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Mehanika tverdogo tela. Lekcii.

V.A.Aleshkevich, L.G.Dedenko, V.A.Karavaev (Fizicheskii fakul'tet MGU)
Izdatel'stvo Fizicheskogo fakul'teta MGU, 1997 g. Soderzhanie

Centr udara.

Opyt pokazyvaet, chto esli telo, zakreplennoe na osi vrasheniya, ispytyvaet udar, to deistvie udara v obshem sluchae peredaetsya i na os'. Pri etom velichina i napravlenie sily, prilozhennoi k osi, zavisyat ot togo, v kakuyu tochku tela nanesen udar.

Rassmotrim sploshnoi odnorodnyi sterzhen' AV, podveshennyi v tochke A na gorizontal'noi, zakreplennoi v podshipnikah osi OO' (ris. 3.9). Esli udar (korotkodeistvuyushaya sila F ( nanesen blizko k osi vrasheniya, to os' progibaetsya v napravlenii deistviya sily F (ris. 3.9a). Esli udar nanesen po nizhnemu koncu sterzhnya, vblizi tochki V, to os' progibaetsya v protivopolozhnom napravlenii (ris. 3.9b). Nakonec, esli udar nanesen v strogo opredelennuyu tochku sterzhnya, nazyvaemuyu centrom udara (ris. 3.9v, tochka S), to os' ne ispytyvaet nikakih dopolnitel'nyh nagruzok, svyazannyh s udarom. Ochevidno, v etom sluchae skorost' postupatel'nogo dvizheniya, priobretaemogo tochnoi A vmeste s centrom mass O, budet kompensirovat'sya lineinoi skorost'yu vrashatel'nogo dvizheniya vokrug centra mass O (oba eti dvizheniya iniciiruyutsya siloi F i proishodyat odnovremenno).

Ris. 3.9.

Vychislim, na kakom rasstoyanii $\ell$ ot tochki podvesa sterzhnya nahoditsya centr udara. Uravnenie momentov otnositel'no osi vrasheniya OO' daet

$J \cdot {\displaystyle \frac{\displaystyle {\displaystyle d\omega}}{\displaystyle {\displaystyle dt}}} = F \cdot \ell .$

(3.15)

Sil reakcii so storony osi, kak predpolagaetsya, pri udare ne voznikaet, poetomu na osnovanii teoremy o dvizhenii centra mass mozhno zapisat'

$m \cdot {\displaystyle \frac{\displaystyle {\displaystyle dv_{0} }}{\displaystyle {\displaystyle dt}}} = F,$

(3.16)

gde $m$ - massa tela, $v_{0}$ - skorost' centra mass. Esli $a$ - rasstoyanie ot osi do centra mass tela, to

$v_{0} = \omega a,$

(3.17)

i v rezul'tate iz uravneniya momentov i uravneniya dvizheniya centra mass nahodim

$\ell = {\displaystyle \frac{\displaystyle {\displaystyle J}}{\displaystyle {\displaystyle ma}}}.$

(3.18)

Pri etom tochka C (centr udara) sovpadaet s tak nazyvaemym centrom kachaniya dannogo fizicheskogo mayatnika - tochkoi, gde nado sosredotochit' vsyu massu tverdogo tela, chtoby poluchennyi matematicheskii mayatnik imel takoi zhe period kolebanii, kak i dannyi fizicheskii.

V sluchae sploshnogo odnorodnogo sterzhnya dlinoi $L$ imeem:

$a = {\displaystyle \frac{\displaystyle {\displaystyle L}}{\displaystyle {\displaystyle 2}}}, \quad J = {\displaystyle \frac{\displaystyle {\displaystyle mL^{2}}}{\displaystyle {\displaystyle 3}}},\quad i \quad \ell = {\displaystyle \frac{\displaystyle {\displaystyle 2}}{\displaystyle {\displaystyle 3}}}L.$

Zamechanie. Poluchennoe vyrazhenie dlya $\ell$ (3.18) spravedlivo i dlya proizvol'nogo tverdogo tela. Pri etom nado tol'ko imet' v vidu, chto tochka podvesa tela A i centr mass O dolzhny lezhat' na odnoi vertikali, a os' vrasheniya dolzhna sovpadat' s odnoi iz glavnyh osei inercii tela, prohodyashih cherez tochku A.

Primer 1. Pri udarah palkoi dlinoi $L$ po prepyatstviyu ruka "ne chuvstvuet" udara (ne ispytyvaet otdachi) v tom sluchae, esli udar prihoditsya v tochku, raspolozhennuyu na rasstoyanii $L - \ell = L - {\displaystyle \frac{\displaystyle {\displaystyle 2}}{\displaystyle {\displaystyle 3}}}L = {\displaystyle \frac{\displaystyle {\displaystyle 1}}{\displaystyle {\displaystyle 3}}}L$ svobodnogo konca palki.

Primer 2. Pri gorizontal'nom udare kiem po bil'yardnomu sharu (ris. 3.10) shar nachinaet kachenie bez proskal'zyvaniya v tom sluchae, ecli udar nanesen v tochku, nahodyashuyusya na vysote

$h = {\displaystyle \frac{\displaystyle {\displaystyle J}}{\displaystyle {\displaystyle ma}}} = {\displaystyle \frac{\displaystyle {\displaystyle {\displaystyle \frac{\displaystyle {\displaystyle 7}}{\displaystyle {\displaystyle 5}}}mR^{2}}}{\displaystyle {\displaystyle mR}}} = {\displaystyle \frac{\displaystyle {\displaystyle 7}}{\displaystyle {\displaystyle 5}}}R$

ot poverhnosti bil'yarda, to est' na $h - R = {\displaystyle \frac{\displaystyle {\displaystyle 2}}{\displaystyle {\displaystyle 5}}}R$ vyshe centra shara. Esli udar budet nanesen nizhe, kachenie budet soprovozhdat'sya skol'zheniem v napravlenii dvizhenii shara. Esli udar nanesen vyshe, to shar v tochke kasaniya s bil'yardnym stolom budet proskal'zyvat' nazad.

Ris. 3.10.

Rassmotrennye primery formal'no ne otnosyatsya k vrasheniyu tverdogo tela vokrug nepodvizhnoi osi, odnako vse privedennye vyshe soobrazheniya o centre udara, ochevidno, ostayutsya v sile i v etih sluchayah.

II. Ploskoe dvizhenie tverdogo tela.

Napomnim, chto pri ploskom dvizhenii vse tochki tela dvizhutsya v ploskostyah, parallel'nyh nekotoroi nepodvizhnoi ploskosti, poetomu dostatochno rassmotret' dvizhenie odnogo iz secheniya tela, naprimer, togo, v kotorom lezhit centr mass. Pri razlozhenii ploskogo dvizheniya na postupatel'noe i vrashatel'noe skorost' postupatel'nogo dvizheniya opredelena neodnoznachno - ona zavisit ot vybora osi vrasheniya, odnako uglovaya skorost' vrashatel'nogo dvizheniya okazyvaetsya odnoi i toi zhe (sm. lekciyu 1).

Esli v kachestve osi vrasheniya vybrat' os', prohodyashuyu cherez centr mass, to uravneniyami dvizheniya tverdogo tela budut:

1. Uravnenie dvizheniya centra mass

$m{\displaystyle \frac{\displaystyle {\displaystyle d{\displaystyle \bf v}_{0} }}{\displaystyle {\displaystyle dt}}} = {\displaystyle \bf F}_{0} .$

(3.19)

2. Uravnenie momentov otnositel'no osi, prohodyashei cherez centr mass

$J_{0} {\displaystyle \frac{\displaystyle {\displaystyle d\omega}}{\displaystyle {\displaystyle dt}}} = {\displaystyle \bf M}_{0} .$

(3.20)

Osobennost'yu ploskogo dvizheniya yavlyaetsya to, chto os' vrasheniya sohranyaet svoyu orientaciyu v prostranstve i ostaetsya perpendikulyarnoi ploskosti, v kotoroi dvizhetsya centr mass. Eshe raz podcherknem, chto uravnenie momentov (3.20) zapisano otnositel'no, v obshem sluchae, uskorenno dvizhushegosya centra mass, odnako, kak bylo otmecheno v nachale lekcii, ono imeet takoi zhe vid, kak i uravnenie momentov otnositel'no nepodvizhnoi tochki.

V kachestve primera rassmotrim zadachu o skatyvanii cilindra s naklonnoe ploskosti. Privedem dva sposoba resheniya etoi zadachi s ispol'zovaniem uravnenii dinamiki tverdogo tela.

Pervyi sposob. Rassmatrivaetsya vrashenie cilindra otnositel'no osi, prohodyashee cherez centr mass (ris. 3.11).

Ris. 3.11.

Sistema uravnenii (3.19 - 3.20) imeet vid:

$\left\{\displaystyle {\displaystyle \begin{array}{l} {\displaystyle m \cdot {\displaystyle \frac{\displaystyle {\displaystyle d{\displaystyle \bf v}_{0} }}{\displaystyle {\displaystyle dt}}}m{\displaystyle \bf g} + {\displaystyle \bf F}_{tr} + {\displaystyle \bf N};} \\ {\displaystyle J_{0} {\displaystyle \frac{\displaystyle {\displaystyle d\omega}}{\displaystyle {\displaystyle dt}}} = {\displaystyle \bf R}\times {\displaystyle \bf F}_{tr} .} \\ \end{array}} \right.} \quad \begin{array}{l} (3.21) \\ (3.22) \\ \end{array$

K etoi sisteme neobhodimo dobavit' uravnenie kinematicheskoi svyazi

${\displaystyle \frac{\displaystyle {\displaystyle d{\displaystyle \bf v}_{0} }}{\displaystyle {\displaystyle dt}}} = {\displaystyle \bf R}\times {\displaystyle \frac{\displaystyle {\displaystyle d\omega}}{\displaystyle {\displaystyle dt}}}.$

(3.23)

Poslednee uravnenie poluchaetsya iz usloviya, chto cilindr skatyvaetsya bez proskal'zyvaniya, to est' skorost' tochki M cilindra ravna nulyu.

Uravnenie dvizheniya centra mass (3.1) zapishem dlya proekcii uskoreniya i sil na os' x vdol' naklonnoi ploskosti, a uravnenie momentov (3.22) - dlya proekcii uglovogo uskoreniya i momenta sily treniya na os' y , sovpadayushuyu s os'yu cilindra. Napravleniya osei x i u vybrany soglasovanno, v tom smysle, chto polozhitel'nomu lineinomu uskoreniyu osi cilindra sootvetstvuet polozhitel'noe zhe uglovoe uskorenie vrasheniya vokrug etoi osi. V itoge poluchim:

$\left\{\displaystyle {\displaystyle \begin{array}{l} {\displaystyle ma = mg\sin \alpha - F_{tr} ;} \\ {\displaystyle J_{0} {\displaystyle \frac{\displaystyle {\displaystyle d\omega }}{\displaystyle {\displaystyle dt}}} = F_{tr} \cdot R;} \\ {\displaystyle a = {\displaystyle \frac{\displaystyle {\displaystyle d\omega }}{\displaystyle {\displaystyle dt}}} \cdot R.} \\ \end{array}} \right.} \quad \begin{array}{l} (3.24) \\ (3.25) \\ (3.26) \\ \end{array$

otkuda

$a = {\displaystyle \frac{\displaystyle {\displaystyle g\sin \alpha }}{\displaystyle {\displaystyle 1 + {\displaystyle \frac{\displaystyle {\displaystyle J_{0} }}{\displaystyle {\displaystyle mR^{2}}}}}}}.$

(3.27)

Sleduet podcherknut', chto $F_{tr}$ - sila treniya scepleniya - mozhet prinimat' lyuboe znachenie v intervale ot O do $\left( {\displaystyle F_{tr} } \right)_{maks}$ (sila treniya skol'zheniya) v zavisimosti ot parametrov zadachi. Rabotu eta sila ne sovershaet, no obespechivaet uskorennoe vrashenie cilindra pri ego skatyvanii s naklonnoi ploskosti. V dannom sluchae

$F_{tr} = {\displaystyle \frac{\displaystyle {\displaystyle J_{0} }}{\displaystyle {\displaystyle R^{2}}}} \cdot {\displaystyle \frac{\displaystyle {\displaystyle g\sin \alpha }}{\displaystyle {\displaystyle 1 + {\displaystyle \frac{\displaystyle {\displaystyle J_{0} }}{\displaystyle {\displaystyle mR^{2}}}}}}}.$

(3.28)

Esli cilindr sploshnoi, to

$J_{0} = {\displaystyle \frac{\displaystyle {\displaystyle 1}}{\displaystyle {\displaystyle 2}}}mR^{2}; \quad a = {\displaystyle \frac{\displaystyle {\displaystyle 2}}{\displaystyle {\displaystyle 3}}}g\sin \alpha ; \quad F_{tr} = {\displaystyle \frac{\displaystyle {\displaystyle 1}}{\displaystyle {\displaystyle 3}}}mg\sin \alpha .$

(3.29)

Kachenie bez proskal'zyvaniya opredelyaetsya usloviem

$F_{tr} \le kN,$

(3.30)

gde $k$ - koefficient treniya skol'zheniya, $N = mg\cos \alpha$ - sila reakcii opory. Eto uslovie svoditsya k sleduyushemu:

${\displaystyle \frac{\displaystyle {\displaystyle 1}}{\displaystyle {\displaystyle 3}}}mg\sin \alpha \le kmg\cos \alpha ,$

(3.31)

ili

$tg\alpha \le 3k.$

(3.32)

Vtoroi sposob. Rassmatrivaetsya vrashenie cilindra otnositel'no nepodvizhnoi osi, sovpadayushei v dannyi moment vremeni s mgnovennoi os'yu vrasheniya (ris. 3.12).

Ris. 3.12.

Mgnovennaya os' vrasheniya prohodit cherez tochku soprikosnoveniya cilindra i ploskosti (tochku M). Pri takom podhode otpadaet neobhodimost' v uravnenii dvizhenii centra mass i uravnenii kinematicheskoi svyazi. Uravnenie momentov otnositel'no mgnovennoi osi imeet vid:

$J \cdot {\displaystyle \frac{\displaystyle {\displaystyle d\omega}}{\displaystyle {\displaystyle dt}}} = {\displaystyle \bf R}\times \left( {\displaystyle m{\displaystyle \bf g}} \right).$

(3.33)

Zdes'

$J = J_{0} + mR^{2}.$

(3.34)

V proekcii na os' vrasheniya (os' y)

$J \cdot {\displaystyle \frac{\displaystyle {\displaystyle d\omega }}{\displaystyle {\displaystyle dt}}} = Rmg \cdot \sin \left( {\displaystyle 180^{0} - \alpha } \right) = Rmg\sin \alpha .$

(3.35)

Uskorenie centra mass vyrazhaetsya cherez uglovoe uskorenie

$a = {\displaystyle \frac{\displaystyle {\displaystyle d\omega }}{\displaystyle {\displaystyle dt}}}R = {\displaystyle \frac{\displaystyle {\displaystyle g\sin \alpha }}{\displaystyle {\displaystyle 1 + {\displaystyle \frac{\displaystyle {\displaystyle J_{0} }}{\displaystyle {\displaystyle mR^{2}}}}}}}.$

(3.36)

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