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Question Number 164009 by ajfour last updated on 12/Jan/22
Answered by mr W last updated on 13/Jan/22
Commented by mr W last updated on 13/Jan/22
C=center of rotation G=center of mass of rod y=(x^2 /a) with a=1 (dy/dx)=((2x)/a) at B with x=(b/2) tan φ=(2/a)×(b/2)=(b/a) Δx=h Δy=(2/a)×(b/2)Δx=(b/a)Δx=((bh)/a) θ=((2Δy)/b)=((2bh)/(ab))=((2h)/a) α=(d^2 θ/dt^2 )=(2/a)×(d^2 h/dt^2 ) I_C =((mb^2 )/(12))+m((b/(2 tan φ)))^2 =((mb^2 )/(12))+m(((ba)/(2b)))^2 I_C =((m(b^2 +3a^2 ))/(12)) I_C α=−mgh ((m(b^2 +3a^2 ))/(12))×(2/a)×(d^2 h/dt^2 )=−mgh (d^2 h/dt^2 )=−((6ag)/((b^2 +3a^2 )))h=−ω^2 h (d^2 h/dt^2 )+ω^2 h=0 ⇒ harmonic motion ω^2 =((6ag)/(b^2 +3a^2 )) T=((2π)/ω)=((2π)/( (√((6ag)/(b^2 +3a^2 )))))=2π(√((b^2 +3a^2 )/(6ag))) T=2π(√((b^2 +3a^2 )/(6ag)))
$${C}={center}\:{of}\:{rotation} \\ $$$${G}={center}\:{of}\:{mass}\:{of}\:{rod} \\ $$$${y}=\frac{{x}^{\mathrm{2}} }{{a}}\:{with}\:{a}=\mathrm{1} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{2}{x}}{{a}} \\ $$$${at}\:{B}\:{with}\:{x}=\frac{{b}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\phi=\frac{\mathrm{2}}{{a}}×\frac{{b}}{\mathrm{2}}=\frac{{b}}{{a}} \\ $$$$\Delta{x}={h} \\ $$$$\Delta{y}=\frac{\mathrm{2}}{{a}}×\frac{{b}}{\mathrm{2}}\Delta{x}=\frac{{b}}{{a}}\Delta{x}=\frac{{bh}}{{a}} \\ $$$$\theta=\frac{\mathrm{2}\Delta{y}}{{b}}=\frac{\mathrm{2}{bh}}{{ab}}=\frac{\mathrm{2}{h}}{{a}} \\ $$$$\alpha=\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }=\frac{\mathrm{2}}{{a}}×\frac{{d}^{\mathrm{2}} {h}}{{dt}^{\mathrm{2}} } \\ $$$${I}_{{C}} =\frac{{mb}^{\mathrm{2}} }{\mathrm{12}}+{m}\left(\frac{{b}}{\mathrm{2}\:\mathrm{tan}\:\phi}\right)^{\mathrm{2}} =\frac{{mb}^{\mathrm{2}} }{\mathrm{12}}+{m}\left(\frac{{ba}}{\mathrm{2}{b}}\right)^{\mathrm{2}} \\ $$$${I}_{{C}} =\frac{{m}\left({b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} \right)}{\mathrm{12}} \\ $$$${I}_{{C}} \alpha=−{mgh} \\ $$$$\frac{{m}\left({b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} \right)}{\mathrm{12}}×\frac{\mathrm{2}}{{a}}×\frac{{d}^{\mathrm{2}} {h}}{{dt}^{\mathrm{2}} }=−{mgh} \\ $$$$\frac{{d}^{\mathrm{2}} {h}}{{dt}^{\mathrm{2}} }=−\frac{\mathrm{6}{ag}}{\left({b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} \right)}{h}=−\omega^{\mathrm{2}} {h} \\ $$$$\frac{{d}^{\mathrm{2}} {h}}{{dt}^{\mathrm{2}} }+\omega^{\mathrm{2}} {h}=\mathrm{0}\:\:\Rightarrow\:{harmonic}\:{motion} \\ $$$$\omega^{\mathrm{2}} =\frac{\mathrm{6}{ag}}{{b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} } \\ $$$${T}=\frac{\mathrm{2}\pi}{\omega}=\frac{\mathrm{2}\pi}{\:\sqrt{\frac{\mathrm{6}{ag}}{{b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} }}}=\mathrm{2}\pi\sqrt{\frac{{b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{6}{ag}}} \\ $$$${T}=\mathrm{2}\pi\sqrt{\frac{{b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{6}{ag}}} \\ $$
Commented by Tawa11 last updated on 12/Jan/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 13/Jan/22
is answer correct? i′m not sure.
$${is}\:{answer}\:{correct}?\:{i}'{m}\:{not}\:{sure}. \\ $$
Commented by ajfour last updated on 13/Jan/22
must be, sir. I could not find a way. I will learn n check your solution only now.. .^( (√(−−))) _( ⌣) .
$${must}\:{be},\:{sir}.\:{I}\:{could}\:{not}\:{find}\:{a}\:{way}. \\ $$$${I}\:{will}\:{learn}\:{n}\:{check}\:{your}\:{solution} \\ $$$${only}\:{now}..\:\:\:\underset{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\smile} {\overset{\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{−−}} {.}}\:. \\ $$
Commented by mr W last updated on 13/Jan/22
the horizontal position of the rod is not the only possible equilibrium position of the rod in the bowl. it is even not a stable equilibrium position as we can image. i have opened a new post with the question to find other possible equilibrium positions which the rod may have in the bowl. please give your try to solve it sir!
$${the}\:{horizontal}\:{position}\:{of}\:{the}\:{rod} \\ $$$${is}\:{not}\:{the}\:{only}\:{possible}\:{equilibrium} \\ $$$${position}\:{of}\:{the}\:{rod}\:{in}\:{the}\:{bowl}.\:{it}\:{is} \\ $$$${even}\:{not}\:{a}\:{stable}\:{equilibrium}\:{position} \\ $$$${as}\:{we}\:{can}\:{image}.\:{i}\:{have}\:{opened}\:{a} \\ $$$${new}\:{post}\:{with}\:{the}\:{question}\:{to}\:{find} \\ $$$${other}\:{possible}\:{equilibrium}\:{positions} \\ $$$${which}\:{the}\:{rod}\:{may}\:{have}\:{in}\:{the}\:{bowl}. \\ $$$${please}\:{give}\:{your}\:{try}\:{to}\:{solve}\:{it}\:{sir}! \\ $$
Answered by ajfour last updated on 13/Jan/22
mgdy=Iωdω+mvdv tan θ=(dy/dx)=2x∣_(x=b/2) =((ωb/2)/v) ⇒ gb△x=((b^2 ω^2 )/(24))+(ω^2 /8) ⇒ ω^2 =(((24bg)/(b^2 +3)))△x ⇒ T=2π(√((b^2 +3)/(24bg)))
$${mgdy}={I}\omega{d}\omega+{mvdv} \\ $$$$\mathrm{tan}\:\theta=\frac{{dy}}{{dx}}=\mathrm{2}{x}\mid_{{x}={b}/\mathrm{2}} =\frac{\omega{b}/\mathrm{2}}{{v}} \\ $$$$\Rightarrow\:\:{gb}\bigtriangleup{x}=\frac{{b}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{24}}+\frac{\omega^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\Rightarrow\:\:\:\omega^{\mathrm{2}} =\left(\frac{\mathrm{24}{bg}}{{b}^{\mathrm{2}} +\mathrm{3}}\right)\bigtriangleup{x} \\ $$$$\Rightarrow\:\:{T}=\mathrm{2}\pi\sqrt{\frac{{b}^{\mathrm{2}} +\mathrm{3}}{\mathrm{24}{bg}}} \\ $$
Commented by mr W last updated on 14/Jan/22
to be honest, i havn′t understood your workings yet.
$${to}\:{be}\:{honest},\:{i}\:{havn}'{t}\:{understood}\:{your} \\ $$$${workings}\:{yet}. \\ $$

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