Question Number 227353 by fantastic2 last updated on 17/Jan/26
Answered by mr W last updated on 19/Jan/26
Commented by mr W last updated on 19/Jan/26
![θ=(π/3)=60° I=((ml^2 )/3) Iα=mg×((l sin θ)/2) ((ml^2 )/3)α=mg×(l/2)×((√3)/2) ⇒α=((3(√3)g)/(4l)) (a) ⇒wrong [((ml^2 )/(12))+m(((lcos θ)/2))^2 ]α=N×((l sin θ)/2) ⇒N=((7mg)/(16)) (b) ⇒wrong a_r =ω^2 (l/2)=((3g)/(2l))×(l/2)=((3g)/4) a_θ =α×(l/2)=((3(√3)g)/8) (c) ⇒correct (1/2)Iω^2 =mg×((l(1−cos θ))/2) ⇒ω=(√((3g)/(2l))) (d) ⇒correct](https://www.tinkutara.com/question/Q227365.png)
$$\theta=\frac{\pi}{\mathrm{3}}=\mathrm{60}° \\ $$$${I}=\frac{{ml}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${I}\alpha={mg}×\frac{{l}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\frac{{ml}^{\mathrm{2}} }{\mathrm{3}}\alpha={mg}×\frac{{l}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\frac{\mathrm{3}\sqrt{\mathrm{3}}{g}}{\mathrm{4}{l}} \\ $$$$\left({a}\right)\:\Rightarrow{wrong} \\ $$$$ \\ $$$$\left[\frac{{ml}^{\mathrm{2}} }{\mathrm{12}}+{m}\left(\frac{{l}\mathrm{cos}\:\theta}{\mathrm{2}}\right)^{\mathrm{2}} \right]\alpha={N}×\frac{{l}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\Rightarrow{N}=\frac{\mathrm{7}{mg}}{\mathrm{16}} \\ $$$$\left({b}\right)\:\Rightarrow{wrong} \\ $$$$ \\ $$$${a}_{{r}} =\omega^{\mathrm{2}} \frac{{l}}{\mathrm{2}}=\frac{\mathrm{3}{g}}{\mathrm{2}{l}}×\frac{{l}}{\mathrm{2}}=\frac{\mathrm{3}{g}}{\mathrm{4}} \\ $$$${a}_{\theta} =\alpha×\frac{{l}}{\mathrm{2}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}{g}}{\mathrm{8}} \\ $$$$\left({c}\right)\:\Rightarrow{correct} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} ={mg}×\frac{{l}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{2}} \\ $$$$\Rightarrow\omega=\sqrt{\frac{\mathrm{3}{g}}{\mathrm{2}{l}}} \\ $$$$\left({d}\right)\:\Rightarrow{correct} \\ $$
Commented by fantastic2 last updated on 19/Jan/26
$${only}\:{the}\:{first}\:{one}\:{is}\:{wrong}.\:{others}\:{are}\:{correct} \\ $$
Commented by mr W last updated on 19/Jan/26
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Answered by fantastic2 last updated on 20/Jan/26
Commented by fantastic2 last updated on 20/Jan/26
$${mg}−{N}={mA} \\ $$$${N}={m}\left({g}−{A}\right) \\ $$$${N}={m}\left({g}−\left({a}_{{r}} \mathrm{cos}\:\theta+{a}_{{t}} \mathrm{sin}\:\theta\right)\right) \\ $$$${a}_{{r}} =\omega^{\mathrm{2}} \frac{{l}}{\mathrm{2}}=\frac{\mathrm{3}{g}}{\mathrm{4}} \\ $$$${a}_{{t}} =\alpha\frac{{l}}{\mathrm{2}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}{g}}{\mathrm{8}} \\ $$$${N}={m}\left({g}−\left(\frac{\mathrm{3}{g}}{\mathrm{8}}+\frac{\mathrm{9}{g}}{\mathrm{16}}\right)={mg}\frac{\mathrm{1}}{\mathrm{16}}\right. \\ $$