Question Number 70831 by ajfour last updated on 08/Oct/19
Commented by ajfour last updated on 08/Oct/19
$${Stick}\:{of}\:{mass}\:{m},\:{length}\:{b}\:{is} \\ $$$${released}\:{as}\:{shown}.\:{As}\:{it}\:{slides} \\ $$$${down}\:{and}\:{breaks}\:{contact}\:{with} \\ $$$${hemisphere},\:{the}\:{top}\:{end}\:{is}\:{at} \\ $$$${angular}\:{position}\:\beta.\:{Assume} \\ $$$${no}\:{friction},\:{find}\:\beta. \\ $$
Answered by ajfour last updated on 08/Oct/19
Answered by mr W last updated on 08/Oct/19
Commented by ajfour last updated on 10/Oct/19
$${so}\:{far},\:{so}\:{good},\:{how}\:{do}\:{we} \\ $$$${utilise}\:\omega\:{in}\:{determining}\:\beta,\:{Sir}? \\ $$$$\left({you}\:{like}\:{the}\:{question},\:{sir}?\right) \\ $$
Commented by mr W last updated on 09/Oct/19
![OA=R[cos β+(√((b^2 /R^2 )−sin^2 β))] AC=OA tan β=R[sin β+tan β(√((b^2 /R^2 )−sin^2 β))] BC=((OA)/(cos β))−R=(R/(cos β))(√((b^2 /R^2 )−sin^2 β)) CD^2 =((AC^2 +BC^2 )/2)−(b^2 /4) CD^2 =(R^2 /2){(1/(cos^2 β))((b^2 /R^2 )−sin^2 β)+sin^2 β[1+(1/(cos β))(√((b^2 /R^2 )−sin^2 β))]^2 }−(b^2 /4) I_C =I_0 +mCD^2 I_C =((mb^2 )/(12))+((mR^2 )/2){(1/(cos^2 β))((b^2 /R^2 )−sin^2 β)+sin^2 β[1+(1/(cos β))(√((b^2 /R^2 )−sin^2 β))]^2 }−((mb^2 )/4) I_C =((mR^2 )/2){(1/(cos^2 β))((b^2 /R^2 )−sin^2 β)+sin^2 β[1+(1/(cos β))(√((b^2 /R^2 )−sin^2 β))]^2 }−((mb^2 )/6) Δh=(R/2)((b/R) cos α−sin β) (1/2)I_C ω^2 =mgΔh ((R/2){(1/(cos^2 β))((b^2 /R^2 )−sin^2 β)+sin^2 β[1+(1/(cos β))(√((b^2 /R^2 )−sin^2 β))]^2 }−(b^2 /(6R^2 )))ω^2 =g((b/R) cos α−sin β) let λ=(b/R) ((1/2){(1/(cos^2 β))(λ^2 −sin^2 β)+sin^2 β[1+(1/(cos β))(√(λ^2 −sin^2 β))]^2 }−(λ^2 /6))ω^2 =(g/R)(λ cos α−sin β) ⇒ω=(√(g/R))(√((6(λ cos α−sin β))/(3{(1/(cos^2 β))(λ^2 −sin^2 β)+sin^2 β[1+(1/(cos β))(√(λ^2 −sin^2 β))]^2 }−λ^2 ))) ....](https://www.tinkutara.com/question/Q70928.png)
$${OA}={R}\left[\mathrm{cos}\:\beta+\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta}\right] \\ $$$${AC}={OA}\:\mathrm{tan}\:\beta={R}\left[\mathrm{sin}\:\beta+\mathrm{tan}\:\beta\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta}\right] \\ $$$${BC}=\frac{{OA}}{\mathrm{cos}\:\beta}−{R}=\frac{{R}}{\mathrm{cos}\:\beta}\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta} \\ $$$${CD}^{\mathrm{2}} =\frac{{AC}^{\mathrm{2}} +{BC}^{\mathrm{2}} }{\mathrm{2}}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${CD}^{\mathrm{2}} =\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\beta}\left(\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta\right)+\mathrm{sin}^{\mathrm{2}} \:\beta\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\:\beta}\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta}\right]^{\mathrm{2}} \right\}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${I}_{{C}} ={I}_{\mathrm{0}} +{mCD}^{\mathrm{2}} \\ $$$${I}_{{C}} =\frac{{mb}^{\mathrm{2}} }{\mathrm{12}}+\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\beta}\left(\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta\right)+\mathrm{sin}^{\mathrm{2}} \:\beta\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\:\beta}\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta}\right]^{\mathrm{2}} \right\}−\frac{{mb}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${I}_{{C}} =\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\beta}\left(\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta\right)+\mathrm{sin}^{\mathrm{2}} \:\beta\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\:\beta}\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta}\right]^{\mathrm{2}} \right\}−\frac{{mb}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Delta{h}=\frac{{R}}{\mathrm{2}}\left(\frac{{b}}{{R}}\:\mathrm{cos}\:\alpha−\mathrm{sin}\:\beta\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}_{{C}} \omega^{\mathrm{2}} ={mg}\Delta{h} \\ $$$$\left(\frac{{R}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\beta}\left(\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta\right)+\mathrm{sin}^{\mathrm{2}} \:\beta\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\:\beta}\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta}\right]^{\mathrm{2}} \right\}−\frac{{b}^{\mathrm{2}} }{\mathrm{6}{R}^{\mathrm{2}} }\right)\omega^{\mathrm{2}} ={g}\left(\frac{{b}}{{R}}\:\mathrm{cos}\:\alpha−\mathrm{sin}\:\beta\right) \\ $$$${let}\:\lambda=\frac{{b}}{{R}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\beta}\left(\lambda^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\beta\right)+\mathrm{sin}^{\mathrm{2}} \:\beta\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\:\beta}\sqrt{\lambda^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\beta}\right]^{\mathrm{2}} \right\}−\frac{\lambda^{\mathrm{2}} }{\mathrm{6}}\right)\omega^{\mathrm{2}} =\frac{{g}}{{R}}\left(\lambda\:\mathrm{cos}\:\alpha−\mathrm{sin}\:\beta\right) \\ $$$$\Rightarrow\omega=\sqrt{\frac{{g}}{{R}}}\sqrt{\frac{\mathrm{6}\left(\lambda\:\mathrm{cos}\:\alpha−\mathrm{sin}\:\beta\right)}{\mathrm{3}\left\{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\beta}\left(\lambda^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\beta\right)+\mathrm{sin}^{\mathrm{2}} \:\beta\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\:\beta}\sqrt{\lambda^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\beta}\right]^{\mathrm{2}} \right\}−\lambda^{\mathrm{2}} }} \\ $$$$…. \\ $$
Commented by mr W last updated on 10/Oct/19
$${nice}\:{question}\:{sir}!\:{but}\:{not}\:{easy}. \\ $$