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Question-163053




Question Number 163053 by ajfour last updated on 03/Jan/22
Commented by ajfour last updated on 03/Jan/22
Q.162949  (revisit)
$${Q}.\mathrm{162949}\:\:\left({revisit}\right) \\ $$
Answered by ajfour last updated on 04/Jan/22
  e=((2R)/π)=((4a)/π)        (2a=R)    r^2 =(2acos φ)^2 +(3a+2asin φ)^2   dm_(sc) =2λadφ  I_0 =((2aλ(6a)^2 )/3)+∫r^2 dm_(sc)     =24λa^3 +2λa^3 ∫_0 ^( π) (13+12sin φ)dφ    =24λa^3 +2λa^3 (13π+24)   say  m_(rod) =m=6λa  ⇒  I_0 =ma^2 (12+((13π)/3))=(((36+13π)ma^2 )/3)  M=(((π+3)m)/3)  ⇒ I_0 =(((36+13π)MR^2 )/(4(π+3)))          ✓  for center of mass:  2πaλ(((4a)/π))=(2πaλ+6aλ)s  s=((4a)/(π+3))  p^2 =s^2 +9a^2   tan β=(s/(3a))=(4/(3(π+3)))  Mgpcos (θ+β)=I_0 (((ωdω)/dθ))  ω^2 =2Mgp{sin (θ+β)−sin β}  T=(√(I_0 /(2Mgp)))∫_0 ^(  (π/2)−ϕ) (dθ/( (√(sin (θ+β)−sin β))))  where  ϕ=sin^(−1) (2/3)  T=(√((a(36+13π))/(4(3+π)g(√(((4/(π+3)))^2 +9)))))×∫★  ∫★=∫_0 ^(  cos^(−1) (2/3)) (dθ/( (√(sin (θ+tan^(−1) (4/(3(π+3))))−(4/( (√(16+9(π+3)^2 ))))))))  Sir, please  check  the answer this  yields  with yours..i will try too..
$$\:\:{e}=\frac{\mathrm{2}{R}}{\pi}=\frac{\mathrm{4}{a}}{\pi}\:\:\:\:\:\:\:\:\left(\mathrm{2}{a}={R}\right) \\ $$$$\:\:{r}^{\mathrm{2}} =\left(\mathrm{2}{a}\mathrm{cos}\:\phi\right)^{\mathrm{2}} +\left(\mathrm{3}{a}+\mathrm{2}{a}\mathrm{sin}\:\phi\right)^{\mathrm{2}} \\ $$$${dm}_{{sc}} =\mathrm{2}\lambda{ad}\phi \\ $$$${I}_{\mathrm{0}} =\frac{\mathrm{2}{a}\lambda\left(\mathrm{6}{a}\right)^{\mathrm{2}} }{\mathrm{3}}+\int{r}^{\mathrm{2}} {dm}_{{sc}} \\ $$$$\:\:=\mathrm{24}\lambda{a}^{\mathrm{3}} +\mathrm{2}\lambda{a}^{\mathrm{3}} \int_{\mathrm{0}} ^{\:\pi} \left(\mathrm{13}+\mathrm{12sin}\:\phi\right){d}\phi \\ $$$$\:\:=\mathrm{24}\lambda{a}^{\mathrm{3}} +\mathrm{2}\lambda{a}^{\mathrm{3}} \left(\mathrm{13}\pi+\mathrm{24}\right) \\ $$$$\:{say}\:\:{m}_{{rod}} ={m}=\mathrm{6}\lambda{a} \\ $$$$\Rightarrow\:\:{I}_{\mathrm{0}} ={ma}^{\mathrm{2}} \left(\mathrm{12}+\frac{\mathrm{13}\pi}{\mathrm{3}}\right)=\frac{\left(\mathrm{36}+\mathrm{13}\pi\right){ma}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${M}=\frac{\left(\pi+\mathrm{3}\right){m}}{\mathrm{3}} \\ $$$$\Rightarrow\:{I}_{\mathrm{0}} =\frac{\left(\mathrm{36}+\mathrm{13}\pi\right){MR}^{\mathrm{2}} }{\mathrm{4}\left(\pi+\mathrm{3}\right)} \\ $$$$ \\ $$$$\: \\ $$$$\:\:\:\checkmark \\ $$$${for}\:{center}\:{of}\:{mass}: \\ $$$$\mathrm{2}\pi{a}\lambda\left(\frac{\mathrm{4}{a}}{\pi}\right)=\left(\mathrm{2}\pi{a}\lambda+\mathrm{6}{a}\lambda\right){s} \\ $$$${s}=\frac{\mathrm{4}{a}}{\pi+\mathrm{3}} \\ $$$${p}^{\mathrm{2}} ={s}^{\mathrm{2}} +\mathrm{9}{a}^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\beta=\frac{{s}}{\mathrm{3}{a}}=\frac{\mathrm{4}}{\mathrm{3}\left(\pi+\mathrm{3}\right)} \\ $$$${Mgp}\mathrm{cos}\:\left(\theta+\beta\right)={I}_{\mathrm{0}} \left(\frac{\omega{d}\omega}{{d}\theta}\right) \\ $$$$\omega^{\mathrm{2}} =\mathrm{2}{Mgp}\left\{\mathrm{sin}\:\left(\theta+\beta\right)−\mathrm{sin}\:\beta\right\} \\ $$$${T}=\sqrt{\frac{{I}_{\mathrm{0}} }{\mathrm{2}{Mgp}}}\int_{\mathrm{0}} ^{\:\:\frac{\pi}{\mathrm{2}}−\varphi} \frac{{d}\theta}{\:\sqrt{\mathrm{sin}\:\left(\theta+\beta\right)−\mathrm{sin}\:\beta}} \\ $$$${where}\:\:\varphi=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}} \\ $$$${T}=\sqrt{\frac{{a}\left(\mathrm{36}+\mathrm{13}\pi\right)}{\mathrm{4}\left(\mathrm{3}+\pi\right){g}\sqrt{\left(\frac{\mathrm{4}}{\pi+\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{9}}}}×\int\bigstar \\ $$$$\int\bigstar=\int_{\mathrm{0}} ^{\:\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}} \frac{{d}\theta}{\:\sqrt{\mathrm{sin}\:\left(\theta+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}\left(\pi+\mathrm{3}\right)}\right)−\frac{\mathrm{4}}{\:\sqrt{\mathrm{16}+\mathrm{9}\left(\pi+\mathrm{3}\right)^{\mathrm{2}} }}}} \\ $$$${Sir},\:{please}\:\:{check}\:\:{the}\:{answer}\:{this} \\ $$$${yields}\:\:{with}\:{yours}..{i}\:{will}\:{try}\:{too}.. \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 04/Jan/22
i got with your formula T≈1.3626(√(R/g))  that differs from my result.
$${i}\:{got}\:{with}\:{your}\:{formula}\:{T}\approx\mathrm{1}.\mathrm{3626}\sqrt{\frac{{R}}{{g}}} \\ $$$${that}\:{differs}\:{from}\:{my}\:{result}. \\ $$

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