Question Number 225474 by mr W last updated on 27/Oct/25
Commented by mr W last updated on 27/Oct/25
$${find}\:{the}\:{time}\:{the}\:{uniform}\:{rod} \\ $$$${takes}\:{to}\:{strike}\:{the}\:{ground}\:{as}\:{shown}. \\ $$$${wall}\:{and}\:{ground}\:{are}\:{frictionless}. \\ $$
Commented by mr W last updated on 28/Oct/25
Answered by mr W last updated on 28/Oct/25
Commented by mr W last updated on 29/Oct/25
![Method 2: we take D as (0,0). ((AD)/(sin (θ−φ)))=(L/(sin (π−θ))) ⇒AD=((L sin (θ−φ))/(sin θ)) x_C =AD−AC×cos φ=((L sin (θ−φ))/(sin θ))−((L cos φ)/2) y_C =((L sin φ)/2) let ω=−(dθ/dt) v_(Cx) =ωL(((cos (θ−φ))/(sin θ))−((sin φ)/2)) v_(Cy) =−((ωL cos φ)/2) v_C ^2 =ω^2 L^2 [(((cos (θ−φ))/(sin θ))−((sin φ)/2))^2 +(((cos φ)/2))^2 ] v_C ^2 =ω^2 L^2 [((cos^2 (θ−φ))/(sin^2 θ))−((cos (θ−φ) sin φ)/(sin θ))+(1/4)] v_C ^2 =ω^2 L^2 [((cos (θ−φ)cos θ cos φ)/(sin^2 θ))+(1/4)] (1/2)Iω^2 +(1/2)Mv_C ^2 =Mg(L/2)(sin θ−sin φ) ((ML^2 )/(12))ω^2 +Mω^2 L^2 [((cos (θ−φ)cos θ cos φ)/(sin^2 θ))+(1/4)]=MgL(sin θ−sin φ) ω^2 L[((3 cos (θ−φ)cos θ cos φ)/(sin^2 θ))+1]=3g(sin θ−sin φ) ⇒ω=(√((3g(sin θ−sin φ))/(L[1+((3cos (θ−φ)cos θ cos φ)/(sin^2 θ))])))](https://www.tinkutara.com/question/Q225487.png)
$${Method}\:\mathrm{2}: \\ $$$${we}\:{take}\:{D}\:{as}\:\left(\mathrm{0},\mathrm{0}\right). \\ $$$$\frac{{AD}}{\mathrm{sin}\:\left(\theta−\phi\right)}=\frac{{L}}{\mathrm{sin}\:\left(\pi−\theta\right)} \\ $$$$\Rightarrow{AD}=\frac{{L}\:\mathrm{sin}\:\left(\theta−\phi\right)}{\mathrm{sin}\:\theta} \\ $$$${x}_{{C}} ={AD}−{AC}×\mathrm{cos}\:\phi=\frac{{L}\:\mathrm{sin}\:\left(\theta−\phi\right)}{\mathrm{sin}\:\theta}−\frac{{L}\:\mathrm{cos}\:\phi}{\mathrm{2}} \\ $$$${y}_{{C}} =\frac{{L}\:\mathrm{sin}\:\phi}{\mathrm{2}} \\ $$$${let}\:\omega=−\frac{{d}\theta}{{dt}} \\ $$$${v}_{{Cx}} =\omega{L}\left(\frac{\mathrm{cos}\:\left(\theta−\phi\right)}{\mathrm{sin}\:\theta}−\frac{\mathrm{sin}\:\phi}{\mathrm{2}}\right) \\ $$$${v}_{{Cy}} =−\frac{\omega{L}\:\mathrm{cos}\:\phi}{\mathrm{2}} \\ $$$${v}_{{C}} ^{\mathrm{2}} =\omega^{\mathrm{2}} {L}^{\mathrm{2}} \left[\left(\frac{\mathrm{cos}\:\left(\theta−\phi\right)}{\mathrm{sin}\:\theta}−\frac{\mathrm{sin}\:\phi}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{cos}\:\phi}{\mathrm{2}}\right)^{\mathrm{2}} \right] \\ $$$${v}_{{C}} ^{\mathrm{2}} =\omega^{\mathrm{2}} {L}^{\mathrm{2}} \left[\frac{\mathrm{cos}^{\mathrm{2}} \:\left(\theta−\phi\right)}{\mathrm{sin}^{\mathrm{2}} \:\theta}−\frac{\mathrm{cos}\:\left(\theta−\phi\right)\:\mathrm{sin}\:\phi}{\mathrm{sin}\:\theta}+\frac{\mathrm{1}}{\mathrm{4}}\right] \\ $$$${v}_{{C}} ^{\mathrm{2}} =\omega^{\mathrm{2}} {L}^{\mathrm{2}} \left[\frac{\mathrm{cos}\:\left(\theta−\phi\right)\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi}{\mathrm{sin}^{\mathrm{2}} \:\theta}+\frac{\mathrm{1}}{\mathrm{4}}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{Mv}_{{C}} ^{\mathrm{2}} ={Mg}\frac{{L}}{\mathrm{2}}\left(\mathrm{sin}\:\theta−\mathrm{sin}\:\phi\right) \\ $$$$\frac{{ML}^{\mathrm{2}} }{\mathrm{12}}\omega^{\mathrm{2}} +{M}\omega^{\mathrm{2}} {L}^{\mathrm{2}} \left[\frac{\mathrm{cos}\:\left(\theta−\phi\right)\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi}{\mathrm{sin}^{\mathrm{2}} \:\theta}+\frac{\mathrm{1}}{\mathrm{4}}\right]={MgL}\left(\mathrm{sin}\:\theta−\mathrm{sin}\:\phi\right) \\ $$$$\omega^{\mathrm{2}} {L}\left[\frac{\mathrm{3}\:\mathrm{cos}\:\left(\theta−\phi\right)\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi}{\mathrm{sin}^{\mathrm{2}} \:\theta}+\mathrm{1}\right]=\mathrm{3}{g}\left(\mathrm{sin}\:\theta−\mathrm{sin}\:\phi\right) \\ $$$$\Rightarrow\omega=\sqrt{\frac{\mathrm{3}{g}\left(\mathrm{sin}\:\theta−\mathrm{sin}\:\phi\right)}{{L}\left[\mathrm{1}+\frac{\mathrm{3cos}\:\left(\theta−\phi\right)\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi}{\mathrm{sin}^{\mathrm{2}} \:\theta}\right]}} \\ $$
Commented by mr W last updated on 29/Oct/25
![the rod rotates about point O. AOBD are cyclic, say with radius R. OD=2R=(L/(sin θ)) p=2R cos (θ−φ)=((L cos (θ−φ))/(sin θ)) CD^2 =d^2 =((L/2))^2 +p^2 −Lp sin φ d^2 =(L^2 /4)+((L^2 cos^2 (θ−φ))/(sin^2 θ))−((L^2 cos (θ−φ) sin φ)/(sin θ)) d^2 =(L^2 /4)+((L^2 cos (θ−φ) cos θ cos φ)/(sin^2 θ)) I_O =((ML^2 )/(12))+Md^2 I_O =((ML^2 )/(12))+((ML^2 )/4)+((ML^2 cos (θ−φ) cos θ cos φ)/(sin^2 θ))] I_O =((ML^2 )/3)[1+((3 cos (θ−φ) cos θ cos φ)/(sin^2 θ))] (1/2)I_O ω^2 =Mg(L/2)(sin θ−sin φ) ((ML^2 )/3)[1+((3 cos (θ−φ) cos θ cos φ)/(sin^2 θ))]ω^2 =MgL(sin θ−sin φ) ⇒ω=(√((3g(sin θ−sin φ))/(L[1+((3 cos (θ−φ) cos θ cos φ)/(sin^2 θ))]))) at φ=0: ω=(√((3g sin θ)/(L(1+(3/(tan^2 θ)))))) ✓ (dφ/dt)=−ω=−(√((3g(sin θ−sin φ))/(L[1+((3 cos (θ−φ) cos θ cos φ)/(sin^2 θ))]))) dt=−(1/(sin θ))(√(L/g))(√((sin^2 θ+3 cos (θ−φ) cos θ cos φ)/(3(sin θ−sin φ))))dφ ⇒T=(1/(sin θ))(√(L/g))∫_0 ^θ (√((sin^2 θ+3 cos (θ−φ) cos θ cos φ)/(3(sin θ−sin φ))))dφ example: θ=60° T=2(√(L/g))∫_0 ^(π/3) (√((1+(cos φ+(√3) sin φ)cos φ)/( 6((√3)−2 sin φ))))dφ ≈2.25914944(√(L/g))](https://www.tinkutara.com/question/Q225484.png)
$${the}\:{rod}\:{rotates}\:{about}\:{point}\:{O}. \\ $$$${AOBD}\:{are}\:{cyclic},\:{say}\:{with}\:{radius}\:{R}. \\ $$$${OD}=\mathrm{2}{R}=\frac{{L}}{\mathrm{sin}\:\theta} \\ $$$${p}=\mathrm{2}{R}\:\mathrm{cos}\:\left(\theta−\phi\right)=\frac{{L}\:\mathrm{cos}\:\left(\theta−\phi\right)}{\mathrm{sin}\:\theta} \\ $$$${CD}^{\mathrm{2}} ={d}^{\mathrm{2}} =\left(\frac{{L}}{\mathrm{2}}\right)^{\mathrm{2}} +{p}^{\mathrm{2}} −{Lp}\:\mathrm{sin}\:\phi \\ $$$${d}^{\mathrm{2}} =\frac{{L}^{\mathrm{2}} }{\mathrm{4}}+\frac{{L}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\left(\theta−\phi\right)}{\mathrm{sin}^{\mathrm{2}} \:\theta}−\frac{{L}^{\mathrm{2}} \:\mathrm{cos}\:\left(\theta−\phi\right)\:\mathrm{sin}\:\phi}{\mathrm{sin}\:\theta} \\ $$$${d}^{\mathrm{2}} =\frac{{L}^{\mathrm{2}} }{\mathrm{4}}+\frac{{L}^{\mathrm{2}} \:\mathrm{cos}\:\left(\theta−\phi\right)\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi}{\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$${I}_{{O}} =\frac{{ML}^{\mathrm{2}} }{\mathrm{12}}+{Md}^{\mathrm{2}} \\ $$$$\left.{I}_{{O}} =\frac{{ML}^{\mathrm{2}} }{\mathrm{12}}+\frac{{ML}^{\mathrm{2}} }{\mathrm{4}}+\frac{{ML}^{\mathrm{2}} \:\mathrm{cos}\:\left(\theta−\phi\right)\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi}{\mathrm{sin}^{\mathrm{2}} \:\theta}\right] \\ $$$${I}_{{O}} =\frac{{ML}^{\mathrm{2}} }{\mathrm{3}}\left[\mathrm{1}+\frac{\mathrm{3}\:\mathrm{cos}\:\left(\theta−\phi\right)\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi}{\mathrm{sin}^{\mathrm{2}} \:\theta}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}_{{O}} \omega^{\mathrm{2}} ={Mg}\frac{{L}}{\mathrm{2}}\left(\mathrm{sin}\:\theta−\mathrm{sin}\:\phi\right) \\ $$$$\frac{{ML}^{\mathrm{2}} }{\mathrm{3}}\left[\mathrm{1}+\frac{\mathrm{3}\:\mathrm{cos}\:\left(\theta−\phi\right)\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi}{\mathrm{sin}^{\mathrm{2}} \:\theta}\right]\omega^{\mathrm{2}} ={MgL}\left(\mathrm{sin}\:\theta−\mathrm{sin}\:\phi\right) \\ $$$$\Rightarrow\omega=\sqrt{\frac{\mathrm{3}{g}\left(\mathrm{sin}\:\theta−\mathrm{sin}\:\phi\right)}{{L}\left[\mathrm{1}+\frac{\mathrm{3}\:\mathrm{cos}\:\left(\theta−\phi\right)\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi}{\mathrm{sin}^{\mathrm{2}} \:\theta}\right]}} \\ $$$${at}\:\phi=\mathrm{0}: \\ $$$$\omega=\sqrt{\frac{\mathrm{3}{g}\:\mathrm{sin}\:\theta}{{L}\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{tan}^{\mathrm{2}} \:\theta}\right)}}\:\:\:\checkmark \\ $$$$\frac{{d}\phi}{{dt}}=−\omega=−\sqrt{\frac{\mathrm{3}{g}\left(\mathrm{sin}\:\theta−\mathrm{sin}\:\phi\right)}{{L}\left[\mathrm{1}+\frac{\mathrm{3}\:\mathrm{cos}\:\left(\theta−\phi\right)\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi}{\mathrm{sin}^{\mathrm{2}} \:\theta}\right]}} \\ $$$${dt}=−\frac{\mathrm{1}}{\mathrm{sin}\:\theta}\sqrt{\frac{{L}}{{g}}}\sqrt{\frac{\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{3}\:\mathrm{cos}\:\left(\theta−\phi\right)\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi}{\mathrm{3}\left(\mathrm{sin}\:\theta−\mathrm{sin}\:\phi\right)}}{d}\phi \\ $$$$\Rightarrow{T}=\frac{\mathrm{1}}{\mathrm{sin}\:\theta}\sqrt{\frac{{L}}{{g}}}\int_{\mathrm{0}} ^{\theta} \sqrt{\frac{\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{3}\:\mathrm{cos}\:\left(\theta−\phi\right)\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi}{\mathrm{3}\left(\mathrm{sin}\:\theta−\mathrm{sin}\:\phi\right)}}{d}\phi \\ $$$${example}: \\ $$$$\theta=\mathrm{60}° \\ $$$${T}=\mathrm{2}\sqrt{\frac{{L}}{{g}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \sqrt{\frac{\mathrm{1}+\left(\mathrm{cos}\:\phi+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\phi\right)\mathrm{cos}\:\phi}{\:\mathrm{6}\left(\sqrt{\mathrm{3}}−\mathrm{2}\:\mathrm{sin}\:\phi\right)}}{d}\phi \\ $$$$\:\:\approx\mathrm{2}.\mathrm{25914944}\sqrt{\frac{{L}}{{g}}} \\ $$