Question Number 227386 by fantastic2 last updated on 19/Jan/26
Answered by mr W last updated on 23/Jan/26
Commented by mr W last updated on 23/Jan/26
$${since}\:{no}\:{horizontal}\:{force},\:{COM} \\ $$$${moves}\:{only}\:{in}\:{vertical}\:{direction}. \\ $$$${CC}'=\frac{{l}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\left({a}\right)\:\Rightarrow{correct} \\ $$$$\left({b}\right)\:\Rightarrow\:{correct} \\ $$$$ \\ $$$$\tau_{{B}} ={mg}\frac{{l}}{\mathrm{2}}\mathrm{sin}\:\theta \\ $$$$\left({c}\right)\:\Rightarrow\:{correct} \\ $$$$ \\ $$$${x}_{{A}} =\frac{{l}}{\mathrm{2}}\:\mathrm{cos}\:\theta \\ $$$${y}_{{A}} ={l}\:\mathrm{sin}\:\theta \\ $$$${x}_{{A}} ^{\mathrm{2}} +\left(\frac{{y}_{{A}} }{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{{l}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${i}.{e}. \\ $$$${x}^{\mathrm{2}} +\frac{{y}^{\mathrm{2}} }{\mathrm{4}}=\frac{{l}^{\mathrm{2}} }{\mathrm{4}}\: \\ $$$${the}\:{locus}\:{of}\:{point}\:{A}\:{is}\:{an}\:{elipse}. \\ $$$$\left({d}\right)\:\Rightarrow{wrong} \\ $$
Commented by mr W last updated on 24/Jan/26
![ω=(dθ/dt) h=C′C=((l(1−cos θ))/2) v=(dh/dt)=((lω sin θ)/2) (1/2)(((ml^2 )/(12)))ω^2 +(m/2)(((lω sin θ)/2))^2 =mg((l(1−cos θ))/2) ⇒ω^2 =((12g(1−cos θ))/(l(1+3 sin^2 θ))) 2ω(dω/dθ)=((12g)/l)[((sin θ)/(1+3 sin^2 θ))−((6 sin θ cos θ(1−cos θ))/((1+3 sin^2 θ)^2 ))] ω(dω/dθ)=((6g sin θ(4−6 cos θ+3 cos^2 θ))/(l(1+3 sin^2 θ)^2 )) a=(dv/dt)=((lω)/2)(ω cos θ+sin θ (dω/dθ)) =(l/2)(ω^2 cos θ+sin θ ((ωdω)/dθ)) =(l/2)[((12g cos θ(1−cos θ))/(l(1+3 sin^2 θ)))+((6g sin^2 θ(4−6 cos θ+3 cos^2 θ))/(l(1+3 sin^2 θ)^2 ))] =((3g(1−cos θ)(4+6 cos θ−3 cos^2 θ−3 cos^3 θ))/((4−3 cos^2 θ)^2 )) mg−N=ma N=mg[1−((3(1−cos θ)(4+6 cos θ−3 cos^2 θ−3 cos^3 θ))/((4−3 cos^2 θ)^2 ))] N_(min) ≈0.1654 mg at θ≈1.0739 (61.822°)](https://www.tinkutara.com/question/Q227413.png)
$$\omega=\frac{{d}\theta}{{dt}} \\ $$$${h}={C}'{C}=\frac{{l}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{2}} \\ $$$${v}=\frac{{dh}}{{dt}}=\frac{{l}\omega\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{ml}^{\mathrm{2}} }{\mathrm{12}}\right)\omega^{\mathrm{2}} +\frac{{m}}{\mathrm{2}}\left(\frac{{l}\omega\:\mathrm{sin}\:\theta}{\mathrm{2}}\right)^{\mathrm{2}} ={mg}\frac{{l}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{2}} \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{12}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{l}\left(\mathrm{1}+\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right)} \\ $$$$\mathrm{2}\omega\frac{{d}\omega}{{d}\theta}=\frac{\mathrm{12}{g}}{{l}}\left[\frac{\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\theta}−\frac{\mathrm{6}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\left(\mathrm{1}+\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} }\right] \\ $$$$\omega\frac{{d}\omega}{{d}\theta}=\frac{\mathrm{6}{g}\:\mathrm{sin}\:\theta\left(\mathrm{4}−\mathrm{6}\:\mathrm{cos}\:\theta+\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)}{{l}\left(\mathrm{1}+\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} } \\ $$$${a}=\frac{{dv}}{{dt}}=\frac{{l}\omega}{\mathrm{2}}\left(\omega\:\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\:\frac{{d}\omega}{{d}\theta}\right) \\ $$$$\:\:=\frac{{l}}{\mathrm{2}}\left(\omega^{\mathrm{2}} \:\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\:\frac{\omega{d}\omega}{{d}\theta}\right) \\ $$$$\:\:=\frac{{l}}{\mathrm{2}}\left[\frac{\mathrm{12}{g}\:\mathrm{cos}\:\theta\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{l}\left(\mathrm{1}+\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right)}+\frac{\mathrm{6}{g}\:\mathrm{sin}^{\mathrm{2}} \:\theta\left(\mathrm{4}−\mathrm{6}\:\mathrm{cos}\:\theta+\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)}{{l}\left(\mathrm{1}+\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} }\right] \\ $$$$\:\:=\frac{\mathrm{3}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left(\mathrm{4}+\mathrm{6}\:\mathrm{cos}\:\theta−\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{3}\:\mathrm{cos}^{\mathrm{3}} \:\theta\right)}{\left(\mathrm{4}−\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} } \\ $$$${mg}−{N}={ma} \\ $$$${N}={mg}\left[\mathrm{1}−\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left(\mathrm{4}+\mathrm{6}\:\mathrm{cos}\:\theta−\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{3}\:\mathrm{cos}^{\mathrm{3}} \:\theta\right)}{\left(\mathrm{4}−\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} }\right] \\ $$$${N}_{{min}} \approx\mathrm{0}.\mathrm{1654}\:{mg}\:{at}\:\theta\approx\mathrm{1}.\mathrm{0739}\:\left(\mathrm{61}.\mathrm{822}°\right) \\ $$
Commented by mr W last updated on 24/Jan/26
