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 04/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} p=(q/(2(M+m))){((my)/( (√(a^2 −y^2 ))))−((My)/( (√(b^2 −y^2 ))))} 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)) ⇒ (q^2 /(12)){((b^2 M)/(b^2 −y^2 ))+((a^2 m)/(a^2 −y^2 ))+3(M+m)} +((m^2 /M)+m)(((qy)/(2(√(a^2 −y^2 ))))−p)^2 =2(M+m)g(y_0 −y) replacing for p ⇒ (1/(12)){((b^2 M)/(b^2 −y^2 ))+((a^2 m)/(a^2 −y^2 ))+3(M+m)} +((mMy^2 )/(4(M+m)))[(1/( (√(a^2 −y^2 ))))+(1/( (√(b^2 −y^2 ))))]^2 =((2(M+m)g(y_0 −y))/q^2 ) q=v_y](https://www.tinkutara.com/question/Q214283.png)
$$\omega{b}\mathrm{sin}\:\theta={Q}={q}\:\left({simply}\right)={sa}\mathrm{sin}\:\phi \\ $$$${b}\mathrm{cos}\:\theta={y}={a}\mathrm{cos}\:\phi \\ $$$$\omega=\frac{{q}}{\:\sqrt{{b}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\:\:\:;\:\:\:{s}=\frac{{q}}{\:\sqrt{{a}^{\mathrm{2}} −{y}^{\mathrm{2}} }} \\ $$$${M}\left\{\omega\left(\frac{{b}}{\mathrm{2}}\right)\mathrm{cos}\:\theta+{p}\right\}={m}\left\{{s}\left(\frac{{a}}{\mathrm{2}}\right)\mathrm{cos}\:\phi+{p}\right\} \\ $$$${M}\left\{\frac{{qy}}{\:\mathrm{2}\sqrt{{b}^{\mathrm{2}} −{y}^{\mathrm{2}} }}+{p}\right\}={m}\left\{\frac{{qy}}{\:\mathrm{2}\sqrt{{a}^{\mathrm{2}} −{y}^{\mathrm{2}} }}−{p}\right\} \\ $$$${p}=\frac{{q}}{\mathrm{2}\left({M}+{m}\right)}\left\{\frac{{my}}{\:\sqrt{{a}^{\mathrm{2}} −{y}^{\mathrm{2}} }}−\frac{{My}}{\:\sqrt{{b}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\right\} \\ $$$$\mathrm{2}\left({M}+{m}\right){g}\left({y}_{\mathrm{0}} −{y}\right)= \\ $$$$\frac{{Mb}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{12}}+{M}\left\{\left({p}+\frac{\omega{b}}{\mathrm{2}}\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\frac{{q}^{\mathrm{2}} }{\mathrm{4}}\right\} \\ $$$$+\frac{{ma}^{\mathrm{2}} {s}^{\mathrm{2}} }{\mathrm{12}}+{m}\left\{\left(\frac{{sa}}{\mathrm{2}}\mathrm{cos}\:\phi−{p}\right)^{\mathrm{2}} +\frac{{q}^{\mathrm{2}} }{\mathrm{4}}\right\} \\ $$$$\Rightarrow \\ $$$$\frac{{Mb}^{\mathrm{2}} }{\mathrm{12}}\left(\frac{{q}^{\mathrm{2}} }{\:{b}^{\mathrm{2}} −{y}^{\mathrm{2}} }\right)+{M}\left\{\left({p}+\frac{{qy}}{\:\mathrm{2}\sqrt{{b}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\right)^{\mathrm{2}} +\frac{{q}^{\mathrm{2}} }{\mathrm{4}}\right\} \\ $$$$+\frac{{ma}^{\mathrm{2}} }{\mathrm{12}}\left(\frac{{q}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{y}^{\mathrm{2}} }\right)+{m}\left\{\left(\frac{{qy}}{\:\mathrm{2}\sqrt{{a}^{\mathrm{2}} −{y}^{\mathrm{2}} }}−{p}\right)^{\mathrm{2}} +\frac{{q}^{\mathrm{2}} }{\mathrm{4}}\right\} \\ $$$$\:\:=\mathrm{2}\left({M}+{m}\right){g}\left({y}_{\mathrm{0}} −{y}\right) \\ $$$${For}\:{y}=\mathrm{0}\:\:{we}\:{have}\:{p}=\mathrm{0}\:\:{hence} \\ $$$$\frac{{Mq}^{\mathrm{2}} }{\mathrm{3}}+\frac{{mq}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{2}\left({M}+{m}\right){g}\left(\frac{{h}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{q}=\sqrt{\mathrm{3}{gh}} \\ $$$$\Rightarrow\:\frac{{q}^{\mathrm{2}} }{\mathrm{12}}\left\{\frac{{b}^{\mathrm{2}} {M}}{{b}^{\mathrm{2}} −{y}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} {m}}{{a}^{\mathrm{2}} −{y}^{\mathrm{2}} }+\mathrm{3}\left({M}+{m}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{m}^{\mathrm{2}} }{{M}}+{m}\right)\left(\frac{{qy}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} −{y}^{\mathrm{2}} }}−{p}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\mathrm{2}\left({M}+{m}\right){g}\left({y}_{\mathrm{0}} −{y}\right) \\ $$$${replacing}\:{for}\:{p} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{12}}\left\{\frac{{b}^{\mathrm{2}} {M}}{{b}^{\mathrm{2}} −{y}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} {m}}{{a}^{\mathrm{2}} −{y}^{\mathrm{2}} }+\mathrm{3}\left({M}+{m}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\frac{{mMy}^{\mathrm{2}} }{\mathrm{4}\left({M}+{m}\right)}\left[\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} −{y}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt{{b}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\right]^{\mathrm{2}} \\ $$$$\:\:\:=\frac{\mathrm{2}\left({M}+{m}\right){g}\left({y}_{\mathrm{0}} −{y}\right)}{{q}^{\mathrm{2}} } \\ $$$${q}={v}_{{y}} \\ $$
Commented by mr W last updated on 04/Dec/24
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Commented by mr W last updated on 04/Dec/24
$${the}\:{hinge}\:{hits}\:{the}\:{ground}\:{with}\:{a} \\ $$$${speed}\:{of}\:\sqrt{\mathrm{3}{gh}},\:{which}\:{is}\:{larger}\:{than} \\ $$$${the}\:{speed}\:{of}\:{free}\:{fall}\:{from}\:{the}\:{same} \\ $$$${hight}.\:{does}\:{it}\:{mean}\:{that}\:{the}\:{hinge} \\ $$$${reaches}\:{the}\:{ground}\:{faster}\:{than}\:{an} \\ $$$${object}\:{in}\:{free}\:{fall}\:{from}\:{the}\:{same} \\ $$$${hight}?\:{if}\:{this}\:{is}\:{true},\:{then}\:{free}\:{fall} \\ $$$${is}\:{not}\:{the}\:{fastest}\:{motion}\:{for}\:{an} \\ $$$${object}\:{from}\:{the}\:{hight}\:{h}.\:{i}\:{don}'{t}\:{think} \\ $$$${this}\:{is}\:{true}\:{and}\:{want}\:{to}\:{check}\:{it}\: \\ $$$${through}\:{Q}\mathrm{214248}. \\ $$
Commented by ajfour last updated on 04/Dec/24
$${I}\:{Think}\:{it}\:{must}\:{be}\:{right}.\:{Its}\:{pulled} \\ $$$${down}.\:{center}\:{of}\:{mass}\:{after}\:{all} \\ $$$${comes}\:{with}\:{speed}\:\frac{\sqrt{\mathrm{3}{gh}}}{\mathrm{2}}\:<\:\sqrt{\mathrm{2}{gh}}\:. \\ $$
Commented by ajfour last updated on 04/Dec/24
Commented by mr W last updated on 04/Dec/24
$${the}\:{speed}\:{can}\:{be}\:{larger}\:{than}\:{free} \\ $$$${fall},\:{i}\:{don}'{t}\:{doubt}\:{this}.\:{i}\:{mean},\: \\ $$$${nevertheless}\:{the}\:{hinge}\:{should}\:{take} \\ $$$${more}\:{time}\:{to}\:{reach}\:{the}\:{ground}\:{than} \\ $$$${in}\:{free}\:{fall}.\:{is}\:{this}\:{true}? \\ $$