Question-214281

Question Number 214281 by ajfour last updated on 03/Dec/24

Commented by ajfour last updated on 03/Dec/24

$${Q}.\:\mathrm{214132} \\ $$

Answered by ajfour last updated on 03/Dec/24

ωbsin θ=Q=q (simply)=sasin φ bcos θ=y=acos φ ω=(q/( (√(b^2 −y^2 )))) ; s=(q/( (√(a^2 −y^2 )))) M{ω((b/2))cos θ+p}=m{s((a/2))cos φ+p} M{((qy)/( 2(√(b^2 −y^2 ))))+p}=m{((qy)/( 2(√(a^2 −y^2 ))))+p} 2(M+m)g(y_0 −y)= ((Mb^2 ω^2 )/(12))+M{(p+((ωb)/2)cos θ)^2 +(q^2 /4)} +((ma^2 s^2 )/(12))+m{(((sa)/2)cos φ−p)^2 +(q^2 /4)} ⇒ ((Mb^2 )/(12))((q^2 /( b^2 −y^2 )))+M{(p+((qy)/( 2(√(b^2 −y^2 )))))^2 +(q^2 /4)} +((ma^2 )/(12))((q^2 /(a^2 −y^2 )))+m{(((qy)/( 2(√(a^2 −y^2 ))))−p)^2 +(q^2 /4)} =2(M+m)g(y_0 −y) For y=0 we have p=0 hence ((Mq^2 )/3)+((mq^2 )/3)=2(M+m)g((h/2)) q=(√(3gh))

$$\omega{b}\mathrm{sin}\:\theta={Q}={q}\:\left({simply}\right)={sa}\mathrm{sin}\:\phi \\ $$$${b}\mathrm{cos}\:\theta={y}={a}\mathrm{cos}\:\phi \\ $$$$\omega=\frac{{q}}{\:\sqrt{{b}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\:\:\:;\:\:\:{s}=\frac{{q}}{\:\sqrt{{a}^{\mathrm{2}} −{y}^{\mathrm{2}} }} \\ $$$${M}\left\{\omega\left(\frac{{b}}{\mathrm{2}}\right)\mathrm{cos}\:\theta+{p}\right\}={m}\left\{{s}\left(\frac{{a}}{\mathrm{2}}\right)\mathrm{cos}\:\phi+{p}\right\} \\ $$$${M}\left\{\frac{{qy}}{\:\mathrm{2}\sqrt{{b}^{\mathrm{2}} −{y}^{\mathrm{2}} }}+{p}\right\}={m}\left\{\frac{{qy}}{\:\mathrm{2}\sqrt{{a}^{\mathrm{2}} −{y}^{\mathrm{2}} }}+{p}\right\} \\ $$$$\mathrm{2}\left({M}+{m}\right){g}\left({y}_{\mathrm{0}} −{y}\right)= \\ $$$$\frac{{Mb}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{12}}+{M}\left\{\left({p}+\frac{\omega{b}}{\mathrm{2}}\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\frac{{q}^{\mathrm{2}} }{\mathrm{4}}\right\} \\ $$$$+\frac{{ma}^{\mathrm{2}} {s}^{\mathrm{2}} }{\mathrm{12}}+{m}\left\{\left(\frac{{sa}}{\mathrm{2}}\mathrm{cos}\:\phi−{p}\right)^{\mathrm{2}} +\frac{{q}^{\mathrm{2}} }{\mathrm{4}}\right\} \\ $$$$\Rightarrow \\ $$$$\frac{{Mb}^{\mathrm{2}} }{\mathrm{12}}\left(\frac{{q}^{\mathrm{2}} }{\:{b}^{\mathrm{2}} −{y}^{\mathrm{2}} }\right)+{M}\left\{\left({p}+\frac{{qy}}{\:\mathrm{2}\sqrt{{b}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\right)^{\mathrm{2}} +\frac{{q}^{\mathrm{2}} }{\mathrm{4}}\right\} \\ $$$$+\frac{{ma}^{\mathrm{2}} }{\mathrm{12}}\left(\frac{{q}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{y}^{\mathrm{2}} }\right)+{m}\left\{\left(\frac{{qy}}{\:\mathrm{2}\sqrt{{a}^{\mathrm{2}} −{y}^{\mathrm{2}} }}−{p}\right)^{\mathrm{2}} +\frac{{q}^{\mathrm{2}} }{\mathrm{4}}\right\} \\ $$$$\:\:=\mathrm{2}\left({M}+{m}\right){g}\left({y}_{\mathrm{0}} −{y}\right) \\ $$$${For}\:{y}=\mathrm{0}\:\:{we}\:{have}\:{p}=\mathrm{0}\:\:{hence} \\ $$$$\frac{{Mq}^{\mathrm{2}} }{\mathrm{3}}+\frac{{mq}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{2}\left({M}+{m}\right){g}\left(\frac{{h}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{q}=\sqrt{\mathrm{3}{gh}} \\ $$$$ \\ $$

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