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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?
$${Great}\:{sir}\:!\:{But}\:{i}\:{need}\:{a}\:{suggestion}.\:{How}\:{do}\:{you}\:{think}\:{this}\:{kind} \\ $$$${solutions}\:\:{easily}? \\ $$
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.
$${it}'{s}\:{not}\:{easy}\:{to}\:{give}\:{a}\:{general}\: \\ $$$${sugestion}.\:{actually}\:{this}\:{is}\:{not}\:{an} \\ $$$${easy}\:{question},\:{because}\:{one}\:{needs}\:{to} \\ $$$${apply}\:{alot}\:{of}\:{things}\:{both}\:{in}\:{physics} \\ $$$${and}\:{in}\:{calculus}. \\ $$
Commented by peter frank last updated on 24/Apr/21
GOD  has bless him with  great  ability maths  and physics.
$${GOD}\:\:{has}\:{bless}\:{him}\:{with} \\ $$$${great}\:\:{ability}\:{maths}\:\:{and}\:{physics}. \\ $$
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!
$${thanks}\:{to}\:{you}\:{both}\:{for}\:{your}\:{heart}− \\ $$$${warming}\:{words}! \\ $$

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