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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

$$C=centerofrotation$$$$G=centerofmassofrod$$$$y=\frac{x^2}{a}witha=1$$$$\frac{dy}{dx}=\frac{2x}{a}$$$$atBwithx=\frac{b}{2}$$$$\tan \varphi =\frac{2}{a}\times \frac{b}{2}=\frac{b}{a}$$$$\mathrm{\Delta }x=h$$$$\mathrm{\Delta }y=\frac{2}{a}\times \frac{b}{2}\mathrm{\Delta }x=\frac{b}{a}\mathrm{\Delta }x=\frac{bh}{a}$$$$\theta =\frac{2\mathrm{\Delta }y}{b}=\frac{2bh}{ab}=\frac{2h}{a}$$$$\alpha =\frac{d^2\theta }{{dt}^2}=\frac{2}{a}\times \frac{d^{2}h}{{dt}^2}$$$$I_C=\frac{{mb}^2}{12}+m{\left(\frac{b}{2\tan \varphi }\right)}^2=\frac{{mb}^2}{12}+m{\left(\frac{ba}{2b}\right)}^2$$$$I_C=\frac{m(b^2+3a^2)}{12}$$$$I_C\alpha =-mgh$$$$\frac{m(b^2+3a^2)}{12}\times \frac{2}{a}\times \frac{d^{2}h}{{dt}^2}=-mgh$$$$\frac{d^{2}h}{{dt}^2}=-\frac{6ag}{(b^2+3a^2)}h=-{\omega }^{2}h$$$$\frac{d^{2}h}{{dt}^2}+{\omega }^{2}h=0\Rightarrow harmonicmotion$$$${\omega }^2=\frac{6ag}{b^2+3a^2}$$$$T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{\frac{6ag}{b^2+3a^2}}}=2\pi \sqrt{\frac{b^2+3a^2}{6ag}}$$$$T=2\pi \sqrt{\frac{b^2+3a^2}{6ag}}$$

Commented by Tawa11 last updated on 12/Jan/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by mr W last updated on 13/Jan/22

$$isanswercorrect?i^{\prime }mnotsure.$$

Commented by ajfour last updated on 13/Jan/22

$$mustbe,sir.Icouldnotfindaway.$$$$Iwilllearnncheckyoursolution$$$$onlynow..\underset{⌣}{\stackrel{\sqrt{--}}{.}}.$$

Commented by mr W last updated on 13/Jan/22

$$thehorizontalpositionoftherod$$$$isnottheonlypossibleequilibrium$$$$positionoftherodinthebowl.itis$$$$evennotastableequilibriumposition$$$$aswecanimage.ihaveopeneda$$$$newpostwiththequestiontofind$$$$otherpossibleequilibriumpositions$$$$whichtherodmayhaveinthebowl.$$$$pleasegiveyourtrytosolveitsir!$$

Answered by ajfour last updated on 13/Jan/22

$$mgdy=I\omega d\omega +mvdv$$$$\tan \theta =\frac{dy}{dx}=2x{\mid }_{x=b/2}=\frac{\omega b/2}{v}$$$$\Rightarrow gb△x=\frac{b^2{\omega }^2}{24}+\frac{{\omega }^2}{8}$$$$\Rightarrow {\omega }^2=\left(\frac{24bg}{b^2+3}\right)△x$$$$\Rightarrow T=2\pi \sqrt{\frac{b^2+3}{24bg}}$$

Commented by mr W last updated on 14/Jan/22

$$tobehonest,i{havn}^{\prime }tunderstoodyour$$$$workingsyet.$$

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