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(1)(2)=KK0<strong>10</strong>2=Mg hMg h12= 1siK01= K02⇒hh12= 1( même hauteur)b) Les mêmes vitesses linéaires, donc angulaires aussi:• Sphère :K = Mg h1201Mv2+12• Disque :K = Mg h1202Mv2+1212I ω122I ω2(1)= Mg h(1)1= Mg h2• Les 2 ont les mêmes v et Sphère :1Mv212127<strong>10</strong>v2 Disque :1Mv221 2+ I1ω2= g h1ω =vR= Mg h12(1)2 1 ⎛ 2 2 ⎞⎛v ⎞Mv + ⎜ M R ⎟⎜⎟ = Mg h12 ⎝ 5 ⎠⎝R ⎠2⎛ 2 ⎞Mv ⎜1+ ⎟ = Mg h1⎝ 5 ⎠21 2+ I2ω= Mg h222(3)(2)1 2 1 ⎛ 1 2 ⎞⎛v ⎞Mv + ⎜ M R ⎟⎜⎟ = Mg h22 2 ⎝ 2 ⎠⎝R ⎠1 2 ⎛ 1 ⎞Mv ⎜1+ ⎟ = Mg h22 ⎝ 2 ⎠3 2v = g h2(4)419
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