Question-139167

Question Number 139167 by Dwaipayan Shikari last updated on 23/Apr/21

Answered by mr W last updated on 23/Apr/21

Commented by mr W last updated on 23/Apr/21

y=(x^2 /(2a)) tan θ=y′=(x/a)=λ, say y′′=(1/a) r=(((1+y′^2 )^(3/2) )/(y′′))=a(1+(x^2 /a^2 ))^(3/2) =a(1+λ^2 )^(3/2) u=velocity at (x,y) (1/2)mv^2 =(1/2)mu^2 +mgy ⇒u^2 =v^2 −((gx^2 )/a)=v^2 −agλ^2 at highest position u=0: v^2 −agλ_(max) ^2 =0 ⇒λ_(max) =(v/( (√(ag)))) R−mgcos θ=((mu^2 )/r)=((m(v^2 −agλ^2 ))/(a(1+λ^2 )^(3/2) )) R=((mg)/( (√(1+t^2 ))))+((mg((v^2 /(ag))−λ^2 ))/((1+λ^2 )^(3/2) ))=((mg(1+(v^2 /(ag))))/( (1+λ^2 )^(3/2) )) such that the block doesn′t slip, R sin θ≤μ(Mg+R cos θ) R (sin θ−μ cos θ)≤μMg ((m(1+(v^2 /(ag)))(λ−μ))/( (1+λ^2 )^2 ))≤μM ((μM)/(m(1+(v^2 /(ag)))))≥((λ−μ)/( (1+λ^2 )^2 ))=f(λ) say for maximum f(λ): ((df(λ))/dλ)=(1/((1+λ^2 )^2 ))−((4(λ−μ)λ)/((1+λ^2 )^3 ))=0 ((4(λ−μ)λ)/(1+λ^2 ))=1 3λ^2 −4μλ−1=0 ⇒λ=((2μ+(√(4μ^2 +3)))/3) with μ=(1/2) ⇒λ=((2×(1/2)+(√(4×(1/4)+3)))/3)=1 f(λ)_(max) =((1−(1/2))/((1+1)^2 ))=(1/8) ((μM)/(m(1+(v^2 /(ag)))))≥(1/8) (((4M−m)/m))ag≥(v^2 /(ag)) v≤(√(((4M−m)/m)ag)) λ_(max) =(v/( (√(ag))))=(√(4×(M/m)−1))>(√(4×(1/2)−1))=1 that means λ=1 is reached before the highest position is reached. i.e. v_(max) =(√(((4M−m)/m)ag)) is valid. ⇒α=4, β=1, γ=1 ⇒α+β+γ=6

$${y}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}{a}} \\ $$$$\mathrm{tan}\:\theta={y}'=\frac{{x}}{{a}}=\lambda,\:{say} \\ $$$${y}''=\frac{\mathrm{1}}{{a}} \\ $$$${r}=\frac{\left(\mathrm{1}+{y}'^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }{{y}''}={a}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)^{\mathrm{3}/\mathrm{2}} ={a}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${u}={velocity}\:{at}\:\left({x},{y}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mu}^{\mathrm{2}} +{mgy} \\ $$$$\Rightarrow{u}^{\mathrm{2}} ={v}^{\mathrm{2}} −\frac{{gx}^{\mathrm{2}} }{{a}}={v}^{\mathrm{2}} −{ag}\lambda^{\mathrm{2}} \\ $$$${at}\:{highest}\:{position}\:{u}=\mathrm{0}: \\ $$$${v}^{\mathrm{2}} −{ag}\lambda_{{max}} ^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\lambda_{{max}} =\frac{{v}}{\:\sqrt{{ag}}} \\ $$$${R}−{mg}\mathrm{cos}\:\theta=\frac{{mu}^{\mathrm{2}} }{{r}}=\frac{{m}\left({v}^{\mathrm{2}} −{ag}\lambda^{\mathrm{2}} \right)}{{a}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$${R}=\frac{{mg}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}+\frac{{mg}\left(\frac{{v}^{\mathrm{2}} }{{ag}}−\lambda^{\mathrm{2}} \right)}{\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\frac{{mg}\left(\mathrm{1}+\frac{{v}^{\mathrm{2}} }{{ag}}\right)}{\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$${such}\:{that}\:{the}\:{block}\:{doesn}'{t}\:{slip}, \\ $$$${R}\:\mathrm{sin}\:\theta\leqslant\mu\left({Mg}+{R}\:\mathrm{cos}\:\theta\right) \\ $$$${R}\:\left(\mathrm{sin}\:\theta−\mu\:\mathrm{cos}\:\theta\right)\leqslant\mu{Mg} \\ $$$$\frac{{m}\left(\mathrm{1}+\frac{{v}^{\mathrm{2}} }{{ag}}\right)\left(\lambda−\mu\right)}{\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\mathrm{2}} }\leqslant\mu{M} \\ $$$$\frac{\mu{M}}{{m}\left(\mathrm{1}+\frac{{v}^{\mathrm{2}} }{{ag}}\right)}\geqslant\frac{\lambda−\mu}{\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\mathrm{2}} }={f}\left(\lambda\right)\:{say} \\ $$$${for}\:{maximum}\:{f}\left(\lambda\right): \\ $$$$\frac{{df}\left(\lambda\right)}{{d}\lambda}=\frac{\mathrm{1}}{\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{\mathrm{4}\left(\lambda−\mu\right)\lambda}{\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\mathrm{3}} }=\mathrm{0} \\ $$$$\frac{\mathrm{4}\left(\lambda−\mu\right)\lambda}{\mathrm{1}+\lambda^{\mathrm{2}} }=\mathrm{1} \\ $$$$\mathrm{3}\lambda^{\mathrm{2}} −\mathrm{4}\mu\lambda−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{2}\mu+\sqrt{\mathrm{4}\mu^{\mathrm{2}} +\mathrm{3}}}{\mathrm{3}} \\ $$$${with}\:\mu=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{4}×\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{3}}}{\mathrm{3}}=\mathrm{1} \\ $$$${f}\left(\lambda\right)_{{max}} =\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}{\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\frac{\mu{M}}{{m}\left(\mathrm{1}+\frac{{v}^{\mathrm{2}} }{{ag}}\right)}\geqslant\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\left(\frac{\mathrm{4}{M}−{m}}{{m}}\right){ag}\geqslant\frac{{v}^{\mathrm{2}} }{{ag}} \\ $$$${v}\leqslant\sqrt{\frac{\mathrm{4}{M}−{m}}{{m}}{ag}} \\ $$$$\lambda_{{max}} =\frac{{v}}{\:\sqrt{{ag}}}=\sqrt{\mathrm{4}×\frac{{M}}{{m}}−\mathrm{1}}>\sqrt{\mathrm{4}×\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}}=\mathrm{1} \\ $$$${that}\:{means}\:\lambda=\mathrm{1}\:{is}\:{reached}\:{before}\:{the} \\ $$$${highest}\:{position}\:{is}\:{reached}. \\ $$$${i}.{e}.\:{v}_{{max}} =\sqrt{\frac{\mathrm{4}{M}−{m}}{{m}}{ag}}\:{is}\:{valid}. \\ $$$$\Rightarrow\alpha=\mathrm{4},\:\beta=\mathrm{1},\:\gamma=\mathrm{1} \\ $$$$\Rightarrow\alpha+\beta+\gamma=\mathrm{6} \\ $$

Commented by Dwaipayan Shikari last updated on 23/Apr/21

$${Great}\:{sir}\:!\:{But}\:{i}\:{need}\:{a}\:{suggestion}.\:{How}\:{do}\:{you}\:{think}\:{this}\:{kind} \\ $$$${solutions}\:\:{easily}? \\ $$

Commented by mr W last updated on 24/Apr/21

$${it}'{s}\:{not}\:{easy}\:{to}\:{give}\:{a}\:{general}\: \\ $$$${sugestion}.\:{actually}\:{this}\:{is}\:{not}\:{an} \\ $$$${easy}\:{question},\:{because}\:{one}\:{needs}\:{to} \\ $$$${apply}\:{alot}\:{of}\:{things}\:{both}\:{in}\:{physics} \\ $$$${and}\:{in}\:{calculus}. \\ $$

Commented by peter frank last updated on 24/Apr/21

$${GOD}\:\:{has}\:{bless}\:{him}\:{with} \\ $$$${great}\:\:{ability}\:{maths}\:\:{and}\:{physics}. \\ $$

Commented by otchereabdullai@gmail.com last updated on 24/Apr/21

Is very true sir Frank i really love this man. For me he is the world best! Because he looks very different in solving mathematics and physics questions. And he has idea in almost all questions even if the question is wrong he can tell you the quetion is wrong and tell the reason and how the question should look like even if is a very difficult question! I wish he is my father! God bless you prof W ! long life is what we pray for you !

$$\mathrm{Is}\:\mathrm{very}\:\mathrm{true}\:\mathrm{sir}\:\mathrm{Frank}\:\mathrm{i}\:\mathrm{really}\:\mathrm{love}\:\mathrm{this} \\ $$$$\mathrm{man}.\:\mathrm{For}\:\mathrm{me}\:\mathrm{he}\:\mathrm{is}\:\mathrm{the}\:\mathrm{world}\:\mathrm{best}!\: \\ $$$$\mathrm{Because}\:\mathrm{he}\:\mathrm{looks}\:\mathrm{very}\:\mathrm{different}\:\mathrm{in}\: \\ $$$$\mathrm{solving}\:\mathrm{mathematics}\:\mathrm{and}\:\mathrm{physics}\: \\ $$$$\mathrm{questions}.\:\mathrm{And}\:\mathrm{he}\:\mathrm{has}\:\mathrm{idea}\:\mathrm{in}\:\mathrm{almost} \\ $$$$\mathrm{all}\:\mathrm{questions}\:\mathrm{even}\:\mathrm{if}\:\mathrm{the}\:\mathrm{question} \\ $$$$\mathrm{is}\:\mathrm{wrong}\:\mathrm{he}\:\mathrm{can}\:\mathrm{tell}\:\mathrm{you}\:\mathrm{the}\:\mathrm{quetion}\:\mathrm{is} \\ $$$$\mathrm{wrong}\:\mathrm{and}\:\mathrm{tell}\:\mathrm{the}\:\mathrm{reason}\:\mathrm{and}\:\mathrm{how}\:\mathrm{the} \\ $$$$\mathrm{question}\:\mathrm{should}\:\mathrm{look}\:\mathrm{like}\:\mathrm{even}\:\mathrm{if}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{very}\:\mathrm{difficult}\:\mathrm{question}!\:\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{wish}}\:\boldsymbol{\mathrm{he}}\:\boldsymbol{\mathrm{is}} \\ $$$$\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{father}}!\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\boldsymbol{\mathrm{prof}}\:\boldsymbol{\mathrm{W}}\:!\:\boldsymbol{\mathrm{long}} \\ $$$$\boldsymbol{\mathrm{life}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{pray}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{you}}\:! \\ $$

Commented by mr W last updated on 24/Apr/21

$${thanks}\:{to}\:{you}\:{both}\:{for}\:{your}\:{heart}− \\ $$$${warming}\:{words}! \\ $$

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