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Question Number 134678 by mr W last updated on 07/Mar/21

Commented by mr W last updated on 06/Mar/21

$$attempttosolveQ134376$$

Commented by mr W last updated on 06/Mar/21

$$someusefuldataaboutsolidcones:$$

Commented by mr W last updated on 06/Mar/21

Commented by mr W last updated on 07/Mar/21

Commented by mr W last updated on 07/Mar/21

$$OC=axisofcone$$$$OD=contactlineofconeonplane$$$$G=centerofmassofcone$$$$OC=h$$$$OG=\frac{3}{4}h$$$$OD=\sqrt{h^2+r^2}$$$$\tan \rho =\frac{r}{h}\Rightarrow \rho ={\tan }^{-1}\frac{r}{h}$$$$GE\mathrm{\perp }OD$$$$GE=e=\frac{OG}{OD}\times r=\frac{3hr}{4\sqrt{h^2+r^2}}$$$$OE=f=\frac{OG}{OD}\times h=\frac{3h^2}{4\sqrt{h^2+r^2}}$$$$$$$$thepositionoftheconeisdescribed$$$$through\theta with-90°⩽\theta ⩽90°,since$$$$theconeisreleasedfromrestat$$$$\theta =-90°.$$$$sincethesurfaceisroughenough$$$$suchthattheconecanonlyroll$$$$ontheplanewithoutslipping,$$$$theconerotatesaboutitstippointO$$$$alongtheplane.besidesitrotates$$$$aboutitsownaxisOC.saythe$$$$angularspeedofthemotionaboutO$$$$is{\omega }_{P}andthataboutitsaxisis{\omega }_a.$$$${\omega }_P=\frac{d\theta }{dt}=\omega$$$$sincethereisnoslippingonthe$$$$contact,$$$$OD\times {\omega }_P=r\times {\omega }_a$$$$\Rightarrow {\omega }_a=\frac{OD}{r}{\omega }_P=\frac{\omega }{\sin \rho }$$$${\omega }_{P,s}={\omega }_P\cos \rho =\omega \cos \rho$$$${\omega }_{P,a}=-{\omega }_P\sin \rho =-\omega \sin \rho$$$$wesee{\omega }_{P,a}isoppositeto{\omega }_a,therefore$$$$thenegativesign.$$$$I_a=\frac{3{mr}^2}{10}$$$$I_s=\frac{3m}{20}(r^2+4h^2)$$$$thekineticenergyofconeat\theta :$$$${KE}_t=\frac{1}{2}{I}_a{({\omega }_a+{\omega }_{P,a})}^2+\frac{1}{2}{I}_s{\omega }_{P,s}^2$$$${KE}_t=\frac{1}{2}\times \frac{3{mr}^2}{10}\times {(\frac{1}{\sin \rho }-\sin \rho )}^2{\omega }^2+\frac{1}{2}\times \frac{3m}{20}(r^2+4h^2)\times {\cos }^2\rho \times {\omega }^2$$$${KE}_t=\frac{3m}{40}[(\frac{2}{\sin \rho }-\sin \rho )(\frac{1}{\sin \rho }-\sin \rho )r^2+4{\cos }^2\rho h^2]{\omega }^2$$$${KE}_t=\frac{3{mh}^2}{40}[(\frac{2}{\sin \rho }-\sin \rho )(\frac{1}{\sin \rho }-\sin \rho ){\tan }^2\rho +4{\cos }^2\rho ]{\omega }^2$$$${KE}_t=\frac{3{mh}^2}{40}(1+5{\cos }^2\rho ){\omega }^2$$$$let{z}_O=0$$$$att=0and\theta =-90°:$$$$z_{E,0}=0$$$$at\theta :$$$$z_{E,t}=-OE\times \cos \theta \times \sin \varphi =-f\cos \theta \sin \varphi$$$${KE}_t=mg(z_{E,0}-z_{E,t})$$$$\frac{3{mh}^2}{40}(1+5{\cos }^2\rho ){\omega }^2=mgf\cos \theta \sin \varphi$$$$(1+5{\cos }^2\rho ){\omega }^2=\frac{10g}{\sqrt{h^2+r^2}}\times \cos \theta \sin \varphi$$$${\omega }^2=\frac{10g\cos \rho }{h(1+5{\cos }^2\rho )}\times \cos \theta \sin \varphi$$$$with\xi =\sqrt{\frac{10g\cos \rho }{h(1+5{\cos }^2\rho )}}$$$$\Rightarrow \omega =\xi \sqrt{\sin \varphi \cos \theta }$$$$\frac{d\theta }{dt}=\xi \sqrt{\sin \varphi \cos \theta }$$$$\int \frac{d\theta }{\sqrt{\cos \theta }}=\xi \sqrt{\sin \varphi }\int dt$$$$letT=period$$$${\int }_0^{\frac{\pi }{2}}\frac{d\theta }{\sqrt{\cos \theta }}=\frac{\xi \sqrt{\sin \varphi }T}{4}$$$$\Rightarrow T=\frac{4}{\xi \sqrt{\sin \varphi }}{\int }_0^{\frac{\pi }{2}}\frac{d\theta }{\sqrt{\cos \theta }}=\frac{2B(\frac{1}{4},\frac{1}{2})}{\xi \sqrt{\sin \varphi }}$$$$\Rightarrow T\approx \frac{10.48823}{\xi \sqrt{\sin \varphi }}=3.31667\sqrt{\frac{1+5{\cos }^2\rho }{\cos \rho \sin \varphi }}\sqrt{\frac{h}{g}}$$$$$$$$example:$$$$\rho =15°=\frac{\pi }{12}$$$$\varphi =30°=\frac{\pi }{6}$$$$T\approx 11.359\sqrt{\frac{h}{g}}$$

Commented by physicstutes last updated on 06/Mar/21

$$\mathrm{sir}\mathrm{W}\mathrm{please}\mathrm{reference}\mathrm{me}\mathrm{to}\mathrm{an}\mathrm{e}-\mathrm{book}\mathrm{where}$$$$\mathrm{i}\mathrm{can}\mathrm{learn}\mathrm{basic}\mathrm{moment}\mathrm{of}\mathrm{inertia},\mathrm{especially}\mathrm{things}$$$$\mathrm{like}\mathrm{basic}\mathrm{integral}\mathrm{derivation}\mathrm{of}\mathrm{moment}\mathrm{of}\mathrm{inertia}$$$$\mathrm{for}\mathrm{certain}\mathrm{shapes}(\mathrm{like}\mathrm{rods},\mathrm{rectangles},\mathrm{rings}\mathrm{etc})\mathrm{and}$$$$\mathrm{appling}\mathrm{these}\mathrm{knowledge}\mathrm{to}\mathrm{compound}\mathrm{pendulum}\mathrm{and}\mathrm{swinging}$$$$\mathrm{masses}.\mathrm{please}\mathrm{i}\mathrm{need}\mathrm{help}\mathrm{with}\mathrm{those}.$$

Commented by mr W last updated on 06/Mar/21

$$ireally{don}^{\prime }tknowsuchabookwhich$$$$icanrecommend.nowadayswheni$$$$needtolooksomething,ijustgoogle.$$

Commented by I want to learn more last updated on 06/Mar/21

$$\mathrm{weldone}\mathrm{sir}.$$

Commented by physicstutes last updated on 06/Mar/21

$$\mathrm{i}\mathrm{tried}\mathrm{googling}\mathrm{but}\mathrm{could}\mathrm{not}\mathrm{find}\mathrm{a}\mathrm{suitable}$$$$\mathrm{pdf}\mathrm{or}\mathrm{even}\mathrm{a}\mathrm{book}\mathrm{for}\mathrm{the}\mathrm{assesment}.$$

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