Question-108133

Question Number 108133 by ajfour last updated on 14/Aug/20

Commented by ajfour last updated on 14/Aug/20

$${Rod}\:{AB}\:{inside}\:{a}\:{smooth}\:{sphere} \\ $$$${of}\:{radius}\:{R},\:{Find}\:{Normal} \\ $$$${reactions}\:{at}\:{A}\:{and}\:{B}. \\ $$

Answered by mr W last updated on 15/Aug/20

Commented by mr W last updated on 15/Aug/20

let λ=(m/M) AC=((M×((a+b)/2)+m×a)/(M+m))=(((1+2λ)a+b)/(2(1+λ))) CB=((a+(1+2λ)b)/(2(1+λ))) cos γ=((a+b)/(2R)) ⇒γ=cos^(−1) {((a+b)/(2R))} ((sin (α+γ))/(sin α))=((AO)/(AC))=((2(1+λ)R)/((1+2λ)a+b)) cos γ+((sin γ)/(tan α))=((2(1+λ)R)/((1+2λ)a+b)) ⇒α=tan^(−1) {((√(1−(((a+b)/(2R)))^2 ))/(((2(1+λ)R)/((1+2λ)a+b))−((a+b)/(2R))))} ⇒β=tan^(−1) {((√(1−(((a+b)/(2R)))^2 ))/(((2(1+λ)R)/(a+(1+2λ)b))−((a+b)/(2R))))} (N_1 /(sin β))=(((M+m)g)/(sin 2γ)) ⇒N_1 =((sin β (M+m)g)/(sin 2γ)) ⇒N_2 =((sin α (M+m)g)/(sin 2γ))

$${let}\:\lambda=\frac{{m}}{{M}} \\ $$$${AC}=\frac{{M}×\frac{{a}+{b}}{\mathrm{2}}+{m}×{a}}{{M}+{m}}=\frac{\left(\mathrm{1}+\mathrm{2}\lambda\right){a}+{b}}{\mathrm{2}\left(\mathrm{1}+\lambda\right)} \\ $$$${CB}=\frac{{a}+\left(\mathrm{1}+\mathrm{2}\lambda\right){b}}{\mathrm{2}\left(\mathrm{1}+\lambda\right)} \\ $$$$\mathrm{cos}\:\gamma=\frac{{a}+{b}}{\mathrm{2}{R}} \\ $$$$\Rightarrow\gamma=\mathrm{cos}^{−\mathrm{1}} \left\{\frac{{a}+{b}}{\mathrm{2}{R}}\right\} \\ $$$$\frac{\mathrm{sin}\:\left(\alpha+\gamma\right)}{\mathrm{sin}\:\alpha}=\frac{{AO}}{{AC}}=\frac{\mathrm{2}\left(\mathrm{1}+\lambda\right){R}}{\left(\mathrm{1}+\mathrm{2}\lambda\right){a}+{b}} \\ $$$$\mathrm{cos}\:\gamma+\frac{\mathrm{sin}\:\gamma}{\mathrm{tan}\:\alpha}=\frac{\mathrm{2}\left(\mathrm{1}+\lambda\right){R}}{\left(\mathrm{1}+\mathrm{2}\lambda\right){a}+{b}} \\ $$$$\Rightarrow\alpha=\mathrm{tan}^{−\mathrm{1}} \left\{\frac{\sqrt{\mathrm{1}−\left(\frac{{a}+{b}}{\mathrm{2}{R}}\right)^{\mathrm{2}} }}{\frac{\mathrm{2}\left(\mathrm{1}+\lambda\right){R}}{\left(\mathrm{1}+\mathrm{2}\lambda\right){a}+{b}}−\frac{{a}+{b}}{\mathrm{2}{R}}}\right\} \\ $$$$\Rightarrow\beta=\mathrm{tan}^{−\mathrm{1}} \left\{\frac{\sqrt{\mathrm{1}−\left(\frac{{a}+{b}}{\mathrm{2}{R}}\right)^{\mathrm{2}} }}{\frac{\mathrm{2}\left(\mathrm{1}+\lambda\right){R}}{{a}+\left(\mathrm{1}+\mathrm{2}\lambda\right){b}}−\frac{{a}+{b}}{\mathrm{2}{R}}}\right\} \\ $$$$\frac{{N}_{\mathrm{1}} }{\mathrm{sin}\:\beta}=\frac{\left({M}+{m}\right){g}}{\mathrm{sin}\:\mathrm{2}\gamma} \\ $$$$\Rightarrow{N}_{\mathrm{1}} =\frac{\mathrm{sin}\:\beta\:\left({M}+{m}\right){g}}{\mathrm{sin}\:\mathrm{2}\gamma} \\ $$$$\Rightarrow{N}_{\mathrm{2}} =\frac{\mathrm{sin}\:\alpha\:\left({M}+{m}\right){g}}{\mathrm{sin}\:\mathrm{2}\gamma} \\ $$

Commented by ajfour last updated on 15/Aug/20

$${So}\:{quick},\:{and}\:{superb}\:{solution}, \\ $$$${thanks}\:{a}\:{lot}\:{Sir}. \\ $$

Commented by otchereabdullai@gmail.com last updated on 15/Aug/20

The world mathematician/ physicist I always become very happy when i see your solution to a question

$$\mathrm{The}\:\mathrm{world}\:\mathrm{mathematician}/\:\mathrm{physicist} \\ $$$$\mathrm{I}\:\mathrm{always}\:\mathrm{become}\:\mathrm{very}\:\mathrm{happy}\:\mathrm{when}\:\mathrm{i}\:\mathrm{see} \\ $$$$\mathrm{your}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{a}\:\mathrm{question} \\ $$

Answered by ajfour last updated on 15/Aug/20

Commented by ajfour last updated on 15/Aug/20

Ncos (α+θ)=Fcos (α−θ) ...(i) p=(√(R^2 −(((a+b)/2))^2 )) cos α=((a+b)/(2R)) Nsin (α+θ)+Fsin (α−θ)=(M+m)g ......(ii) Torque balance about centre of sphere mg(((b−a)/2)−ptan θ)cos θ=Mg(psin θ) ⇒ tan θ=((m(b−a))/(2p(M+m))) Now using (i) in (ii) N{sin (α+θ)+sin (α−θ)((cos (α+θ))/(cos (α−θ)))} = (M+m)g ⇒ N=(((M+m)gcos (α−θ))/(sin 2α)) & F=(((M+m)gcos (α+θ))/(sin 2α)) ■

$${N}\mathrm{cos}\:\left(\alpha+\theta\right)={F}\mathrm{cos}\:\left(\alpha−\theta\right)\:\:\:\:…\left({i}\right) \\ $$$${p}=\sqrt{{R}^{\mathrm{2}} −\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{cos}\:\alpha=\frac{{a}+{b}}{\mathrm{2}{R}} \\ $$$${N}\mathrm{sin}\:\left(\alpha+\theta\right)+{F}\mathrm{sin}\:\left(\alpha−\theta\right)=\left({M}+{m}\right){g} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:……\left({ii}\right) \\ $$$${Torque}\:{balance}\:{about}\:{centre}\:{of}\:{sphere} \\ $$$${mg}\left(\frac{{b}−{a}}{\mathrm{2}}−{p}\mathrm{tan}\:\theta\right)\mathrm{cos}\:\theta={Mg}\left({p}\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\:\:\:\mathrm{tan}\:\theta=\frac{{m}\left({b}−{a}\right)}{\mathrm{2}{p}\left({M}+{m}\right)} \\ $$$${Now}\:\:{using}\:\left({i}\right)\:{in}\:\:\left({ii}\right) \\ $$$$\:\:\:{N}\left\{\mathrm{sin}\:\left(\alpha+\theta\right)+\mathrm{sin}\:\left(\alpha−\theta\right)\frac{\mathrm{cos}\:\left(\alpha+\theta\right)}{\mathrm{cos}\:\left(\alpha−\theta\right)}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({M}+{m}\right){g} \\ $$$$\Rightarrow\:\:{N}=\frac{\left({M}+{m}\right){g}\mathrm{cos}\:\left(\alpha−\theta\right)}{\mathrm{sin}\:\mathrm{2}\alpha}\:\:\:\:\& \\ $$$$\:\:\:\:\:\:\:\:{F}=\frac{\left({M}+{m}\right){g}\mathrm{cos}\:\left(\alpha+\theta\right)}{\mathrm{sin}\:\mathrm{2}\alpha}\:\:\:\blacksquare \\ $$

Commented by mr W last updated on 15/Aug/20

$${perfect}\:{solution}! \\ $$

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