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

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