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Question Number 162949 by ajfour last updated on 02/Jan/22
Commented by ajfour last updated on 02/Jan/22
If the semi ring plus rod combina^n , having the same linear mass dens- ity is released as shown, how long does it take to turn around the pivot and hit the ground?
$${If}\:{the}\:{semi}\:{ring}\:{plus}\:{rod}\:{combina}^{{n}} , \\ $$$${having}\:{the}\:{same}\:{linear}\:{mass}\:{dens}- \\ $$$${ity}\:{is}\:{released}\:{as}\:{shown},\:{how}\:{long}\: \\ $$$${does}\:{it}\:{take}\:{to}\:{turn}\:{around}\:{the}\:{pivot} \\ $$$${and}\:{hit}\:{the}\:{ground}? \\ $$
Answered by mr W last updated on 02/Jan/22
Commented by mr W last updated on 02/Jan/22
ρ=mass of unit length m=total mass m=(πR+3R)ρ=(π+3)Rρ x=R cos φ y=((3R)/2)+R sin φ dm=ρRdφ me=∫xdm=∫_(−(π/2)) ^(π/2) R cos φ ρRdφ me=ρR^2 [sin φ]_(−(π/2)) ^(π/2) =2ρR^2 (π+3)Rρe=2ρR^2 ⇒e=((2R)/(π+3)) I_x =(((3Rρ)(3R)^2 )/3)+∫y^2 dm I_x =9ρR^3 +∫_(−(π/2)) ^(π/2) (((3R)/2)+R sin φ)^2 ρRdφ I_x =9ρR^3 +ρR^3 ∫_(−(π/2)) ^(π/2) ((3/2)+sin φ)^2 dφ I_x =9ρR^3 +ρR^3 ∫_(−(π/2)) ^(π/2) ((9/4)+3 sin φ+sin^2 φ)dφ I_x =9ρR^3 +ρR^3 ∫_(−(π/2)) ^(π/2) (((11)/4)+3 sin φ−((cos 2φ)/2))dφ I_x =9ρR^3 +ρR^3 [((11φ)/4)−3 cos φ−((sin 2φ)/4)]_(−(π/2)) ^(π/2) I_x =9ρR^3 +((11πρR^3 )/4)=(9+((11π)/4))ρR^3 ⇒I_x =(((36+11π)mR^2 )/(4(3+π))) I_y =∫x^2 dm I_y =∫_(−(π/2)) ^(π/2) R^2 cos^2 φρRdφ I_y =((ρR^3 )/2)∫_(−(π/2)) ^(π/2) (cos 2φ+1)dφ I_y =((ρR^3 )/2)[((sin 2φ)/2)+φ]_(−(π/2)) ^(π/2) I_y =((πρR^3 )/2) ⇒I_y =((πmR^2 )/(2(3+π))) I_0 =I_x +I_y =(((36+11π)mR^2 )/(4(3+π)))+((πmR^2 )/(2(3+π))) ⇒I_0 =(((36+13π)mR^2 )/(4(3+π)))
$$\rho={mass}\:{of}\:{unit}\:{length} \\ $$$${m}={total}\:{mass} \\ $$$${m}=\left(\pi{R}+\mathrm{3}{R}\right)\rho=\left(\pi+\mathrm{3}\right){R}\rho \\ $$$${x}={R}\:\mathrm{cos}\:\phi \\ $$$${y}=\frac{\mathrm{3}{R}}{\mathrm{2}}+{R}\:\mathrm{sin}\:\phi \\ $$$${dm}=\rho{Rd}\phi \\ $$$${me}=\int{xdm}=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {R}\:\mathrm{cos}\:\phi\:\rho{Rd}\phi \\ $$$${me}=\rho{R}^{\mathrm{2}} \left[\mathrm{sin}\:\phi\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{2}\rho{R}^{\mathrm{2}} \\ $$$$\left(\pi+\mathrm{3}\right){R}\rho{e}=\mathrm{2}\rho{R}^{\mathrm{2}} \\ $$$$\Rightarrow{e}=\frac{\mathrm{2}{R}}{\pi+\mathrm{3}} \\ $$$${I}_{{x}} =\frac{\left(\mathrm{3}{R}\rho\right)\left(\mathrm{3}{R}\right)^{\mathrm{2}} }{\mathrm{3}}+\int{y}^{\mathrm{2}} {dm} \\ $$$${I}_{{x}} =\mathrm{9}\rho{R}^{\mathrm{3}} +\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{3}{R}}{\mathrm{2}}+{R}\:\mathrm{sin}\:\phi\right)^{\mathrm{2}} \rho{Rd}\phi \\ $$$${I}_{{x}} =\mathrm{9}\rho{R}^{\mathrm{3}} +\rho{R}^{\mathrm{3}} \int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{sin}\:\phi\right)^{\mathrm{2}} {d}\phi \\ $$$${I}_{{x}} =\mathrm{9}\rho{R}^{\mathrm{3}} +\rho{R}^{\mathrm{3}} \int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{3}\:\mathrm{sin}\:\phi+\mathrm{sin}^{\mathrm{2}} \:\phi\right){d}\phi \\ $$$${I}_{{x}} =\mathrm{9}\rho{R}^{\mathrm{3}} +\rho{R}^{\mathrm{3}} \int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{11}}{\mathrm{4}}+\mathrm{3}\:\mathrm{sin}\:\phi−\frac{\mathrm{cos}\:\mathrm{2}\phi}{\mathrm{2}}\right){d}\phi \\ $$$${I}_{{x}} =\mathrm{9}\rho{R}^{\mathrm{3}} +\rho{R}^{\mathrm{3}} \left[\frac{\mathrm{11}\phi}{\mathrm{4}}−\mathrm{3}\:\mathrm{cos}\:\phi−\frac{\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{4}}\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$${I}_{{x}} =\mathrm{9}\rho{R}^{\mathrm{3}} +\frac{\mathrm{11}\pi\rho{R}^{\mathrm{3}} }{\mathrm{4}}=\left(\mathrm{9}+\frac{\mathrm{11}\pi}{\mathrm{4}}\right)\rho{R}^{\mathrm{3}} \\ $$$$\Rightarrow{I}_{{x}} =\frac{\left(\mathrm{36}+\mathrm{11}\pi\right){mR}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{3}+\pi\right)} \\ $$$${I}_{{y}} =\int{x}^{\mathrm{2}} {dm} \\ $$$${I}_{{y}} =\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {R}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\phi\rho{Rd}\phi \\ $$$${I}_{{y}} =\frac{\rho{R}^{\mathrm{3}} }{\mathrm{2}}\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{cos}\:\mathrm{2}\phi+\mathrm{1}\right){d}\phi \\ $$$${I}_{{y}} =\frac{\rho{R}^{\mathrm{3}} }{\mathrm{2}}\left[\frac{\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{2}}+\phi\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$${I}_{{y}} =\frac{\pi\rho{R}^{\mathrm{3}} }{\mathrm{2}} \\ $$$$\Rightarrow{I}_{{y}} =\frac{\pi{mR}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{3}+\pi\right)} \\ $$$${I}_{\mathrm{0}} ={I}_{{x}} +{I}_{{y}} =\frac{\left(\mathrm{36}+\mathrm{11}\pi\right){mR}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{3}+\pi\right)}+\frac{\pi{mR}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{3}+\pi\right)} \\ $$$$\Rightarrow{I}_{\mathrm{0}} =\frac{\left(\mathrm{36}+\mathrm{13}\pi\right){mR}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{3}+\pi\right)} \\ $$
Commented by mr W last updated on 02/Jan/22
Commented by mr W last updated on 02/Jan/22
cos ϕ=(R/(1.5R))=(2/3) ⇒ϕ=cos^(−1) (2/3) ω=(dθ/dt) α=(dω/dt)=ω(dω/dθ) I_0 α=mg(1.5R sin θ+e cos θ) (((36+13π)mR^2 )/(4(3+π)))ω(dω/dθ)=mg(((3R)/2) sin θ+((2R)/(3+π)) cos θ) ω(dω/dθ)=((2g)/((36+13π)R))[3(3+π) sin θ+4 cos θ] ∫_0 ^ω ωdω=((2g)/((36+13π)R))∫_0 ^θ [3(3+π) sin θ+4 cos θ]dθ (ω^2 /2)=((2g)/((36+13π)R))[3(3+π)(1−cos θ)+4 sin θ] ω=(√((4g)/((36+13π)R)))×(√(3(3+π)(1−cos θ)+4 sin θ)) (√(((36+13π)R)/(4g)))(dθ/( (√(3(3+π)(1−cos θ)+4 sin θ))))=dt (√(((36+13π)R)/(4g)))∫_0 ^ϕ (dθ/( (√(3(3+π)(1−cos θ)+4 sin θ))))=∫_0 ^T dt T=(√(((36+13π)R)/(4g)))∫_0 ^(cos^(−1) (2/3)) (dθ/( (√(3(3+π)(1−cos θ)+4 sin θ)))) T≈0.751917(√(((36+13π)R)/(4g)))≈3.2956(√(R/g))
$$\mathrm{cos}\:\varphi=\frac{{R}}{\mathrm{1}.\mathrm{5}{R}}=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow\varphi=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\omega=\frac{{d}\theta}{{dt}} \\ $$$$\alpha=\frac{{d}\omega}{{dt}}=\omega\frac{{d}\omega}{{d}\theta} \\ $$$${I}_{\mathrm{0}} \alpha={mg}\left(\mathrm{1}.\mathrm{5}{R}\:\mathrm{sin}\:\theta+{e}\:\mathrm{cos}\:\theta\right) \\ $$$$\frac{\left(\mathrm{36}+\mathrm{13}\pi\right){mR}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{3}+\pi\right)}\omega\frac{{d}\omega}{{d}\theta}={mg}\left(\frac{\mathrm{3}{R}}{\mathrm{2}}\:\mathrm{sin}\:\theta+\frac{\mathrm{2}{R}}{\mathrm{3}+\pi}\:\mathrm{cos}\:\theta\right) \\ $$$$\omega\frac{{d}\omega}{{d}\theta}=\frac{\mathrm{2}{g}}{\left(\mathrm{36}+\mathrm{13}\pi\right){R}}\left[\mathrm{3}\left(\mathrm{3}+\pi\right)\:\mathrm{sin}\:\theta+\mathrm{4}\:\mathrm{cos}\:\theta\right] \\ $$$$\int_{\mathrm{0}} ^{\omega} \omega{d}\omega=\frac{\mathrm{2}{g}}{\left(\mathrm{36}+\mathrm{13}\pi\right){R}}\int_{\mathrm{0}} ^{\theta} \left[\mathrm{3}\left(\mathrm{3}+\pi\right)\:\mathrm{sin}\:\theta+\mathrm{4}\:\mathrm{cos}\:\theta\right]{d}\theta \\ $$$$\frac{\omega^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{2}{g}}{\left(\mathrm{36}+\mathrm{13}\pi\right){R}}\left[\mathrm{3}\left(\mathrm{3}+\pi\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right)+\mathrm{4}\:\mathrm{sin}\:\theta\right] \\ $$$$\omega=\sqrt{\frac{\mathrm{4}{g}}{\left(\mathrm{36}+\mathrm{13}\pi\right){R}}}×\sqrt{\mathrm{3}\left(\mathrm{3}+\pi\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right)+\mathrm{4}\:\mathrm{sin}\:\theta} \\ $$$$\sqrt{\frac{\left(\mathrm{36}+\mathrm{13}\pi\right){R}}{\mathrm{4}{g}}}\frac{{d}\theta}{\:\sqrt{\mathrm{3}\left(\mathrm{3}+\pi\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right)+\mathrm{4}\:\mathrm{sin}\:\theta}}={dt} \\ $$$$\sqrt{\frac{\left(\mathrm{36}+\mathrm{13}\pi\right){R}}{\mathrm{4}{g}}}\int_{\mathrm{0}} ^{\varphi} \frac{{d}\theta}{\:\sqrt{\mathrm{3}\left(\mathrm{3}+\pi\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right)+\mathrm{4}\:\mathrm{sin}\:\theta}}=\int_{\mathrm{0}} ^{{T}} {dt} \\ $$$${T}=\sqrt{\frac{\left(\mathrm{36}+\mathrm{13}\pi\right){R}}{\mathrm{4}{g}}}\int_{\mathrm{0}} ^{\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}} \frac{{d}\theta}{\:\sqrt{\mathrm{3}\left(\mathrm{3}+\pi\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right)+\mathrm{4}\:\mathrm{sin}\:\theta}} \\ $$$${T}\approx\mathrm{0}.\mathrm{751917}\sqrt{\frac{\left(\mathrm{36}+\mathrm{13}\pi\right){R}}{\mathrm{4}{g}}}\approx\mathrm{3}.\mathrm{2956}\sqrt{\frac{{R}}{{g}}} \\ $$
Commented by ajfour last updated on 02/Jan/22
OmG ! gimme time, sir..
$${OmG}\:!\:{gimme}\:{time},\:{sir}.. \\ $$
Commented by mr W last updated on 02/Jan/22
do you have the answer sir?
$${do}\:{you}\:{have}\:{the}\:{answer}\:{sir}? \\ $$
Commented by ajfour last updated on 02/Jan/22
no, my creation.., sir!
$${no},\:{my}\:{creation}..,\:{sir}! \\ $$
Commented by mr W last updated on 02/Jan/22
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Commented by Ar Brandon last updated on 02/Jan/22
Is this Lekh diagram, Sir?
Commented by mr W last updated on 02/Jan/22
for this one i didn′t use LEKH, but an any image app in my smart phone. certainly you can use LEKH.
$${for}\:{this}\:{one}\:{i}\:{didn}'{t}\:{use}\:{LEKH},\:{but} \\ $$$${an}\:{any}\:{image}\:{app}\:{in}\:{my}\:{smart}\:{phone}. \\ $$$${certainly}\:{you}\:{can}\:{use}\:{LEKH}. \\ $$
Commented by Tawa11 last updated on 03/Jan/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by Ar Brandon last updated on 03/Jan/22
OK
Answered by ajfour last updated on 02/Jan/22
Commented by ajfour last updated on 03/Jan/22
Mgrcos (θ+β)+mgLcos θ=I(((ωdω)/dθ)) (ω^2 /(2g))=((Mrsin (θ+β)+mLsin θ)/I) ∫_0 ^( T) dt=(√(I/(2g)))∫_0 ^( φ) (dθ/( (√(Mrsin (θ+β)+mLsin θ)))) .... T(√((2g)/I))=∫_0 ^( φ) (dθ/( (√(M(esin θ+((3R)/2)cos θ)+m(((3R)/2)sin θ))))) cos φ=(2/3) (M/m)=((πR)/(3R))=(π/3) M((R/2))+m(0)=(M+m)e e=((R/2)/(1+(3/π)))=((πR)/(2(π+3))) I=? (i doubt, if we can easily find..) T(√((2mgR)/I))=J J=∫_0 ^( φ) (dθ/( (√([((π/6))((π/(π+3)))+(3/2)]sin θ+((π/2))cos θ)))) i can go this far, Sir!
$${Mgr}\mathrm{cos}\:\left(\theta+\beta\right)+{mgL}\mathrm{cos}\:\theta={I}\left(\frac{\omega{d}\omega}{{d}\theta}\right) \\ $$$$\frac{\omega^{\mathrm{2}} }{\mathrm{2}{g}}=\frac{{Mr}\mathrm{sin}\:\left(\theta+\beta\right)+{mL}\mathrm{sin}\:\theta}{{I}} \\ $$$$\int_{\mathrm{0}} ^{\:{T}} {dt}=\sqrt{\frac{{I}}{\mathrm{2}{g}}}\int_{\mathrm{0}} ^{\:\phi} \frac{{d}\theta}{\:\sqrt{{Mr}\mathrm{sin}\:\left(\theta+\beta\right)+{mL}\mathrm{sin}\:\theta}} \\ $$$$…. \\ $$$${T}\sqrt{\frac{\mathrm{2}{g}}{{I}}}=\int_{\mathrm{0}} ^{\:\phi} \frac{{d}\theta}{\:\sqrt{{M}\left({e}\mathrm{sin}\:\theta+\frac{\mathrm{3}{R}}{\mathrm{2}}\mathrm{cos}\:\theta\right)+{m}\left(\frac{\mathrm{3}{R}}{\mathrm{2}}\mathrm{sin}\:\theta\right)}} \\ $$$${cos}\:\phi=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\frac{{M}}{{m}}=\frac{\pi{R}}{\mathrm{3}{R}}=\frac{\pi}{\mathrm{3}} \\ $$$${M}\left(\frac{{R}}{\mathrm{2}}\right)+{m}\left(\mathrm{0}\right)=\left({M}+{m}\right){e} \\ $$$${e}=\frac{{R}/\mathrm{2}}{\mathrm{1}+\frac{\mathrm{3}}{\pi}}=\frac{\pi{R}}{\mathrm{2}\left(\pi+\mathrm{3}\right)} \\ $$$${I}=?\:\left({i}\:{doubt},\:{if}\:{we}\:{can}\:{easily}\:{find}..\right) \\ $$$${T}\sqrt{\frac{\mathrm{2}{mgR}}{{I}}}={J} \\ $$$${J}=\int_{\mathrm{0}} ^{\:\phi} \frac{{d}\theta}{\:\sqrt{\left[\left(\frac{\pi}{\mathrm{6}}\right)\left(\frac{\pi}{\pi+\mathrm{3}}\right)+\frac{\mathrm{3}}{\mathrm{2}}\right]\mathrm{sin}\:\theta+\left(\frac{\pi}{\mathrm{2}}\right)\mathrm{cos}\:\theta}} \\ $$$${i}\:{can}\:{go}\:{this}\:{far},\:{Sir}! \\ $$
Commented by mr W last updated on 03/Jan/22
the COM of a semi ring is not (R/2) apart from the center of the circle, but ((2R)/π). therefore it should be: M(((2R)/π))+m(0)=(M+m)e e=((2R/π)/(1+(3/π)))=((2R)/(π+3)) i got this too.
$${the}\:{COM}\:{of}\:{a}\:{semi}\:{ring}\:{is}\:{not}\:\frac{{R}}{\mathrm{2}} \\ $$$${apart}\:{from}\:{the}\:{center}\:{of}\:{the}\:{circle},\: \\ $$$${but}\:\frac{\mathrm{2}{R}}{\pi}. \\ $$$${therefore}\:{it}\:{should}\:{be}: \\ $$$${M}\left(\frac{\mathrm{2}{R}}{\pi}\right)+{m}\left(\mathrm{0}\right)=\left({M}+{m}\right){e} \\ $$$${e}=\frac{\mathrm{2}{R}/\pi}{\mathrm{1}+\frac{\mathrm{3}}{\pi}}=\frac{\mathrm{2}{R}}{\pi+\mathrm{3}} \\ $$$${i}\:{got}\:{this}\:{too}. \\ $$
Commented by mr W last updated on 03/Jan/22
Commented by mr W last updated on 03/Jan/22
I_0 can also be “easily” obtained in following way: the MoI of a complete ring about its axis is obviously I_p =MR^2 , so the MoI about its diameter is obviously ((MR^2 )/2), since I_P =I_x +I_y and I_x =I_y . on the other side a complete ring consists of two semi rings. ((MR^2 )/2)=2(I_(semi,y,0) +m_(semi) e^2 ) I_(semi,y) =I_(semi,y,0) +m_(semi) e^2 =((MR^2 )/4)=((m_(semi) R^2 )/2) ((MR^2 )/2)=2I_(semi,x,0) I_(semi,x,0) =((MR^2 )/4)=((m_(semi) R^2 )/2) I_(semi,x) =I_(semi,x,0) +m_(semi) (((3R)/2))^2 I_(semi,x) =((m_(semi) R^2 )/2)+((9m_(semi) R^2 )/4)=((11m_(semu) R^2 )/4) I_(rod,y) =0 I_(rod,x) =((m_(rod) (3R)^2 )/3)=3m_(rod) R^2 I_(total,y) =I_(semi,y) +I_(rod,y) =((m_(semi) R^2 )/2) I_(total,x) =I_(semi,x) +I_(rod,x) =((11m_(semi) R^2 )/4)+3m_(rod) R^2 I_(total,0) =I_(total,y) +I_(total,x) I_(total,0) =((m_(semi) R^2 )/2)+((11m_(semi) R^2 )/4)+3m_(rod) R^2 I_(total,0) =((13m_(semi) R^2 )/4)+3m_(rod) R^2 m_(semi) =(π/(3+π))m_(total) m_(rod) =(3/(3+π))m_(total) I_(total,0) =(((13π)/4)+9)((m_(total) R^2 )/(3+π))=(((36+13π)m_(total) R^2 )/(4(3+π)))
$${I}_{\mathrm{0}} \:{can}\:{also}\:{be}\:“{easily}''\:{obtained}\:{in}\: \\ $$$${following}\:{way}: \\ $$$${the}\:{MoI}\:{of}\:{a}\:{complete}\:{ring}\:{about} \\ $$$${its}\:{axis}\:{is}\:{obviously}\:{I}_{{p}} ={MR}^{\mathrm{2}} ,\:{so}\:{the}\: \\ $$$${MoI}\:{about}\:{its}\:{diameter}\:{is}\:{obviously} \\ $$$$\frac{{MR}^{\mathrm{2}} }{\mathrm{2}},\:{since}\:{I}_{{P}} ={I}_{{x}} +{I}_{{y}} \:{and}\:{I}_{{x}} ={I}_{{y}} . \\ $$$${on}\:{the}\:{other}\:{side}\:{a}\:{complete}\:{ring} \\ $$$${consists}\:{of}\:{two}\:{semi}\:{rings}. \\ $$$$\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{2}\left({I}_{{semi},{y},\mathrm{0}} +{m}_{{semi}} {e}^{\mathrm{2}} \right) \\ $$$${I}_{{semi},{y}} ={I}_{{semi},{y},\mathrm{0}} +{m}_{{semi}} {e}^{\mathrm{2}} =\frac{{MR}^{\mathrm{2}} }{\mathrm{4}}=\frac{{m}_{{semi}} {R}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{2}{I}_{{semi},{x},\mathrm{0}} \\ $$$${I}_{{semi},{x},\mathrm{0}} =\frac{{MR}^{\mathrm{2}} }{\mathrm{4}}=\frac{{m}_{{semi}} {R}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}_{{semi},{x}} ={I}_{{semi},{x},\mathrm{0}} +{m}_{{semi}} \left(\frac{\mathrm{3}{R}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${I}_{{semi},{x}} =\frac{{m}_{{semi}} {R}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{9}{m}_{{semi}} {R}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{11}{m}_{{semu}} {R}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${I}_{{rod},{y}} =\mathrm{0} \\ $$$${I}_{{rod},{x}} =\frac{{m}_{{rod}} \left(\mathrm{3}{R}\right)^{\mathrm{2}} }{\mathrm{3}}=\mathrm{3}{m}_{{rod}} {R}^{\mathrm{2}} \\ $$$${I}_{{total},{y}} ={I}_{{semi},{y}} +{I}_{{rod},{y}} =\frac{{m}_{{semi}} {R}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}_{{total},{x}} ={I}_{{semi},{x}} +{I}_{{rod},{x}} =\frac{\mathrm{11}{m}_{{semi}} {R}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{3}{m}_{{rod}} {R}^{\mathrm{2}} \\ $$$${I}_{{total},\mathrm{0}} ={I}_{{total},{y}} +{I}_{{total},{x}} \\ $$$${I}_{{total},\mathrm{0}} =\frac{{m}_{{semi}} {R}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{11}{m}_{{semi}} {R}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{3}{m}_{{rod}} {R}^{\mathrm{2}} \\ $$$${I}_{{total},\mathrm{0}} =\frac{\mathrm{13}{m}_{{semi}} {R}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{3}{m}_{{rod}} {R}^{\mathrm{2}} \\ $$$${m}_{{semi}} =\frac{\pi}{\mathrm{3}+\pi}{m}_{{total}} \\ $$$${m}_{{rod}} =\frac{\mathrm{3}}{\mathrm{3}+\pi}{m}_{{total}} \\ $$$${I}_{{total},\mathrm{0}} =\left(\frac{\mathrm{13}\pi}{\mathrm{4}}+\mathrm{9}\right)\frac{{m}_{{total}} {R}^{\mathrm{2}} }{\mathrm{3}+\pi}=\frac{\left(\mathrm{36}+\mathrm{13}\pi\right){m}_{{total}} {R}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{3}+\pi\right)} \\ $$
Commented by ajfour last updated on 03/Jan/22
I thought first // axis theorem for moment of inertia wd apply for symmetrical bodies only, but really thats the restriction in ⊥ axis theorem, so i had commented such, n yes sir, sorry for the mistake for c.m. of ring..
$${I}\:{thought}\:\:{first}\:\://\:\:{axis}\:{theorem} \\ $$$${for}\:{moment}\:{of}\:{inertia}\:{wd}\:{apply} \\ $$$${for}\:{symmetrical}\:{bodies}\:{only}, \\ $$$${but}\:{really}\:{thats}\:{the}\:{restriction} \\ $$$${in}\:\:\bot\:{axis}\:{theorem},\:{so}\:{i}\:{had} \\ $$$${commented}\:{such},\:{n}\:{yes}\:{sir},\:{sorry} \\ $$$${for}\:{the}\:{mistake}\:{for}\:{c}.{m}.\:{of}\:{ring}.. \\ $$
Answered by mr W last updated on 03/Jan/22
Commented by ajfour last updated on 03/Jan/22
Awezome ..→∞
$${Awezome}\:..\rightarrow\infty \\ $$
Commented by mr W last updated on 03/Jan/22
cos ϕ=(2/3) e=((2R)/(3+π)) I_0 =(((36+13π)mR^2 )/(4(3+π))) r=(√(e^2 +(((3R)/2))^2 ))=(((√(97+54π+9π^2 ))R)/(2(3+π))) cos φ=((3R)/(2r))=((3(3+π))/( (√(97+54π+9π^2 )))) mgr sin θ=I_0 (ω(dω/dθ)) mgr sin θ=(((36+13π)mR^2 )/(4(3+π)))(ω(dω/dθ)) ((2g(√(97+54π+9π^2 )))/((36+13π)R)) sin θ=ω(dω/dθ) ωdω=((2g(√(97+54π+9π^2 )))/((36+13π)R)) sin θ dθ (ω^2 /2)=((2g(√(97+54π+9π^2 )))/((36+13π)R)) [cos (φ)−cos θ] ω^2 =((4g(√(97+54π+9π^2 )))/((36+13π)R)) [((3(3+π))/( (√(97+54π+9π^2 ))))−cos θ] ω^2 =(g/(a^2 R)) (b−cos θ) a=(1/2)(√((36+13π)/( (√(97+54π+9π^2 )))))≈1.009402 b=cos φ=((3(3+π))/( (√(97+54π+9π^2 ))))≈0.977236 ω=(dθ/dt)=((√(b−cos θ))/a)(√(g/R)) (√(R/g))×((adθ)/( (√(b−cos θ))))=dt T=(√(R/g)) ∫_φ ^(ϕ+φ) ((adθ)/( (√(b−cos θ))))≈3.2956(√(R/g))
$$\mathrm{cos}\:\varphi=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${e}=\frac{\mathrm{2}{R}}{\mathrm{3}+\pi} \\ $$$${I}_{\mathrm{0}} =\frac{\left(\mathrm{36}+\mathrm{13}\pi\right){mR}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{3}+\pi\right)} \\ $$$${r}=\sqrt{{e}^{\mathrm{2}} +\left(\frac{\mathrm{3}{R}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{97}+\mathrm{54}\pi+\mathrm{9}\pi^{\mathrm{2}} \:}{R}}{\mathrm{2}\left(\mathrm{3}+\pi\right)} \\ $$$$\mathrm{cos}\:\phi=\frac{\mathrm{3}{R}}{\mathrm{2}{r}}=\frac{\mathrm{3}\left(\mathrm{3}+\pi\right)}{\:\sqrt{\mathrm{97}+\mathrm{54}\pi+\mathrm{9}\pi^{\mathrm{2}} }} \\ $$$${mgr}\:\mathrm{sin}\:\theta={I}_{\mathrm{0}} \left(\omega\frac{{d}\omega}{{d}\theta}\right) \\ $$$${mgr}\:\mathrm{sin}\:\theta=\frac{\left(\mathrm{36}+\mathrm{13}\pi\right){mR}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{3}+\pi\right)}\left(\omega\frac{{d}\omega}{{d}\theta}\right) \\ $$$$\frac{\mathrm{2}{g}\sqrt{\mathrm{97}+\mathrm{54}\pi+\mathrm{9}\pi^{\mathrm{2}} }}{\left(\mathrm{36}+\mathrm{13}\pi\right){R}}\:\mathrm{sin}\:\theta=\omega\frac{{d}\omega}{{d}\theta} \\ $$$$\omega{d}\omega=\frac{\mathrm{2}{g}\sqrt{\mathrm{97}+\mathrm{54}\pi+\mathrm{9}\pi^{\mathrm{2}} }}{\left(\mathrm{36}+\mathrm{13}\pi\right){R}}\:\mathrm{sin}\:\theta\:{d}\theta \\ $$$$\frac{\omega^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{2}{g}\sqrt{\mathrm{97}+\mathrm{54}\pi+\mathrm{9}\pi^{\mathrm{2}} }}{\left(\mathrm{36}+\mathrm{13}\pi\right){R}}\:\left[\mathrm{cos}\:\left(\phi\right)−\mathrm{cos}\:\theta\right] \\ $$$$\omega^{\mathrm{2}} =\frac{\mathrm{4}{g}\sqrt{\mathrm{97}+\mathrm{54}\pi+\mathrm{9}\pi^{\mathrm{2}} }}{\left(\mathrm{36}+\mathrm{13}\pi\right){R}}\:\left[\frac{\mathrm{3}\left(\mathrm{3}+\pi\right)}{\:\sqrt{\mathrm{97}+\mathrm{54}\pi+\mathrm{9}\pi^{\mathrm{2}} }}−\mathrm{cos}\:\theta\right] \\ $$$$\omega^{\mathrm{2}} =\frac{{g}}{{a}^{\mathrm{2}} {R}}\:\left({b}−\mathrm{cos}\:\theta\right) \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{36}+\mathrm{13}\pi}{\:\sqrt{\mathrm{97}+\mathrm{54}\pi+\mathrm{9}\pi^{\mathrm{2}} }}}\approx\mathrm{1}.\mathrm{009402} \\ $$$${b}=\mathrm{cos}\:\phi=\frac{\mathrm{3}\left(\mathrm{3}+\pi\right)}{\:\sqrt{\mathrm{97}+\mathrm{54}\pi+\mathrm{9}\pi^{\mathrm{2}} }}\approx\mathrm{0}.\mathrm{977236} \\ $$$$\omega=\frac{{d}\theta}{{dt}}=\frac{\sqrt{{b}−\mathrm{cos}\:\theta}}{{a}}\sqrt{\frac{{g}}{{R}}} \\ $$$$\sqrt{\frac{{R}}{{g}}}×\frac{{ad}\theta}{\:\sqrt{{b}−\mathrm{cos}\:\theta}}={dt} \\ $$$${T}=\sqrt{\frac{{R}}{{g}}}\:\int_{\phi} ^{\varphi+\phi} \frac{{ad}\theta}{\:\sqrt{{b}−\mathrm{cos}\:\theta}}\approx\mathrm{3}.\mathrm{2956}\sqrt{\frac{{R}}{{g}}} \\ $$
Commented by mr W last updated on 03/Jan/22
not really better ...
$${not}\:{really}\:{better}\:… \\ $$

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