Question Number 134678 by mr W last updated on 07/Mar/21
![](https://www.tinkutara.com/question/19285.png)
Commented by mr W last updated on 06/Mar/21
![attempt to solve Q134376](https://www.tinkutara.com/question/Q134679.png)
$${attempt}\:{to}\:{solve}\:{Q}\mathrm{134376} \\ $$
Commented by mr W last updated on 06/Mar/21
![some useful data about solid cones:](https://www.tinkutara.com/question/Q134680.png)
$${some}\:{useful}\:{data}\:{about}\:{solid}\:{cones}: \\ $$
Commented by mr W last updated on 06/Mar/21
![](https://www.tinkutara.com/question/19240.png)
Commented by mr W last updated on 07/Mar/21
![](https://www.tinkutara.com/question/19281.png)
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=(3/4)h OD=(√(h^2 +r^2 )) tan ρ=(r/h) ⇒ρ=tan^(−1) (r/h) GE⊥OD GE=e=((OG)/(OD))×r=((3hr)/(4(√(h^2 +r^2 )))) OE=f=((OG)/(OD))×h=((3h^2 )/(4(√(h^2 +r^2 )))) the position of the cone is described through θ with −90°≤θ≤90°, since the cone is released from rest at θ=−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 ω_P and that about its axis is ω_a . ω_P =(dθ/dt)=ω since there is no slipping on the contact, OD×ω_P =r×ω_a ⇒ω_a =((OD)/r)ω_P =(ω/(sin ρ)) ω_(P,s) =ω_P cos ρ=ω cos ρ ω_(P,a) =−ω_P sin ρ=−ω sin ρ we see ω_(P,a) is opposite to ω_a , therefore the negative sign. I_a =((3mr^2 )/(10)) I_s =((3m)/(20))(r^2 +4h^2 ) the kinetic energy of cone at θ: KE_t =(1/2)I_a (ω_a +ω_(P,a) )^2 +(1/2)I_s ω_(P,s) ^2 KE_t =(1/2)×((3mr^2 )/(10))×((1/(sin ρ))−sin ρ)^2 ω^2 +(1/2)×((3m)/(20))(r^2 +4h^2 )×cos^2 ρ×ω^2 KE_t =((3m)/(40))[((2/(sin ρ))−sin ρ)((1/(sin ρ))−sin ρ)r^2 +4 cos^2 ρ h^2 ]ω^2 KE_t =((3mh^2 )/(40))[((2/(sin ρ))−sin ρ)((1/(sin ρ))−sin ρ)tan^2 ρ+4 cos^2 ρ]ω^2 KE_t =((3mh^2 )/(40))(1+5 cos^2 ρ)ω^2 let z_O =0 at t=0 and θ=−90°: z_(E,0) =0 at θ: z_(E,t) =−OE×cos θ×sin φ=−f cos θ sin φ KE_t =mg(z_(E,0) −z_(E,t) ) ((3mh^2 )/(40))(1+5 cos^2 ρ)ω^2 =mgf cos θ sin φ (1+5 cos^2 ρ)ω^2 =((10g)/( (√(h^2 +r^2 ))))×cos θ sin φ ω^2 =((10g cos ρ)/( h(1+5 cos^2 ρ)))×cos θ sin φ with ξ=(√((10g cos ρ)/( h(1+5 cos^2 ρ)))) ⇒ω=ξ(√(sin φ cos θ)) (dθ/dt)=ξ(√(sin φ cos θ)) ∫(dθ/( (√(cos θ))))=ξ(√(sin φ))∫dt let T=period ∫_0 ^(π/2) (dθ/( (√(cos θ))))=((ξ(√(sin φ)) T)/4) ⇒T=(4/(ξ(√(sin φ))))∫_0 ^(π/2) (dθ/( (√(cos θ))))=((2B((1/4),(1/2)))/(ξ(√(sin φ)))) ⇒T≈((10.48823)/(ξ(√(sin φ))))=3.31667(√((1+5 cos^2 ρ)/(cos ρ sin φ)))(√(h/g)) example: ρ=15°=(π/(12)) φ=30°=(π/6) T≈11.359(√(h/g))](https://www.tinkutara.com/question/Q134685.png)
$${OC}={axis}\:{of}\:{cone} \\ $$$${OD}={contact}\:{line}\:{of}\:{cone}\:{on}\:{plane} \\ $$$${G}={center}\:{of}\:{mass}\:{of}\:{cone} \\ $$$${OC}={h} \\ $$$${OG}=\frac{\mathrm{3}}{\mathrm{4}}{h} \\ $$$${OD}=\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\rho=\frac{{r}}{{h}}\:\Rightarrow\rho=\mathrm{tan}^{−\mathrm{1}} \frac{{r}}{{h}} \\ $$$${GE}\bot{OD} \\ $$$${GE}={e}=\frac{{OG}}{{OD}}×{r}=\frac{\mathrm{3}{hr}}{\mathrm{4}\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }} \\ $$$${OE}={f}=\frac{{OG}}{{OD}}×{h}=\frac{\mathrm{3}{h}^{\mathrm{2}} }{\mathrm{4}\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }} \\ $$$$ \\ $$$${the}\:{position}\:{of}\:{the}\:{cone}\:{is}\:{described} \\ $$$${through}\:\theta\:{with}\:−\mathrm{90}°\leqslant\theta\leqslant\mathrm{90}°,\:{since} \\ $$$${the}\:{cone}\:{is}\:{released}\:{from}\:{rest}\:{at} \\ $$$$\theta=−\mathrm{90}°. \\ $$$${since}\:{the}\:{surface}\:{is}\:{rough}\:{enough} \\ $$$${such}\:{that}\:{the}\:{cone}\:{can}\:{only}\:{roll} \\ $$$${on}\:{the}\:{plane}\:{without}\:{slipping}, \\ $$$${the}\:{cone}\:{rotates}\:{about}\:{its}\:{tip}\:{point}\:{O} \\ $$$${along}\:{the}\:{plane}.\:{besides}\:{it}\:{rotates} \\ $$$${about}\:{its}\:{own}\:{axis}\:{OC}.\:{say}\:{the} \\ $$$${angular}\:{speed}\:{of}\:{the}\:{motion}\:{about}\:{O} \\ $$$${is}\:\omega_{{P}} \:{and}\:{that}\:{about}\:{its}\:{axis}\:{is}\:\omega_{{a}} . \\ $$$$\omega_{{P}} =\frac{{d}\theta}{{dt}}=\omega \\ $$$${since}\:{there}\:{is}\:{no}\:{slipping}\:{on}\:{the} \\ $$$${contact},\: \\ $$$${OD}×\omega_{{P}} ={r}×\omega_{{a}} \\ $$$$\Rightarrow\omega_{{a}} =\frac{{OD}}{{r}}\omega_{{P}} =\frac{\omega}{\mathrm{sin}\:\rho} \\ $$$$\omega_{{P},{s}} =\omega_{{P}} \mathrm{cos}\:\rho=\omega\:\mathrm{cos}\:\rho \\ $$$$\omega_{{P},{a}} =−\omega_{{P}} \mathrm{sin}\:\rho=−\omega\:\mathrm{sin}\:\rho \\ $$$${we}\:{see}\:\omega_{{P},{a}} \:{is}\:{opposite}\:{to}\:\omega_{{a}} ,\:{therefore} \\ $$$${the}\:{negative}\:{sign}. \\ $$$${I}_{{a}} =\frac{\mathrm{3}{mr}^{\mathrm{2}} }{\mathrm{10}} \\ $$$${I}_{{s}} =\frac{\mathrm{3}{m}}{\mathrm{20}}\left({r}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{2}} \right) \\ $$$${the}\:{kinetic}\:{energy}\:{of}\:{cone}\:{at}\:\theta: \\ $$$${KE}_{{t}} =\frac{\mathrm{1}}{\mathrm{2}}{I}_{{a}} \left(\omega_{{a}} +\omega_{{P},{a}} \right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{I}_{{s}} \omega_{{P},{s}} ^{\mathrm{2}} \\ $$$${KE}_{{t}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}{mr}^{\mathrm{2}} }{\mathrm{10}}×\left(\frac{\mathrm{1}}{\mathrm{sin}\:\rho}−\mathrm{sin}\:\rho\right)^{\mathrm{2}} \omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}{m}}{\mathrm{20}}\left({r}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{2}} \right)×\mathrm{cos}^{\mathrm{2}} \:\rho×\omega^{\mathrm{2}} \\ $$$${KE}_{{t}} =\frac{\mathrm{3}{m}}{\mathrm{40}}\left[\left(\frac{\mathrm{2}}{\mathrm{sin}\:\rho}−\mathrm{sin}\:\rho\right)\left(\frac{\mathrm{1}}{\mathrm{sin}\:\rho}−\mathrm{sin}\:\rho\right){r}^{\mathrm{2}} +\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\rho\:{h}^{\mathrm{2}} \right]\omega^{\mathrm{2}} \\ $$$${KE}_{{t}} =\frac{\mathrm{3}{mh}^{\mathrm{2}} }{\mathrm{40}}\left[\left(\frac{\mathrm{2}}{\mathrm{sin}\:\rho}−\mathrm{sin}\:\rho\right)\left(\frac{\mathrm{1}}{\mathrm{sin}\:\rho}−\mathrm{sin}\:\rho\right)\mathrm{tan}^{\mathrm{2}} \:\rho+\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\rho\right]\omega^{\mathrm{2}} \\ $$$${KE}_{{t}} =\frac{\mathrm{3}{mh}^{\mathrm{2}} }{\mathrm{40}}\left(\mathrm{1}+\mathrm{5}\:\mathrm{cos}^{\mathrm{2}} \:\rho\right)\omega^{\mathrm{2}} \\ $$$${let}\:{z}_{{O}} =\mathrm{0} \\ $$$${at}\:{t}=\mathrm{0}\:{and}\:\theta=−\mathrm{90}°: \\ $$$${z}_{{E},\mathrm{0}} =\mathrm{0} \\ $$$${at}\:\theta: \\ $$$${z}_{{E},{t}} =−{OE}×\mathrm{cos}\:\theta×\mathrm{sin}\:\phi=−{f}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi \\ $$$${KE}_{{t}} ={mg}\left({z}_{{E},\mathrm{0}} −{z}_{{E},{t}} \right) \\ $$$$\frac{\mathrm{3}{mh}^{\mathrm{2}} }{\mathrm{40}}\left(\mathrm{1}+\mathrm{5}\:\mathrm{cos}^{\mathrm{2}} \:\rho\right)\omega^{\mathrm{2}} ={mgf}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi \\ $$$$\left(\mathrm{1}+\mathrm{5}\:\mathrm{cos}^{\mathrm{2}} \:\rho\right)\omega^{\mathrm{2}} =\frac{\mathrm{10}{g}}{\:\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }}×\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi \\ $$$$\omega^{\mathrm{2}} =\frac{\mathrm{10}{g}\:\mathrm{cos}\:\rho}{\:{h}\left(\mathrm{1}+\mathrm{5}\:\mathrm{cos}^{\mathrm{2}} \:\rho\right)}×\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi \\ $$$${with}\:\xi=\sqrt{\frac{\mathrm{10}{g}\:\mathrm{cos}\:\rho}{\:{h}\left(\mathrm{1}+\mathrm{5}\:\mathrm{cos}^{\mathrm{2}} \:\rho\right)}} \\ $$$$\Rightarrow\omega=\xi\sqrt{\mathrm{sin}\:\phi\:\mathrm{cos}\:\theta} \\ $$$$\frac{{d}\theta}{{dt}}=\xi\sqrt{\mathrm{sin}\:\phi\:\mathrm{cos}\:\theta} \\ $$$$\int\frac{{d}\theta}{\:\sqrt{\mathrm{cos}\:\theta}}=\xi\sqrt{\mathrm{sin}\:\phi}\int{dt} \\ $$$${let}\:{T}={period} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\theta}{\:\sqrt{\mathrm{cos}\:\theta}}=\frac{\xi\sqrt{\mathrm{sin}\:\phi}\:{T}}{\mathrm{4}} \\ $$$$\Rightarrow{T}=\frac{\mathrm{4}}{\xi\sqrt{\mathrm{sin}\:\phi}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\theta}{\:\sqrt{\mathrm{cos}\:\theta}}=\frac{\mathrm{2}{B}\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{2}}\right)}{\xi\sqrt{\mathrm{sin}\:\phi}} \\ $$$$\Rightarrow{T}\approx\frac{\mathrm{10}.\mathrm{48823}}{\xi\sqrt{\mathrm{sin}\:\phi}}=\mathrm{3}.\mathrm{31667}\sqrt{\frac{\mathrm{1}+\mathrm{5}\:\mathrm{cos}^{\mathrm{2}} \:\rho}{\mathrm{cos}\:\rho\:\mathrm{sin}\:\phi}}\sqrt{\frac{{h}}{{g}}} \\ $$$$ \\ $$$${example}: \\ $$$$\rho=\mathrm{15}°=\frac{\pi}{\mathrm{12}} \\ $$$$\phi=\mathrm{30}°=\frac{\pi}{\mathrm{6}} \\ $$$${T}\approx\mathrm{11}.\mathrm{359}\sqrt{\frac{{h}}{{g}}} \\ $$
Commented by physicstutes last updated on 06/Mar/21
![sir W please reference me to an e−book where i can learn basic moment of inertia, especially things like basic integral derivation of moment of inertia for certain shapes(like rods,rectangles, rings etc) and appling these knowledge to compound pendulum and swinging masses. please i need help with those.](https://www.tinkutara.com/question/Q134688.png)
$$\mathrm{sir}\:\mathrm{W}\:\mathrm{please}\:\mathrm{reference}\:\mathrm{me}\:\mathrm{to}\:\mathrm{an}\:\mathrm{e}−\mathrm{book}\:\mathrm{where} \\ $$$$\mathrm{i}\:\mathrm{can}\:\mathrm{learn}\:\mathrm{basic}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia},\:\mathrm{especially}\:\mathrm{things} \\ $$$$\mathrm{like}\:\mathrm{basic}\:\mathrm{integral}\:\mathrm{derivation}\:\mathrm{of}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia} \\ $$$$\mathrm{for}\:\mathrm{certain}\:\mathrm{shapes}\left(\mathrm{like}\:\mathrm{rods},\mathrm{rectangles},\:\mathrm{rings}\:\mathrm{etc}\right)\:\mathrm{and} \\ $$$$\mathrm{appling}\:\mathrm{these}\:\mathrm{knowledge}\:\mathrm{to}\:\mathrm{compound}\:\mathrm{pendulum}\:\mathrm{and}\:\mathrm{swinging} \\ $$$$\mathrm{masses}.\:\mathrm{please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{help}\:\mathrm{with}\:\mathrm{those}.\: \\ $$
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.](https://www.tinkutara.com/question/Q134690.png)
$${i}\:{really}\:{don}'{t}\:{know}\:{such}\:{a}\:{book}\:{which} \\ $$$${i}\:{can}\:{recommend}.\:{nowadays}\:{when}\:{i} \\ $$$${need}\:{to}\:{look}\:{something},\:{i}\:{just}\:{google}. \\ $$
Commented by I want to learn more last updated on 06/Mar/21
![weldone sir.](https://www.tinkutara.com/question/Q134706.png)
$$\mathrm{weldone}\:\mathrm{sir}. \\ $$
Commented by physicstutes last updated on 06/Mar/21
![i tried googling but could not find a suitable pdf or even a book for the assesment.](https://www.tinkutara.com/question/Q134724.png)
$$\mathrm{i}\:\mathrm{tried}\:\mathrm{googling}\:\mathrm{but}\:\mathrm{could}\:\mathrm{not}\:\mathrm{find}\:\mathrm{a}\:\mathrm{suitable}\: \\ $$$$\mathrm{pdf}\:\mathrm{or}\:\mathrm{even}\:\mathrm{a}\:\mathrm{book}\:\mathrm{for}\:\mathrm{the}\:\mathrm{assesment}. \\ $$