QR^ dN[T ZaYN%ZaYN QVNZ$ ZR[TTRYV[QV[T QN^V 6 XR 7& ?VXN QVN_aZ_VXN[ _RYNZN O¥YN OR^¥`N_V `VQNX NQN R[R^TV dN[T UVYN[T QN[ OR^aONU ZR[WNQV R[R^TV XNY¥^$ OR^YNXa =aXaZ @RXRXNYN[ :[R^TV& +E6 # +@6 4 +E7 # +@7 B=> # ( 4 ( # ) * BJ* # ) * - * , !a[`aX _VYV[QR^ ]RWNY$ - 4 ) * BF* " B=> 4 ) * BJ* # ) , BF* ! * J F => 4 + , J * 9N^V ]R^_NZNN[ `R^_ROa`$ XRPR]N`N[ _VYV[QR^ ]RWNY XR`VXN _NZ]NV QV QN_N^ OVQN[T ZV^V[T QN]N` QVXR`NUaV$ dNV`a J 4 + , => !-h,)" :BMN>I>>J !+S*'" QN]N` QV`aYV_XN[ QNYNZ OR[`aX aZaZ$ dNV`a J 4 * ) => " @ !-h,*" ENQN ]R^_NZNN[ `R^_ROa`$ @ ZR^a]NXN[ X¥[_`N[`N TR¥ZR`^V dN[T QVZVYVXV ¥YRU OR[QN QN[ QNYNZ OVQN[T `RX[VX _R^V[T QV_ROa` _RONTNV WN^V% WN^V TV^N_V& CVYNV @ V[V OR^TN[`a[T ]NQN WR[V_ OR[QN[dN& J[`aX _VYV[QR^ ]RWNY @ 4 ) * $ a[`aX O¥YN ]RWNY @ 4 * - $ QN[ a[`aX O¥YN OR^¥[TTN @ 4 * + &
Gerak Rotasi dan Kesetimbangan Benda Tegar 131 HROaNU O¥YN ]RWNY OR^ZN__N B QN[ OR^WN^V%WN^V 4 ZR[TTRYV[QV[T `R^UNQN] ]¥^¥_[dN ]NQN OVQN[T ZV^V[T _R]R^`V ]NQN TNZON^& IR[`aXN[YNU XRYNWaN[ O¥YN XR`VXN _NZ]NV QV QN_N^ OVQN[T WVXN QVYR]N_ QN^V XR`V[TTVN[ > !QN^V XRNQNN[ QVNZ"& 6>Q>?0 7¥YN OR^TR^NX ZR[TTRYV[QV[T +E' " +@' & +E( " +@( B=>' " ) * - , ' * #) * BJ' % 4 B=>( " ) * - , ( *# ) * BJ( % B=> # ( # ( 4 ( # ! * * 7 7 ) ) * * - B , " J J[`aX O¥YN ]RWNY$ * * - - B $ 4 QN[ J & 4 , $ _RUV[TTN B=> & * )* ) * * *- * J B4 BJ 4 2 42 4 6 86 8 " 3 53 5 => & ) ) * * - * J J " => 4 / ) * ( )( / JJ = - $ > @RYNWaN[ O¥YN XR`VXN OR^TR^NX ZRYa[Pa^ QR[TN[ OVQN[T QVN[TTN] YVPV[2 +E' # +@' & +E( # +@( B=>' " ( & ( " ) * B J* - J = $ * > & 7N[QV[TXN[ ¥YRU 6[QN& @RYNWaN[ XR`VXN O¥YN OR^TR^NX ZR[TTRYV[QV[T NQNYNU )( / J = $ > QN[ XRYNWaN[ XR`VXN O¥YN OR^TR^NX ZRYa[Pa^ NQNYNU J = $ * > & 9N]N` QV_VZ]aYXN[ ONUcN O¥YN dN[T OR^TR^NX ZR[TTRYV[QV[T YROVU YNZON` QN^V]NQN O¥YN dN[T OR^TR^NX ZRYa[Pa^& =NY `R^_ROa` QV_RONOXN[ TR^NX ^¥`N_V ]NQN O¥YN ZR[WNQV ]R[TUNZON` `R^UNQN] YNWa O¥YN& Contoh 5.6 Kata Kunci • lengan torsi • momen inersia h B A )& ER^UN`VXN[ TNZON^ OR^VXa` V[V& IR[`aXN[YNU Z¥ZR[ TNdN `¥`NY dN[T ORXR^WN ]NQN ^¥QN `R^UNQN] ]¥^¥_ 1$ WVXN 7 4 )( PZ QN[ 8 4 *- PZ& *& HROaNU _VYV[QR^ ]RWNY QR[TN[ WN^V%WN^V 4 QN[ OR^ZN__N / QVTa[NXN[ _RONTNV XN`^¥Y a[`aX ZR[TNZOVY NV^ _aZa^ QR[TN[ _ROaNU RZOR^& BN__N RZOR^ B QN[ ON`N[T _VYV[QR^ QVN[TTN] YVPV[& IR[`aXN[ ]R^PR]N`N[ RZOR^ _NN` WN`aU XR _aZa^& +& :Z]N` OaNU ]N^`VXRY ZN_V[T%ZN_V[T ZN__N[dN BN 4 + XT$ BO 4 * XT$ BP 4 ) XT$ QN[ BQ 4 , XT _R]R^`V `NZ]NX ]NQN TNZON^& IR[`aXN[ Z¥ZR[ V[R^_VN _V_`RZ WVXN _aZOa ]a`N^[dN2 N& ZRYNYaV _aZOa ''!3 O& ZRYNYaV _aZOa ((!& 10 N a O 12 N b 9 N 30° Tes Kompetensi Subbab B 7BMF>G>JH>D A>H>I ?PGP H>OED>J% A D 3 kg B 2 kg C 1 kg 4 kg A B A’ B’ 2 m 2 m 2 m
132 Mudah dan Aktif Belajar Fisika untuk Kelas XI C. Kesetimbangan Benda Tegar 6[QN `RYNU ZR[TR`NUaV ONUcN _ROaNU ]N^`VXRY QVXN`NXN[ QNYNZ XRNQNN[ _R`VZON[T WVXN ]N^`VXRY `R^_ROa` `VQNX ZR[TNYNZV ]R^PR]N`N[$ N^`V[dN WaZYNU bRX`¥^ QN^V TNdN%TNdN dN[T ORXR^WN ]NQN ]N^`VXRY `R^_ROa` NQNYNU [¥Y$ ! F $(& =NY dN[T _NZN OR^YNXa ]aYN a[`aX OR[QN N`Na _V_`RZ ]N^`VXRY$ dNV`a ]a_N` ZN__N a[`aX OR[QN ZRZVYVXV ]R^PR]N`N[ [¥Y WVXN ^R_aY`N[ bRX`¥^ QN^V _RYa^aU TNdN YaN^ dN[T ORXR^WN ]NQN OR[QN NQNYNU [¥Y& =NY `R^_ROa` QVXR[NY _RONTNV _dN^N` ]R^`NZN a[`aX XR_R`VZON[TN[& 9NYNZ OR[`aX X¥Z]¥[R[%X¥Z]¥[R[[dN QV[dN`NXN[ QR[TN[ ]R^_NZNN[ !!! ,,, L MN $$$ ($ ($ ( !-h,+" 6QN]a[ _dN^N` XRQaN dN[T UN^a_ QV]R[aUV NTN^ OR[QN _R`VZON[T NQNYNU OR[QN `VQNX ZRZVYVXV XRPR[QR^a[TN[ a[`aX OR^¥`N_V& HdN^N` `R^_ROa` QVQN_N^V ¥YRU QV[NZVXN TR^NX ^¥`N_V& 9R[TN[ QRZVXVN[$ OR[QN `RTN^ OR^NQN QNYNZ XR_R`VZON[TN[& ?NQV ^R_aY`N[ `¥^_V YaN^ dN[T ORXR^WN ]NQN OR[QN UN^a_ _NZN QR[TN[ [¥Y& =NY `R^_ROa` NQNYNU _dN^N` XRQaN a[`aX XR_R`VZON[TN[& ! + $( !-h,," 9R[TN[ QRZVXVN[$ _aN`a OR[QN `RTN^ QVXN`NXN[ _R`VZON[T WVXN ! F $( QN[ !+ $(& HNYNU _N`a P¥[`¥U XRWNQVN[ XR_R`VZON[TN[ OR[QN `RTN^$ ZV_NY[dN _ROaNU `N[TTN dN[T OR^_N[QN^ ]NQN QV[QV[T& ER^UN`VXN[ 4>I?>M +%()& HROaNU ON`N[T 67 OR^_N[QN^ ]NQN QV[QV[T& JWa[T ONcNU 6 OR^NQN ]NQN YN[`NV dN[T XN_N^ QN[ ONTVN[ N`N_ 7 OR^_N[QN^ ]NQN QV[QV[T bR^`VXNY dN[T WaTN XN_N^& ?VXN ]N[WN[T ON`N[T $ OR^N` ON`N[T K$ YN[`NV QN[ QV[QV[T _NYV[T `RTNX Ya^a_$ _R^`N ON`N[T ZRZOR[`aX _aQa` `R^UNQN] NYN_[dN ZNXN 6[QN QN]N` ZR[a^a[XN[ ]R^_NZNN[ TNdN QN[ Z¥ZR[ TNdN dN[T ZRZR[TN% ^aUV ON`N[T `R^_ROa`& 7N`N[T 67 QNYNZ XRNQNN[ _R`VZON[T _RUV[TTN _dN^N` XR_R`VZON[TN[ NXN[ ZRZR[aUV ]R^_NZNN[ OR^VXa` V[V& !,L 4 ( 07 h < 6 4 ( 07 4 <6 !-h,-" ,& HROaNU _VYV[QR^ ]RWNY OR^ZN__N * XT QVYR`NXXN[ ]NQN _ROaNU OVQN[T ZV^V[T QR[TN[ _aQa` XRZV^V[TN[ .(f& IR[`aXN[YNU ]R^PR]N`N[ dN[T QVNYNZV[dN$ WVXN2 N& `VQNX NQN TR_RX3 O& _VYV[QR^ ZR[TTRYV[QV[T& -& BN__N OR[QN B) 4 )$0 XT QN[ B* 4 * XT QVUaOa[TXN[ QR[TN[ XN`^¥Y _R]R^`V TNZON^& B¥ZR[ V[R^_VN _V_`RZ XN`^¥Y NQNYNU - 4 )$/ XT B* $ QR[TN[ F ) 4 *( PZ QN[ F * 4 -( PZ& IR[`aXN[2 N& ]R^PR]N`N[ _aQa` XN`^¥Y3 O& `RTN[TN[ `NYV 5) QN[ 5* & Gambar 5.23 Sebuah tangga bersandar pada dinding yang kasar dengan lantai yang kasar pula. f B NB A w B f A NA * T T 2 1 m1 m2 r r 2 1 (
Gerak Rotasi dan Kesetimbangan Benda Tegar 133 ! Fy 4 ( < 7 # 06 h K 4 ( < 7 # 06 4 K !-h,." ?VXN QVN_aZ_VXN[ `V`VX 6 ZR[WNQV ]¥^¥_$ ]R^_NZNN[ Z¥ZR[ TNdN[dN NQNYNU !+ 4 (& AR[TN[ `¥^_V 07 - _V[ * & AR[TN[ `¥^_V K - 1 2 P¥_ * & AR[TN[ `¥^_V < 7 - P¥_ * & ?NQV$ 07 _V[ * # <7 P¥_ * 4 1 2 K P¥_ * !-h,/" 9N^V XR`VTN ]R^_NZNN[ `R^_ROa`$ 6[QN QN]N` ZR[TR`NUaV [VYNV QN^V TNdN TR_RX N[`N^N YN[`NV QN[ ON`N[T 67$ TNdN [¥^ZNY YN[`NV$ QN[ TNdN [¥^ZNY QV[QV[T& HROaNU ON`N[T 67 dN[T ]N[WN[T[dN , ZR`R^ OR^_N[QN^ ]NQN QV[QV[T YVPV[ QV `V`VX 7& HaQa` dN[T QV OR[`aX ON`N[T 67 QN^V NYN_[dN -+f$ _RQN[TXN[ OR^N` ON`N[T K& ?VXN ]NQN _NN` ON`N[T 67 `R]N` NXN[ `R^TRYV[PV^$ `R[`aXN[YNU X¥RSV_VR[ TR_RX N[`N^N ON`N[T QN[ NYN_[dN& 6>Q>?0 9VXR`NUaV2 '( 4 A 4 , Z3 * 4 -+f3 OR^N` OR[QN 4 K g !,L4( & 07 h < N 4 ( & 07 4 < N 07 4 's 06 !`R]N` NXN[ `R^TRYV[PV^" &&&&&&&&&&&&&&&&&&&&&&&&!)" g !,M4 ( - 06 h K 4 ( &064 K &&&&&&&&&&&&&&&&&&&&&&&&&&!*" g !+ 4 ($ ]a_N` QV ' 0( P¥_* h K ) * P¥_* 4 ( - 0( P¥_* 4 ) * K P¥_ * 0( P¥_ -+( - ($. 0( 4 ) * K!($."- 0(4 + . K &&&&&&&&&!+" HaO_`V`a_VXN[ ER^_NZNN[ !*" QN[ !+" XR ]R^_NZNN[ !)"$ QV]R^¥YRU ) * K4 ' G K - ' G & ) * & ?NQV$ X¥RSV_VR[ TR_RX[dN 4 ) * & B f B = 0 NB = 4 m NA w A * Contoh 5.7 Keadaan setimbang berarti !F = 0 dan !+ = 0. Ingatlah f A HROaNU ON`N[T U¥^Ve¥[`NY QR[TN[ OR^N` +(( C QN[ ]N[WN[T - Z QVOR^V R[T_RY ]NQN _ROaNU QV[QV[T& JWa[T dN[T _N`a[dN QV`¥]N[T QR[TN[ _Ra`N_ XNcN` QN[ ZRZOR[`aX _aQa` -+f `R^UNQN] U¥^Ve¥[`NY !YVUN` TNZON^"& 6]NOVYN _R_R¥^N[T dN[T OR^N`[dN -(( C OR^QV^V ]NQN WN^NX * Z QN^V QV[QV[T$ `R[`aXN[ `RTN[TN[ XNcN` QN[ TNdN ^RNX_V ]NQN ON`N[T& T 53° 5 m 0 Contoh 5.8 Kata Kunci • momen gaya • lengan torsi Gambar 5.24 Lengan torsi dari setiap gaya yang bekerja pada tangga. f B NB B A + lengan torsi f B lengan torsi w lengan torsi NB W * *
134 Mudah dan Aktif Belajar Fisika untuk Kelas XI Batang homogen adalah batang yang letak titik beratnya tepat di tengah-tengah. Ingatlah Pembahasan Soal Jarak sumbu roda depan dan sumbu roda belakang sebuah truk yang bermassa 1.500 kg adalah 2 m. Pusat massa truk 1,5 m di belakang roda depan. Jika g = 10 m/s2 . Beban yang diterima roda depan adalah .... a. 1.250 N b. 2.500 N c. 3.750 N d. 5.000 N e. 6.250 N Soal UMPTN Tahun 1995 Pembahasan: !+ = 0 mg ! AP – NB ! AB = 0 NB = mg AP ! AB = ! 1.500 10 0,5 ! ! 2m = 3.750 N Jawaban: c 6>Q>?0 <NdN%TNdN `^N[_YN_V2 !,L 4 ( - 4L h 5L 4 ( - 4 P¥_ * h I P¥_ * 4 ( - 4 P¥_ * h I P¥_ -+ 4 ( - 4 P¥_ * 4 ($. I &&&!)" !,M 4 ( - 4M # 5M h K) h K* 4 ( - 4 _V[ * # 5 _V[ -+ 4 -(( # +(( - 4 _V[ * # ($0 5 4 0(( &&&!*" <NdN%TNdN ^¥`N_V2 < TR_RX R[T_RY 4 ( - YR[TN[ Z¥ZR[ 5 3 _V[ * - * 4 -+ YR[TN[ Z¥ZR[ K) 3 _V[ * - * 4 1( YR[TN[ Z¥ZR[ K* 3 _V[ * - * 4 1( !+ 4 ( _V[ !-+" 5 h _V[ * K) h _V[ * K* 4 ( - !-" !($0 5" h !*$-" !)" !+((" h !*" !)" !-((" 4 ( - ,I 4 )&/-( I 4 ,+/$- C &&&!+" HaO_`V`a_VXN[ !+" XR !)" QN[ !*"$ QV]R^¥YRU 4 P¥_ * 4 ($. !,+/$-" 4 *.*$- C 4 _V[ * # ($0 !,+/$-" 4 0(( 4 _V[ * 4 ,-( C `N[ * 4 _V[ P¥_ 4 4 * * 4 ,-( C *.*$- C 4 )$/) * 4 -1$/, ?NQV$ TNdN ^RNX_V R[T_RY NQNYNU 4 4 ,-( C _V[ -1$/, 4 ,-( C ($0. 4 -*+ C& Tes Kompetensi Subbab C 7BMF>G>JH>D A>H>I ?PGP H>OED>J% )& HROaNU ON`N[T 67 ]N[WN[T[dN * Z QN[ OR^N`[dN QVNONVXN[& JWa[T 7 QVVXN` QR[TN[ XNcN` dN[T ON`N_ `RTN[TN[ ZNX_VZaZ[dN ,(( C& HR_R¥^N[T dN[T OR^N`[dN )&((( C OR^QV^V QV N`N_ ON`N[T _RWNaU L$ _R]R^`V TNZON^ QV _NZ]V[T& 7R^N]NXNU WN^NX ZNX_VZaZ QN^V `V`VX 6 _RORYaZ XNcN` ]a`a_5 2 m A T x B w tumpuan T * 2 m w1 w2 = 5 m 1 2 + T Ty Tx w1 w2 * R Ry Rx
Gerak Rotasi dan Kesetimbangan Benda Tegar 135 0,5 m 2,5m T1 T2 A B ( 30° w = 60 N T 30° 0,5 m Sumber Ilmu 1 m h R F 40 cm *& HRON`N[T ONY¥X dN[T ]N[WN[T[dN 0 Z QN[ OR^N`[dN *(( C QVTN[`a[TXN[ _R]R^`V ]NQN TNZON^& ?VXN WN^NX 67 4 ) Z$ 78 4 , Z$ QN[ 89 4 + Z$ `R[`aXN[YNU [VYNV ]R^ON[QV[TN[ `RTN[TN[ `NYV 5) QN[ 5* & T2 T1 4 m 3 m B C D 1 m A +& HROaNU ]N]N[ VXYN[ OR^ZN__N +* XT ZR[RZ]RY ]NQN _RON`N[T ]V]N dN[T OR^N`[dN QVNONVXN[$ QN[ ]N[WN[T% [dN )$- Z& HNYNU _N`a aWa[T[dN QVVXN` QR[TN[ `NYV$ _RQN[TXN[ aWa[T dN[T YNV[ ZR[RZ]RY ]NQN QV[QV[T _R]R^`V `NZ]NX ]NQN TNZON^& IR[`aXN[YNU `RTN[TN[ dN[T QVZVYVXV `NYV& ,& HROaNU ^¥QN OR^ZN__N *( XT UR[QNX QV[NVXXN[ XR _ROaNU a[QNXN[ _R`V[TTV *( PZ& =V`a[TYNU TNdN ZV[VZaZ dN[T UN^a_ QVOR^VXN[ NTN^ ^¥QN QN]N` [NVX$ WVXN WN^V%WN^V ^¥QN -( PZ QN[ ]R^PR]N`N[ T^NbV`N_V )( Z'_* & -& HROaNU ON`N[T 67 ]N[WN[T[dN + Z QN[ OR^N`[dN _ROR_N^ )*( C& JWa[T%aWa[T[dN QVVXN` ¥YRU `NYV _R]R^`V ]NQN TNZON^& ENQN WN^NX ) * ZR`R^ QN^V 6 QVTN[`a[TXN[ _ROaNU ORON[ dN[T OR^N`[dN .( C& IR[`aXN[ ZN_V[T%ZN_V[T TNdN `RTN[T `NYV QN[ OR_N^ _aQa` (& .& HROaNU ON`N[T 67 OR^N`[dN ,(( C& HROaNU R[T_RY QV`RZ]N`XN[ QV `V`VX 6$ QN[ QV `V`VX 8 QVVXN` ]NQN `RZO¥X QR[TN[ _Ra`N_ `NYV `NX OR^ZN__N& ?VXN _V_`RZ _RVZON[T$ UV`a[TYNU2 N& `RTN[TN[ `NYV3 O& OR_N^ TNdN R[T_RY !_V[ -+f 4 ($0("& 1 4 L B A 53° 2.000 N 3 4 L C Rangkuman )& E¥_V_V _aQa` OR[QN `RTN^ dN[T OR^TR^NX ^¥`N_V QV[dN`NXN[ QR[TN[ ]R^_NZNN[ G F * $ *& @RPR]N`N[ _aQa` ^N`N%^N`N OR[QN `RTN^ dN[T OR^TR^NX ^¥`N_V QV[dN`NXN[ QR[TN[ ]R^_NZNN[ H * , $ $ $ 6QN]a[ XRPR]N`N[ _aQa` _R_NN`[dN2 ( YVZ H : H :H * * , $ & $ $ $ $ +& ER^PR]N`N[ _aQa` _R_NN` OR[QN `RTN^ dN[T OR^¥`N_V QV[dN`NXN[ QR[TN[ ]R^_NZNN[2 * * : : :H :H , * ( $ $ ,& B¥ZR[ TNdN N`Na `¥^_V ZR^a]NXN[ OR_N^N[ dN[T ZR[dRONOXN[ _ROaNU OR[QN `RTN^ PR[QR^a[T a[`aX OR^¥`N_V `R^UNQN] ]¥^¥_[dN& + 4 M i 3 4 F, _V[ * 4 ,: QR[TN[ : 4 F _V[ * 4 Z¥ZR[ YR[TN[&
136 Mudah dan Aktif Belajar Fisika untuk Kelas XI -& EN_N[TN[ QaN OaNU TNdN dN[T _RWNWN^$ _NZN OR_N^$ QN[ OR^YNcN[N[ N^NU QV_ROa` X¥]RY& 7R_N^[dN _ROaNU X¥]RY QV[dN`NXN[ QR[TN[ Z¥ZR[ X¥]RY$ dNV`a /4,:$ QR[TN[ : 4 WN^NX N[`N^N QaN TNdN& /^R_aY`N[ 4 !/ .& B¥ZR[ V[R^_VN NQNYNU aXa^N[ XRPR[QR^a[TN[ _aN`a OR[QN a[`aX ZRYNXaXN[ TR^NX ^¥`N_V& B¥ZR[ V[R^_VN QV`R[`aXN[ ¥YRU ZN__N QN[ ]¥YN QV_`^VOa_V ZN__N `R^UNQN] _aZOa ]a`N^& - 4 !B F* J[`aX _ROaNU OR[QN dN[T ZRZVYVXV QV_`^VOa_V ZN__N dN[T X¥[`V[a$ OR^YNXa ]R^_NZNN[ - 4 * F :B # /& B¥ZR[ V[R^_VN OROR^N]N OR[`aX OR[QN QR[TN[ _aZOa ]a`N^ OR^NQN ]NQN ]a_N` ZN__N[dN NQNYNU _RONTNV OR^VXa`& 7N`N[T2 - 4 ) )* /.* $ . 4 ]N[WN[T ON`N[T& HVYV[QR^ `V]V_ OR^¥[TTN2 - 4 /4* $ 4 4 WN^V%WN^V _VYV[QR^& HVYV[QR^ ]RWNY2 - 4 ) , /4*# ) )* /.* $ 4 4 WN^V%WN^V _VYV[QR^& 7¥YN ]RWNY2 - 4 * - /4* $ 4 4 WN^V%WN^V O¥YN& 7¥YN OR^¥[TTN2 - 4 * + /4* $ 4 4 WN^V%WN^V O¥YN& 0& ER^PR]N`N[ `N[TR[_VNY dN[T QVNYNZV OR[QN OR^ZN__N XN^R[N ]R[TN^aU TNdN ZR[a^a` =aXaZ >> CRc`¥[ NQNYNU ,` 4 B 7` ER^_NZNN[ Z¥ZR[ TNdN _ROR_N^ + 4 - ( 1& 7R_N^ Z¥ZR[`aZ _aQa` `R^UNQN] ]a_N` YV[TXN^N[ QN^V _ROaNU OR[QN OR^`V`VX ZN__N B dN[T OR^TR^NX ZRYV[T% XN^ ZRZR[aUV ]R^_NZNN[ .4F i E N`Na . 4 - $, )(& BR[TTRYV[QV[T NQNYNU ]R^V_`VcN OR^TR_RX[dN _ROaNU OR[QN _RPN^N `^N[_YN_V QN[ ^¥`N_V& 6]NOVYN `VQNX `R^WNQV _RYV]$ XRQaN TR^NX OR[QN OR^YN[T_a[T _RPN^N OR^_NZN% N[ QN[ WN^NX dN[T QV`RZ]aU QNYNZ _N`a ]a`N^N[ _NZN QR[TN[ XRYVYV[T OR[QN& :[R^TV XV[R`VX TR^NX ZR[T% TRYV[QV[T NQNYNU +@ 4 +@ `^N[_YN_V # +@ ^¥`N_V 4 ) * BJ* # ) * -, * ))& HaN`a OR[QN `RTN^ OR^NQN QNYNZ XRNQNN[ _R`VZON[T WVXN ^R_aY`N[ TNdN dN[T ORXR^WN ]NQN OR[QN _NZN QR[TN[ [¥Y& !, 4 ( - !,L4 ( QN[ !,M4 ( !+ 4 ( Peta Konsep B¥ZR[ <NdN B¥ZR[ >[R^_VN B¥ZR[`aZ HaQa` 4BM>G ;KO>NE <NdN !3" BN__N !B" B¥ZR[`aZ AV[RN^ !L" @R_R`VZON[TN[ 7R[QN IRTN^ ZRZONUN_ N[NY¥TV N[NY¥TV N[NY¥TV ZR[TUN_VYXN[ _dN^N`[dN WVXN `VQNX NQN TNdN YaN^$ OR^YNXa Setelah mempelajari bab ini, tentu Anda dapat mengetahui konsep momen gaya dan momen inersia serta pengaruhnya terhadap gerak rotasi. Selain itu, Anda juga dapat mengetahui konsep momentum sudut, gerak rotasi, dan gerak translasi. Dari keseluruhan materi, bagian manakah yang sulit dipahami? Coba diskusikan dengan teman atau guru Fisika Anda. Refleksi Membuka mur dengan kunci bertangan panjang lebih mudah dibanding dengan kunci bertangan pendek. Pengetahuan ini diperoleh dengan mempelajari bab ini. Nah, coba Anda sebutkan manfaat lain mempelajari bab ini. =aXaZ @RXRXNYN[ B¥ZR[`aZ HaQa`
Gerak Rotasi dan Kesetimbangan Benda Tegar 137 1% :EHEDH>D N>H>D N>OP F>Q>?>J R>JC L>HEJC OBL>O A>J GBMF>G>JH>D L>A> ?PGP H>OED>J% )& HROaNU OR[QN dN[T OR^OR[`aX _VYV[QR^ ]RWNY OR^WN^V% WN^V 4 QN[ OR^ZN__N B& @RZaQVN[$ OR[QN `R^_ROa` QV]a`N^ ]NQN _aZOa[dN QR[TN[ ]R^V¥QR 5& 7R_N^ R[R^TV XV[R`VX ^¥`N_V[dN NQNYNU &&&& N& " * ** /4 5 Q& " * * * /4 5 O& " * * * *, /4 5 R& " * * * /5 4 P& " * * * * /4 5 *& HROaNU PV[PV[ QR[TN[ ZN__N ($((+ XT QN[ WN^V%WN^V ($- PZ ZR[TTRYV[QV[T QV N`N_ ]R^ZaXNN[ OVQN[T ZV^V[T dN[T ZRZOR[`aX _aQa` +(f `R^UNQN] OVQN[T U¥^Ve¥[`NY& 8V[PV[ `R^_ROa` QVYR]N_XN[ QN^V XRNQNN[ QVNZ ]NQN XR`V[TTVN[ - Z _RPN^N `RTNX Ya^a_ QN^V OVQN[T U¥^Ve¥[`NY& @RPR]N`N[ YV[RN^ PV[PV[ `R^_ROa` _RcNX`a ZR[PN]NV XNXV OVQN[T ZV^V[T NQNYNU &&&& N& )( Z'_ Q& - Z'_ O& - + Z'_ R& *$- Z'_ P& - * Z'_ +& HROaNU O¥YN ]RWNY OR^`^N[_YN_V QN[ OR^¥`N_V QR[TN[ XRPR]N`N[ YV[RN^ QN[ XRPR]N`N[ _aQa` ZN_V[T%ZN_V[T J QN[ ,& :[R^TV XV[R`VX `¥`NY O¥YN ]RWNY `R^_ROa` NQNYNU &&&& N& ) * BJ* Q& 1 )( BJ* O& * - BJ* R& - * BJ* P& / )( BJ* ,& HROaNU ON`N[T dN[T ]N[WN[T[dN A QN[ ZN__N[dN / QN]N` OR^]a`N^ ORON_ ]NQN _NYNU _N`a aWa[T[dN& ?VXN ON`N[T `R^_ROa` QVYR]N_ QN^V XRNQNN[ QVNZ QR[TN[ ]¥_V_V ZR[QN`N^$ OR_N^ ]R^PR]N`N[ _aQa` ON`N[T NQNYNU &&&& N& + , = Q& " += A O& + A = R& + * = A P& * += A -& HROaNU _VYV[QR^ ]RWNY dN[T ZN__N[dN B QN[ WN^V%WN^V[dN 4 QVYR]N_ `N[]N XRPR]N`N[ NcNY QN^V ]a[PNX OVQN[T ZV^V[T dN[T XN_N^$ QR[TN[ _aQa` XRZV^V[TN[ * `R^UNQN] OVQN[T U¥^Ve¥[`NY& ?VXN ]R^PR]N`N[ T^NbV`N_V OaZV NQNYNU =$ _VYV[QR^ `R^_ROa` NXN[ &&&& N& ZRYa[Pa^ QR[TN[ ]R^PR]N`N[ = _V[ * O& ZRYa[Pa^ QR[TN[ ]R^PR]N`N[ + * = _V[ * P& ZR[TTRYV[QV[T QR[TN[ ]R^PR]N`N[ = _V[ * Q& ZR[TTRYV[QV[T QR[TN[ ]R^PR]N`N[ ) * = _V[ * R& ZR[TTRYV[QV[T QR[TN[ ]R^PR]N`N[ * + = _V[ * .& 9aN OR[QN 6 QN[ 7 ZN_V[T%ZN_V[T ZN__N[dN - XT$ QVUaOa[TXN[ ¥YRU ON`N[T dN[T ]N[WN[T[dN ) Z !ZN__N ON`N[T QVNONVXN["& ?VXN ]a_N` ON`N[T QVTa[NXN[ _RONTNV _aZOa ]a`N^$ Z¥ZR[ V[R^_VN ]NQN ]a_N` ON`N[T `R^_ROa` NQNYNU &&&& N& *$- XTZ* Q& ,$( XTZ* O& +$( XTZ* R& -$( XTZ* P& +$- XTZ* /& HROaNU ^¥QN QN]N` ZR[T% TRYV[QV[T `N[]N _RYV] ]NQN _ROaNU OVQN[T QN`N^ dN[T XN_N^& BN__N ^¥QN ($- XT QN[ WN^V%WN^V[dN *( PZ& G¥QN QV`N^VX QR[TN[ TNdN , 4 * C _RUV[TTN OR^TR^NX QR[TN[ ]R^PR]N`N[ _aQa` X¥[_`N[ _R]R^`V `NZ]NX ]NQN TNZON^& 7R_N^ Z¥ZR[ TNdN dN[T ORXR^WN ]NQN ^¥QN NQNYNU &&&& N& [¥Y Q& ($- CZ O& ($* CZ R& ($0 CZ P& ($, CZ 0& HROaNU ^¥QN OR^OR[`aX _VYV[QR^ OR^¥[TTN ZRZVYVXV WN^V% WN^V *( PZ QN[ ZN__N - XT& ENQN ^¥QN ORXR^WN Z¥ZR[ TNdN _ROR_N^ ). CZ& 7R_N^ ]R^PR]N`N[ _aQa` ^¥QN `R^_ROa` NQNYNU &&&& N& ($. ^NQ'_* Q& *, ^NQ'_* O& . ^NQ'_* R& 0( ^NQ'_* P& )* ^NQ'_* 1& ER^UN`VXN[ TNZON^ OR^VXa`& IVTN OaNU ]N^`VXRY QR[TN[ ZN__N B$ *B$ QN[ +B QV% ]N_N[T ]NQN aWa[T XR^N[TXN dN[T ZN__N[dN QVNONVXN[& HV_`RZ `R^YR`NX ]NQN OVQN[T LM& ?VXN _V_`RZ QV]a`N^ `R^UNQN] _aZOa-M# Z¥ZR[ V[R^_VN _V_`RZ QV `V`VX 1 NQNYNU &&&& N& / B7 Q& . B7* O& - B7 R& / B7* P& - B7* )(& ER^UN`VXN[ TNZON^ OR^VXa`& y 2m 3m a a m x O 2a F = 2 N ( m1 m2 Tes Kompetensi Bab 5
138 Mudah dan Aktif Belajar Fisika untuk Kelas XI 2% 6>Q>?H>D LBMO>JR>>J ?BMEGPO EJE ABJC>J OBL>O% )& HROaNU ^¥QN OR^ZN__N -(( T QR[TN[ Z¥ZR[ V[R^_VN ($()- XT Z* ZaYN%ZaYN OR^]a`N^ QR[TN[ OR_N^ XRPR]N`N[ +( ]a`N^N['_RX¥[& HR`RYNU ^¥QN ZR[RZ]aU ).+ ]a`N^N[$ ^¥QN `R^_ROa` OR^UR[`V& 7R^N]NXNU OR_N^ `¥^_V dN[T ZR[TUR[`VXN[ ]a`N^N[ ^¥QN `R^_ROa`5 *& HROaNU ^¥QN dN[T ZN__N[dN )0( XT QN[ WN^V%WN^V[dN )$* Z ZRYNXaXN[ - ]a`N^N['_& 7R^N]NXNU R[R^TV XV[R`VX ^¥`N_V ^¥QN `R^_ROa`5 +& HROaNU PV[PV[ `RONY 4 YaN^[dN -( PZ QN[ 4 QNYNZ[dN ,( PZ$ _RQN[TXN[ ZN__N[dN - XT& ?VXN PV[PV[ `R^_ROa` OR^¥`N_V ]NQN _aZOa[dN$ OR^N]NXNU Z¥ZR[ V[R^_VN[dN5 ,& HaN`a `V`VX ZN`R^V ZRYNXaXN[ TR^NX ^¥`N_V QR[TN[ ]R^PR]N`N[ _aQa`( 4 .H ^NQ'_* & ?VXN WN^V%WN^V YV[T% XN^N[[dN , PZ$ UV`a[TYNU2 N& ]R^PR]N`N[ `N[TR[_VNY ]NQN _NN` H 4 *_3 O& XRPR]N`N[ YV[RN^ ]NQN _NN` H 4 *_3 P& ]N[WN[T YV[`N_N[ dN[T QV`RZ]aU ]NQN _NN` H 4 *_& -& HROaNU XN`^¥Y dN[T OR^a]N ^¥QN ]RWNY U¥Z¥TR[$ QVTN[`a[T ]NQN _aZOa[dN& ENQN `R]V ^¥QN QVYVYV`XN[ `NYV$ XRZaQVN[ aWa[T `NYV V`a QV`N^VX QR[TN[ OR_N^ TNdN , 4 . C bR^`VXNY XR ONcNU$ _R]R^`V ]NQN TNZON^& ?VXN ZN__N XN`^¥Y - XT QN[ WN^V%WN^V[dN )* PZ$ UV`a[TYNU2 N& ]R^PR]N`N[ `NYV3 O& Z¥ZR[`aZ _aQa`[dN ]NQN _NN` H 4 ($- _RX¥[3 P& ]N[WN[T `NYV dN[T `R^`N^VX _RYNZN * _& .& HROaNU _VYV[QR^ OR^YaON[T OR^ZN__N . XT QN[ WN^V% WN^V[dN ,- PZ OR^]a`N^ QR[TN[ XRPR]N`N[ *(( ^]Z !FDH7H?DC E;F B?CIH;"& IR[`aXN[YNU Z¥ZR[ V[R^_VN QN[ R[R^TV XV[R`VX ^¥`N_V ^¥QN `R^_ROa`5 /& HROaNU ^¥QN TVYN QR[TN[ Z¥ZR[ V[R^_VN +$0 XTZ* ¥YRU _aN`a `¥^_V QV]R^PR]N` UV[TTN QNYNZ . ]a`N^N[ _NWN& @RPR]N`N[ _aQa`[dN OR^aONU QN^V * ]a`N^N['_ ZR[WNQV - ]a`N^N['_& 7R^N]NXNU OR_N^ `¥^_V ^¥QN `R^_ROa`5 0& IR[`aXN[YNU `¥^_V dN[T UN^a_ QVOR^VXN[ NTN^ QNYNZ cNX`a )( QR`VX QN]N` ZRZOR^V XRPR]N`N[ _aQa` _ROR_N^ +(( ^]Z ]NQN ^¥QN TVYN -( XT QR[TN[ WN^V%WN^V ,( PZ& !- 4 /4* " 1& HROaNU ^¥QN OR^ZN__N , XT QR[TN[ WN^V%WN^V *( PZ OR^]a`N^ QR[TN[ XRPR]N`N[ +.( ^]Z& ENQN ^¥QN `R^% _ROa` ORXR^WN TNdN TR_RX dN[T ZR[dRONOXN[ `¥^_V _ROR_N^ ($)* C&Z& 7R^N]N YNZN ^¥QN NXN[ OR^UR[`V5 !- 4 /4* " )(& HROaNU ^¥QN OR^ZN__N *- XT QR[TN[ WN^V%WN^V ** PZ OR^]a`N^ QR[TN[ XRPR]N`N[ . ]a`N^N['QR`VX& 7R^N]N% XNU R[R^TV XV[R`VX ^¥`N_V ^¥QN `R^_ROa`5 F B A T1 T2 C licin 9aN OaNU OR[QN dN[T ZN_V[T%ZN_V[T ZRZVYVXV ZN__N B) 4 , XT QN[ B* 4 * XT QVUaOa[TXN[ QR[TN[ XN`^¥Y ]RWNY OR^ZN__N , XT& ?VXN ]R^PR]N`N[ T^NbV`N_V = 4 )( Z'_* $ ]R^PR]N`N[ dN[T QVNYNZV B) QN[ B* NQNYNU &&&& N& )( Z'_* Q& *$- Z'_* O& - Z'_* R& * Z'_* P& +$++ Z'_* ))& 9aN OR[QN$ ZN_V[T% ZN_V[T OR^ZN__N B)4 , XT QN[ B* 4 , XT QVUaOa[T% XN[ QR[TN[ XN`^¥Y ]RWNY dN[T ZN__N[dN , XT _R]R^`V `NZ]NX ]NQN TNZON^& ?VXN ]R^ZaXNN[ OVQN[T ZV^V[T 67 YVPV[ QN[ XN`^¥Y `VQNX _RYV]$ OR_N^ ]R^PR]N`N[ OR[QN B) $ QN[ B* NQNYNU &&&& != 4 )( Z'_* " N& )$( Z'_* Q& *$* Z'_* O& )$- Z'_* R& *$- Z'_* P& *$( Z'_* )*& ?VXN QaN ^¥QN ZN_V[T%ZN_V[T OR^WN^V%WN^V 4) QN[ 4* QV]a`N^ QR[TN[ `NYV QV `V`VX _V[TTa[T[dN$ XRPR]N`N[ _aQa` !,"$ ]R^V¥QR !5"$ S^RXaR[_V !<"$ QN[ XRPR]N`N[ YV[RN^ !J" ZRZVYVXV UaOa[TN[ QR[TN[ WN^V%WN^V 4 _RONTNV OR^VXa`$ @;9I7A? &&&& N& , , ) * 4 * ) 4 4 Q& $ ) ) * * J 4 J 4 O& $ ) * * ) 5 4 5 4 R& R , , $ ) * * * 4 P& $ ) * * ) < 4 < 4 )+& ENQN TNZON^ QV _NZ]V[T$ ZN__N ONY¥X 6$ ZN__N ORON[ 7$ QN[ ZN__N ^¥QN XN`^¥Y OR^¥[TTN 8 ZN_V[T%ZN_V[T NQNYNU * XT$ / XT$ QN[ ) XT& ?VXN = 4)( Z'_* $ `RTN[TN[ `NYV 5) NQNYNU &&&& N& +( C Q& 0 C O& *( C R& / C P& ), C ),& HROaNU PNX^NZ QR[TN[ WN^V%WN^V 4 OR^]a`N^ OR^N`a^N[ ZRYNYaV _aZOa U¥^Ve¥[`NY& IV`VX 2 `R^YR`NX ]NQN `R]V PNX^NZ QN[ 3 ]NQN ]R^`R[TNUN[ N[`N^N ]a_N` QN[ `V`VX 2& ER^[dN`NN[ OR^VXa` dN[T `VQNX OR[N^ NQNYNU $$$$ N& XRPR]N`N[ _aQa`[dN _NZN O& ]R^PR]N`N[ _aQa` XRQaN[dN _NZN QR[TN[ [¥Y P& XRPR]N`N[ YV[RN^ 2 QaN XNYV XRPR]N`N[ YV[RN^ 3 Q& ]R^PR]N`N[ `N[TR[_VNY XRQaN[dN _NZN QR[TN[ [¥Y R& XRPR]N`N[ YV[RN^ XRQaN[dN _NZN )-& HROaNU O¥YN ]RWNY ZR[TTRYV[QV[T `N[]N _RYV] ]NQN OVQN[T ZV^V[T QR[TN[ _aQa` XRZV^V[TN[ ! * & ER^PR]N`N[ YV[RN^ O¥YN `R^_ROa` WVXN QVXR`NUaV ]R^PR]N`N[ T^NbV`N_V[dN = NQNYNU &&&& N& * _V[ + = * Q& - _V[ / = * O& ) _V[ , = * R& = _V[* P& ) _V[ + = * A B m1 30° m2
139 A. Fluida Statis B. Viskositas Fluida C. Fluida Dinamis JYebU¥_U¥ ;bXU VYegUbkU' aYb[UdU dYfUjUg XUdUg gYeVUb[: JYfUjUg XUdUg gYeVUb[ _UeYbU aYb[U`Ua] [UkU Ub[_Ug kUb[ VYfUe) AUkU Ub[_Ug dUXU dYfUjUg X]fYVUV_Ub c`Y¥ _cbfgeh_f] fUkUd dYfUjUg kUb[ aYbkYVUV_Ub UXUbkU dYeVYXUUb _YWYdUgUb U`]e XUe] Z`h]XU hXUeU X] UgUf fUkUd XYb[Ub X] VUjU¥ fUkUd) Je]bf]d gYefYVhg aYehdU_Ub Ud`]_Uf] XUe] dYefUaUUb <Yebch``]) >U`Ua VUV ]b]' ;bXU U_Ub VY`U^Ue gYbgUb[ Z`h]XU fgUg]f XUb Z`h]XU X]bUa]f) EcbfYd gYbgUb[ Z`h]XU VUbkU_ X]Ud`]_Uf]_Ub XU`Ua _Y¥]XhdUb fY¥Ue](¥Ue]) Fluida Pesawat dapat terbang karena mengalami gaya angkat yang bebas pada sayapnya. Sumber:www.chez.com menganalisis hukum-hukum yang berhubungan dengan fluida statis dan dinamis serta penerapannya dalam kehidupan sehari-hari. Setelah mempelajari bab ini, Anda harus mampu: menerapkan konsep dan prinsip mekanika klasik sistem kontinu dalam menyelesaikan masalah. Hasil yang harus Anda capai: Bab 6
140 Mudah dan Aktif Belajar Fisika untuk Kelas XI A. Fluida Statis @`h]XU aYehdU_Ub lUg kUb[ XUdUg aYb[U`]e) =cbgc¥ Z`h]XU X] UbgUeUbkU U]e' a]bkU_' fhfh' VYbf]b' fc`Ue' hUd' [Uf' XUb UfUd) @`h]XU gYeX]e] UgUf XhU aUWUa' kU]gh Z`h]XU kUb[ XUdUg X]aUadUg_Ub UgUh ?CADE9F=69@ XUb kUb[ g]XU_ XUdUg X]aUadUg_Ub UgUh =B?CADE9F=69@# 1. Tekanan =cVU ;bXU gY_Ub fYVhU¥ dh`dYb _Y XU`Ua gUbU¥' _YahX]Ub VUbX]b[( _Ub g]b[_Ug _Yfh_UeUbbkU XYb[Ub ^Ueha kUb[ X]gY_Ub _Y XU`Ua gUbU¥) Jh`dYb U_Ub `YV]¥ fh_Ue aUfh_ _Y XU`Ua gUbU¥ XUe]dUXU ^Ueha kUb[ X]aUfh__Ub _Y XU`Ua gUbU¥) GYb[UdU XYa]_]Ub: N]b[_Ug _Yfh`]gUb fYfYceUb[ XU`Ua aYbY_Ub fhUgh VYbXU' VYe¥hVhb[Ub XYb[Ub `hUf XUYeU¥ kUb[ X]gY_Ub) MYaU_]b `hUf XUYeU¥ kUb[ X]gY_Ub' fYaU_]b _YW]` gY_UbUb kUb[ X]¥Uf]`_Ub) 29?5B5B A9EHD5?5B 69F5E ;5L5 L5B; 69?9E>5 D585 FH5GH D9EAH?55B 8=65;= 89B;5B @H5F D9EAH?55B G9EF96HG) D]_U [UkU , VY_Ye^U gY[U_ `hehf dUXU dYeah_UUb VYbXU fY`hUf (' VYfUebkU gY_UbUb fYWUeU aUgYaUg]f X]gh`]f_Ub fYVU[U] VYe]_hg) $ , D ( #1o,$ EYgYeUb[Ub5 D 8 gY_UbUb #H*a- 8 dUfWU`$ , 8 [UkU #H$ ( 8 `hUf V]XUb[ gY_Ub #a- $ 2. Tekanan Hidrostatik JYe¥Ug]_Ub 6=H>=L -&)) JUXU [UaVUe gYefYVhg gYe`]¥Ug fYVhU¥ gUVhb[ VYe]f] lUg WU]e VYeaUffU ^Yb]f ) ' _YXU`UaUb <' XUb `hUf dYbUadUb[ () RUg WU]e kUb[ VYeUXU X] XU`Ua VY^UbU aYa]`]_] [UkU VYeUg P kUb[ aYbY_Ub XUfUe gUVhb[) MYaU_]b g]b[[] dYeah_UUb lUg WU]e' fYaU_]b VYfUe gY_UbUb kUb[ X]¥Uf]`_Ub dUXU XUfUe gUVhb[) MYWUeU aUgYaUg]f' ¥hVhb[Ub UbgUeU VYfUe gY_UbUb kUb[ X]¥Uf]`_Ub XUb _Yg]b[[]Ub lUg WU]e X]gh`]f_Ub fYVU[U] VYe]_hg) ! ) $$ $ A; 3 J D (( ( ) $ ;<( ; ( 8 );< #1o-$ <YeXUfUe_Ub :ALM=H==I !-S)"' XUdUg X]f]adh`_Ub VU¥jU dYefUaUUb gY_UbUb ¥]XecfgUg]_ UXU`U¥ D ; $ ) < #1o.$ EYgYeUb[Ub5 D 8 gY_UbUb ¥]XecfgUg]_ #H*a- $ UgUh JU ) 8 aUffU ^Yb]f lUg WU]e #_[*a. $ Gambar 6.2 Tekanan hidrostatik pada dasar tabung. Gambar 6.1 Tekanan yang dilakukan seorang anak terhadap meja. w A h ,) JYebU¥_U¥ ;bXU XUgUb[ _Y gYadUg WhW] acV]`: Bh_ha UdU_U¥ kUb[ X]gYeUd_Ub XU`Ua U`Ug dYb[( Ub[_Ug acV]` X] fUbU: <U[U]aUbU WUeU _Ye^U U`Ug dYb[( Ub[_Ug acV]` gYefYVhg: -) JYeWUkU_U¥ ;bXU ^]_U UXU gYaUb ;bXU kUb[ aYbkUgU_Ub VU¥jU X]U V]fU aYbYbgh_Ub _YUf`]Ub `c[Ua ah`]U YaUf ¥UbkU XYb[Ub aYbWY`hd_UbbkU _Y XU`Ua U]e: .) GYb[UdU bkUah_ XUdUg VYe^U`Ub X] UgUf dYeah_UUb U]e: /) DY`Uf_Ub XYb[Ub VU¥UfUa ;bXU fYbX]e] WUeU _Ye^U XUe] iYbghe]aYgYe XUb d]dU d]gcg) ;A>AGOH HAHKAG=E=LD FJIMAK 5GOD@=$ FALE=F=IG=C MJ=G%MJ=G >ALDFON @=G=H >OFO G=NDC=I& Tes Kompetensi Awal
Fluida 141 < 8 _YXU`UaUb lUg WU]e #a$ ; 8 dYeWYdUgUb [eUi]gUf] #a*f- $ >Ue] dYefUaUUb gYefYVhg' XUdUg X]dY`U^Ue] VU¥jU gY_UbUb ¥]XecfgUg]_ fUb[Ug X]dYb[Ueh¥] c`Y¥ [UkU [eUi]gUf] XUb _YXU`UaUb lUg WU]e) JYe¥Ug]_Ub 6=H>=L -&*) JUXU _Ufhf fYdYeg] ]b] #hbgh_ lUg WU]e kUb[ fY^Yb]f$' gY_UbUb lUg WU]e g]XU_ VYe[Ubghb[ dUXU `hUf dYbUadUb[ XUb VYbgh_ VY^UbU' gYgUd] VYe[Ubghb[ dUXU _YXU`UaUb lUg WU]e) DUX]' gY_UbUb ¥]XecfgUg]_ dUXU g]g]_ ;' <' XUb = UXU`U¥ fUaU VYfUe) I`Y¥ _UeYbU ]gh' VYeXUfUe_Ub :ALM=H==I !-S)" U_Ub X]XUdUg D ; 8 D < 8 D = 8 ) ;< #1o/$ JYe¥Ug]_Ub 6=H>=L -&+) Gh`U(ah`U VY^UbU dUXU [UaVUe gYefYVhg X]]f] lUg WU]e dYegUaU kUb[ VYeaUffU ^Yb]f )1) EYahX]Ub' _Y XU`Ua ah`hg VY^UbU fYVY`U¥ _UbUb X]aUfh__Ub lUg WU]e _YXhU kUb[ VYeaUffU ^Yb]f )2) N]g]_ < VYeUXU dUXU dYeVUgUfUb _YXhU lUg WU]e gYefYVhg XUb X]gY_Ub c`Y¥ lUg WU]e _YXhU fYg]b[[] <- )N]g]_ ; VYeUXU dUXU lUg WU]e dYegUaU XUb X]gY_Ub c`Y¥ lUg WU]e dYegUaU fYg]b[[] <, ) N]g]_ ; XUb < VYeUXU dUXU fUgh [Ue]f) MYfhU] XYb[Ub Bh_ha B]XecfgUg]_U' _YXhU g]g]_ gYefYVhg aYa]`]_] gY_UbUb kUb[ fUaU) ;_Ub gYgUd]' gY_UbUb dUXU g]g]_ = XUb > g]XU_ fUaU _UeYbU ^Yb]f lUg WU]e X] _YXhU g]g]_ gYefYVhg VYeVYXU) D ; 8 D < ), <, B 8 )- <- B #1o0$ Obgh_ `YV]¥ aYaU¥Ua] gYbgUb[ gY_UbUb ¥]XecfgUg]_U' `U_h_Ub`U¥ ;_g]i]gUf @]f]_U VYe]_hg ]b]) Gambar 6.4 Pipa U diisi dua zat cair berbeda, tekanan di A sama dengan tekanan di B. A C B h2 h1 air D Aktivitas Fisika 6.1 Tekanan Hidrostatik Tujuan Percobaan Memahami tekanan hidrostatika Alat-Alat Percobaan 1. Sebuah ember 2. dua buah gelas plastik bening Langkah-Langkah Percobaan 1. Isilah ember dengan air. 2. Jawablah pertanyaan berikut sebelum melakukan percobaan. a. Apa yang akan terjadi jika sebuah gelas plastik dicelupkan perlahan ke dalam air dalam keadaan terbalik? Jelaskan apa yang akan terjadi, berikut alasannya. b. Lakukan hal yang sama, namun dengan gelas yang diberi lubang pada dasar gelasnya. Jelaskan apa yang akan terjadi, berikut alasannya. 3. Buktikanlah jawaban Anda dengan melakukan percobaan. 4. Buatlah hukum tentang tekanan menurut versi Anda, berdasarkan jawaban dan hasil percobaan yang Anda peroleh. 5. Bandingkanlah dengan hukum hidrostatika yang telah Anda pelajari. 6. Buatlah kesimpulan dari percobaan tersebut. MYVhU¥ gYadUg U]e VYeVYbgh_ _hVhf aYa]`]_] dUb^Ub[ ehfh_ 1+ Wa X]]f] ,3+ `]gYe U]e #aUffU ^Yb]f U]e 8 ,+. _[*a. $) D]_U ; 8 ,+ a*f- ' gYbgh_Ub`U¥5 U) gY_UbUb ¥]XecfgUg]_ dUXU XUfUe _hVhf6 V) [UkU ¥]XecfgUg]_ dUXU XUfUe _hVhf6 XUb W) [UkU ¥]XecfgUg]_ dUXU g]g]_ < kUb[ VYe^UeU_ +'-0 a XUe] dYeah_UUb U]e) Contoh 6.1 Gambar 6.3 Tekanan hidrostatik pada titik A, B, dan C adalah sama. A B C h