# Question-134678

Question Number 134678 by mr W last updated on 07/Mar/21
Commented by mr W last updated on 06/Mar/21
$${attempt}\:{to}\:{solve}\:{Q}\mathrm{134376} \\$$
Commented by mr W last updated on 06/Mar/21
$${some}\:{useful}\:{data}\:{about}\:{solid}\:{cones}: \\$$
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
$${OC}={axis}\:{of}\:{cone} \\$$$${OD}={contact}\:{line}\:{of}\:{cone}\:{on}\:{plane} \\$$$${G}={center}\:{of}\:{mass}\:{of}\:{cone} \\$$$${OC}={h} \\$$$${OG}=\frac{\mathrm{3}}{\mathrm{4}}{h} \\$$$${OD}=\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} } \\$$$$\mathrm{tan}\:\rho=\frac{{r}}{{h}}\:\Rightarrow\rho=\mathrm{tan}^{−\mathrm{1}} \frac{{r}}{{h}} \\$$$${GE}\bot{OD} \\$$$${GE}={e}=\frac{{OG}}{{OD}}×{r}=\frac{\mathrm{3}{hr}}{\mathrm{4}\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }} \\$$$${OE}={f}=\frac{{OG}}{{OD}}×{h}=\frac{\mathrm{3}{h}^{\mathrm{2}} }{\mathrm{4}\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }} \\$$$$\\$$$${the}\:{position}\:{of}\:{the}\:{cone}\:{is}\:{described} \\$$$${through}\:\theta\:{with}\:−\mathrm{90}°\leqslant\theta\leqslant\mathrm{90}°,\:{since} \\$$$${the}\:{cone}\:{is}\:{released}\:{from}\:{rest}\:{at} \\$$$$\theta=−\mathrm{90}°. \\$$$${since}\:{the}\:{surface}\:{is}\:{rough}\:{enough} \\$$$${such}\:{that}\:{the}\:{cone}\:{can}\:{only}\:{roll} \\$$$${on}\:{the}\:{plane}\:{without}\:{slipping}, \\$$$${the}\:{cone}\:{rotates}\:{about}\:{its}\:{tip}\:{point}\:{O} \\$$$${along}\:{the}\:{plane}.\:{besides}\:{it}\:{rotates} \\$$$${about}\:{its}\:{own}\:{axis}\:{OC}.\:{say}\:{the} \\$$$${angular}\:{speed}\:{of}\:{the}\:{motion}\:{about}\:{O} \\$$$${is}\:\omega_{{P}} \:{and}\:{that}\:{about}\:{its}\:{axis}\:{is}\:\omega_{{a}} . \\$$$$\omega_{{P}} =\frac{{d}\theta}{{dt}}=\omega \\$$$${since}\:{there}\:{is}\:{no}\:{slipping}\:{on}\:{the} \\$$$${contact},\: \\$$$${OD}×\omega_{{P}} ={r}×\omega_{{a}} \\$$$$\Rightarrow\omega_{{a}} =\frac{{OD}}{{r}}\omega_{{P}} =\frac{\omega}{\mathrm{sin}\:\rho} \\$$$$\omega_{{P},{s}} =\omega_{{P}} \mathrm{cos}\:\rho=\omega\:\mathrm{cos}\:\rho \\$$$$\omega_{{P},{a}} =−\omega_{{P}} \mathrm{sin}\:\rho=−\omega\:\mathrm{sin}\:\rho \\$$$${we}\:{see}\:\omega_{{P},{a}} \:{is}\:{opposite}\:{to}\:\omega_{{a}} ,\:{therefore} \\$$$${the}\:{negative}\:{sign}. \\$$$${I}_{{a}} =\frac{\mathrm{3}{mr}^{\mathrm{2}} }{\mathrm{10}} \\$$$${I}_{{s}} =\frac{\mathrm{3}{m}}{\mathrm{20}}\left({r}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{2}} \right) \\$$$${the}\:{kinetic}\:{energy}\:{of}\:{cone}\:{at}\:\theta: \\$$$${KE}_{{t}} =\frac{\mathrm{1}}{\mathrm{2}}{I}_{{a}} \left(\omega_{{a}} +\omega_{{P},{a}} \right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{I}_{{s}} \omega_{{P},{s}} ^{\mathrm{2}} \\$$$${KE}_{{t}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}{mr}^{\mathrm{2}} }{\mathrm{10}}×\left(\frac{\mathrm{1}}{\mathrm{sin}\:\rho}−\mathrm{sin}\:\rho\right)^{\mathrm{2}} \omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}{m}}{\mathrm{20}}\left({r}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{2}} \right)×\mathrm{cos}^{\mathrm{2}} \:\rho×\omega^{\mathrm{2}} \\$$$${KE}_{{t}} =\frac{\mathrm{3}{m}}{\mathrm{40}}\left[\left(\frac{\mathrm{2}}{\mathrm{sin}\:\rho}−\mathrm{sin}\:\rho\right)\left(\frac{\mathrm{1}}{\mathrm{sin}\:\rho}−\mathrm{sin}\:\rho\right){r}^{\mathrm{2}} +\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\rho\:{h}^{\mathrm{2}} \right]\omega^{\mathrm{2}} \\$$$${KE}_{{t}} =\frac{\mathrm{3}{mh}^{\mathrm{2}} }{\mathrm{40}}\left[\left(\frac{\mathrm{2}}{\mathrm{sin}\:\rho}−\mathrm{sin}\:\rho\right)\left(\frac{\mathrm{1}}{\mathrm{sin}\:\rho}−\mathrm{sin}\:\rho\right)\mathrm{tan}^{\mathrm{2}} \:\rho+\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\rho\right]\omega^{\mathrm{2}} \\$$$${KE}_{{t}} =\frac{\mathrm{3}{mh}^{\mathrm{2}} }{\mathrm{40}}\left(\mathrm{1}+\mathrm{5}\:\mathrm{cos}^{\mathrm{2}} \:\rho\right)\omega^{\mathrm{2}} \\$$$${let}\:{z}_{{O}} =\mathrm{0} \\$$$${at}\:{t}=\mathrm{0}\:{and}\:\theta=−\mathrm{90}°: \\$$$${z}_{{E},\mathrm{0}} =\mathrm{0} \\$$$${at}\:\theta: \\$$$${z}_{{E},{t}} =−{OE}×\mathrm{cos}\:\theta×\mathrm{sin}\:\phi=−{f}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi \\$$$${KE}_{{t}} ={mg}\left({z}_{{E},\mathrm{0}} −{z}_{{E},{t}} \right) \\$$$$\frac{\mathrm{3}{mh}^{\mathrm{2}} }{\mathrm{40}}\left(\mathrm{1}+\mathrm{5}\:\mathrm{cos}^{\mathrm{2}} \:\rho\right)\omega^{\mathrm{2}} ={mgf}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi \\$$$$\left(\mathrm{1}+\mathrm{5}\:\mathrm{cos}^{\mathrm{2}} \:\rho\right)\omega^{\mathrm{2}} =\frac{\mathrm{10}{g}}{\:\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }}×\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi \\$$$$\omega^{\mathrm{2}} =\frac{\mathrm{10}{g}\:\mathrm{cos}\:\rho}{\:{h}\left(\mathrm{1}+\mathrm{5}\:\mathrm{cos}^{\mathrm{2}} \:\rho\right)}×\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi \\$$$${with}\:\xi=\sqrt{\frac{\mathrm{10}{g}\:\mathrm{cos}\:\rho}{\:{h}\left(\mathrm{1}+\mathrm{5}\:\mathrm{cos}^{\mathrm{2}} \:\rho\right)}} \\$$$$\Rightarrow\omega=\xi\sqrt{\mathrm{sin}\:\phi\:\mathrm{cos}\:\theta} \\$$$$\frac{{d}\theta}{{dt}}=\xi\sqrt{\mathrm{sin}\:\phi\:\mathrm{cos}\:\theta} \\$$$$\int\frac{{d}\theta}{\:\sqrt{\mathrm{cos}\:\theta}}=\xi\sqrt{\mathrm{sin}\:\phi}\int{dt} \\$$$${let}\:{T}={period} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\theta}{\:\sqrt{\mathrm{cos}\:\theta}}=\frac{\xi\sqrt{\mathrm{sin}\:\phi}\:{T}}{\mathrm{4}} \\$$$$\Rightarrow{T}=\frac{\mathrm{4}}{\xi\sqrt{\mathrm{sin}\:\phi}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\theta}{\:\sqrt{\mathrm{cos}\:\theta}}=\frac{\mathrm{2}{B}\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{2}}\right)}{\xi\sqrt{\mathrm{sin}\:\phi}} \\$$$$\Rightarrow{T}\approx\frac{\mathrm{10}.\mathrm{48823}}{\xi\sqrt{\mathrm{sin}\:\phi}}=\mathrm{3}.\mathrm{31667}\sqrt{\frac{\mathrm{1}+\mathrm{5}\:\mathrm{cos}^{\mathrm{2}} \:\rho}{\mathrm{cos}\:\rho\:\mathrm{sin}\:\phi}}\sqrt{\frac{{h}}{{g}}} \\$$$$\\$$$${example}: \\$$$$\rho=\mathrm{15}°=\frac{\pi}{\mathrm{12}} \\$$$$\phi=\mathrm{30}°=\frac{\pi}{\mathrm{6}} \\$$$${T}\approx\mathrm{11}.\mathrm{359}\sqrt{\frac{{h}}{{g}}} \\$$
Commented by physicstutes last updated on 06/Mar/21
$$\mathrm{sir}\:\mathrm{W}\:\mathrm{please}\:\mathrm{reference}\:\mathrm{me}\:\mathrm{to}\:\mathrm{an}\:\mathrm{e}−\mathrm{book}\:\mathrm{where} \\$$$$\mathrm{i}\:\mathrm{can}\:\mathrm{learn}\:\mathrm{basic}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia},\:\mathrm{especially}\:\mathrm{things} \\$$$$\mathrm{like}\:\mathrm{basic}\:\mathrm{integral}\:\mathrm{derivation}\:\mathrm{of}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia} \\$$$$\mathrm{for}\:\mathrm{certain}\:\mathrm{shapes}\left(\mathrm{like}\:\mathrm{rods},\mathrm{rectangles},\:\mathrm{rings}\:\mathrm{etc}\right)\:\mathrm{and} \\$$$$\mathrm{appling}\:\mathrm{these}\:\mathrm{knowledge}\:\mathrm{to}\:\mathrm{compound}\:\mathrm{pendulum}\:\mathrm{and}\:\mathrm{swinging} \\$$$$\mathrm{masses}.\:\mathrm{please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{help}\:\mathrm{with}\:\mathrm{those}.\: \\$$
Commented by mr W last updated on 06/Mar/21
$${i}\:{really}\:{don}'{t}\:{know}\:{such}\:{a}\:{book}\:{which} \\$$$${i}\:{can}\:{recommend}.\:{nowadays}\:{when}\:{i} \\$$$${need}\:{to}\:{look}\:{something},\:{i}\:{just}\:{google}. \\$$
$$\mathrm{weldone}\:\mathrm{sir}. \\$$
$$\mathrm{i}\:\mathrm{tried}\:\mathrm{googling}\:\mathrm{but}\:\mathrm{could}\:\mathrm{not}\:\mathrm{find}\:\mathrm{a}\:\mathrm{suitable}\: \\$$$$\mathrm{pdf}\:\mathrm{or}\:\mathrm{even}\:\mathrm{a}\:\mathrm{book}\:\mathrm{for}\:\mathrm{the}\:\mathrm{assesment}. \\$$