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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 β. Assume  no friction, find β.

$${Stick}\:{of}\:{mass}\:{m},\:{length}\:{b}\:{is} \\ $$$${released}\:{as}\:{shown}.\:{As}\:{it}\:{slides} \\ $$$${down}\:{and}\:{breaks}\:{contact}\:{with} \\ $$$${hemisphere},\:{the}\:{top}\:{end}\:{is}\:{at} \\ $$$${angular}\:{position}\:\beta.\:{Assume} \\ $$$${no}\:{friction},\:{find}\:\beta. \\ $$

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 ω in determining β, Sir?  (you like the question, sir?)

$${so}\:{far},\:{so}\:{good},\:{how}\:{do}\:{we} \\ $$$${utilise}\:\omega\:{in}\:{determining}\:\beta,\:{Sir}? \\ $$$$\left({you}\:{like}\:{the}\:{question},\:{sir}?\right) \\ $$

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 )))  ....

$${OA}={R}\left[\mathrm{cos}\:\beta+\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta}\right] \\ $$$${AC}={OA}\:\mathrm{tan}\:\beta={R}\left[\mathrm{sin}\:\beta+\mathrm{tan}\:\beta\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta}\right] \\ $$$${BC}=\frac{{OA}}{\mathrm{cos}\:\beta}−{R}=\frac{{R}}{\mathrm{cos}\:\beta}\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta} \\ $$$${CD}^{\mathrm{2}} =\frac{{AC}^{\mathrm{2}} +{BC}^{\mathrm{2}} }{\mathrm{2}}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${CD}^{\mathrm{2}} =\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\beta}\left(\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta\right)+\mathrm{sin}^{\mathrm{2}} \:\beta\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\:\beta}\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta}\right]^{\mathrm{2}} \right\}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${I}_{{C}} ={I}_{\mathrm{0}} +{mCD}^{\mathrm{2}} \\ $$$${I}_{{C}} =\frac{{mb}^{\mathrm{2}} }{\mathrm{12}}+\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\beta}\left(\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta\right)+\mathrm{sin}^{\mathrm{2}} \:\beta\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\:\beta}\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta}\right]^{\mathrm{2}} \right\}−\frac{{mb}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${I}_{{C}} =\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\beta}\left(\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta\right)+\mathrm{sin}^{\mathrm{2}} \:\beta\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\:\beta}\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta}\right]^{\mathrm{2}} \right\}−\frac{{mb}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Delta{h}=\frac{{R}}{\mathrm{2}}\left(\frac{{b}}{{R}}\:\mathrm{cos}\:\alpha−\mathrm{sin}\:\beta\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}_{{C}} \omega^{\mathrm{2}} ={mg}\Delta{h} \\ $$$$\left(\frac{{R}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\beta}\left(\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta\right)+\mathrm{sin}^{\mathrm{2}} \:\beta\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\:\beta}\sqrt{\frac{{b}^{\mathrm{2}} }{{R}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\beta}\right]^{\mathrm{2}} \right\}−\frac{{b}^{\mathrm{2}} }{\mathrm{6}{R}^{\mathrm{2}} }\right)\omega^{\mathrm{2}} ={g}\left(\frac{{b}}{{R}}\:\mathrm{cos}\:\alpha−\mathrm{sin}\:\beta\right) \\ $$$${let}\:\lambda=\frac{{b}}{{R}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\beta}\left(\lambda^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\beta\right)+\mathrm{sin}^{\mathrm{2}} \:\beta\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\:\beta}\sqrt{\lambda^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\beta}\right]^{\mathrm{2}} \right\}−\frac{\lambda^{\mathrm{2}} }{\mathrm{6}}\right)\omega^{\mathrm{2}} =\frac{{g}}{{R}}\left(\lambda\:\mathrm{cos}\:\alpha−\mathrm{sin}\:\beta\right) \\ $$$$\Rightarrow\omega=\sqrt{\frac{{g}}{{R}}}\sqrt{\frac{\mathrm{6}\left(\lambda\:\mathrm{cos}\:\alpha−\mathrm{sin}\:\beta\right)}{\mathrm{3}\left\{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\beta}\left(\lambda^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\beta\right)+\mathrm{sin}^{\mathrm{2}} \:\beta\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\:\beta}\sqrt{\lambda^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\beta}\right]^{\mathrm{2}} \right\}−\lambda^{\mathrm{2}} }} \\ $$$$.... \\ $$

Commented by mr W last updated on 10/Oct/19

nice question sir! but not easy.

$${nice}\:{question}\:{sir}!\:{but}\:{not}\:{easy}. \\ $$

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