Question Number 127997 by mr W last updated on 05/Jan/21
Commented by mr W last updated on 05/Jan/21
$${a}\:{paraboloid}\:{bowl}\:{has}\:{a}\:{depth}\:{of}\:{H} \\ $$$${and}\:{an}\:{opening}\:{with}\:{diameter}\:{L}. \\ $$$${a}\:{small}\:{block}\:{of}\:{mass}\:{m}\:{is}\:{released} \\ $$$${from}\:{rest}\:{at}\:{the}\:{rim}\:{of}\:{the}\:{bowl}. \\ $$$${the}\:{friction}\:{coefficient}\:{between}\:{the} \\ $$$${objects}\:{is}\:\mu.\:{assume}\:{the}\:{friction}\:{is} \\ $$$${small}\:{enough}\:{such}\:{that}\:{the}\:{block}\:{can} \\ $$$${pass}\:{through}\:{the}\:{lowest}\:{point}. \\ $$$${find}\:{the}\:{maximum}\:{height}\:{h}\:{the}\:{small} \\ $$$${block}\:{reaches}. \\ $$
Answered by mr W last updated on 04/Jan/21
Commented by mr W last updated on 05/Jan/21
![R=(L/2)=radius of opening let η=(H/R), ξ=(x/R)∈[−1,1] y=H((x/R))^2 =Hξ^2 tan θ=y′=2H(x/R^2 )=2ηξ y′′=2(H/R^2 )=((2η)/R) radius of curvature r=(([1+(y′)^2 ]^(3/2) )/(∣y′′∣)) ⇒r=(((1+4η^2 ξ^2 )^(3/2) R)/(2η)) ds=(√(1+(y′)^2 ))dx=R(√(1+4η^2 ξ^2 ))dξ phase 1: from point A to point O phase 2: from point O to point B phase 1: N=mg cos θ+m(v^2 /r) f=μN=μm(g cos θ+(v^2 /r)) ma=mg sin θ−f a=g sin θ−μ(g cos θ+(v^2 /r)) a=g(sin θ−μ cos θ)−((μv^2 )/r) a=(dv/dt)=(dv/ds)×(ds/dt)=−v(dv/ds) −((vdv)/(R(√(1+4η^2 ξ^2 ))dξ))=((g(2ηξ−μ))/( (√(1+4η^2 ξ^2 ))))−((2μηv^2 )/((1+4η^2 ξ^2 )^(3/2) R)) ((vdv)/( dξ))=−gR(2ηξ−μ)+((2μηv^2 )/(1+4η^2 ξ^2 )) ...(i) ((2ηvdv)/( d(2ηξ)))=−gR(2ηξ−μ)+((2μηv^2 )/(1+4η^2 ξ^2 )) ((vdv)/( du))=−((gR)/(2η))(u−μ)+((μv^2 )/(1+u^2 )) let u=2ηξ, w=v^2 ⇒(dw/( du))−((2μw)/(1+u^2 ))=−((gR)/η)(u−μ) I=−2μ∫(du/(1+u^2 ))=−2μ tan^(−1) u ⇒w=−((gR)/(ηe^(−2μ tan^(−1) u) ))[∫(u−μ)e^(−2μ tan^(−1) u) du−C] ⇒w=−((gR)/(2ηe^(−2μ tan^(−1) u) ))[(1+u^2 )e^(−2μ tan^(−1) u) −C] ⇒v^2 =((gR)/(2η))e^(2μ tan^(−1) (2ηξ)) [C−(1+4η^2 ξ^2 )e^(−2μ tan^(−1) (2ηξ)) ] at ξ=1, v=0: C=(1+4η^2 )e^(−2μ tan^(−1) (2η)) ⇒v^2 =((gR)/(2η))e^(2μ tan^(−1) (2ηξ)) [(1+4η^2 )e^(−2μ tan^(−1) (2η)) −(1+4η^2 ξ^2 )e^(−2μ tan^(−1) (2ηξ)) ] ⇒v^2 =((gR)/(2η)){(1+4η^2 )e^(−2μ[tan^(−1) (2η)−tan^(−1) (2ηξ)]) −(1+4η^2 ξ^2 )} ...(I) at ξ=0, v=v_0 : v_0 ^2 =((gR)/(2η))[(1+4η^2 )e^(−2μ tan^(−1) (2η)) −1] ⇒v_0 =(√(((gR)/(2η))[(1+4η^2 )e^(−2μ tan^(−1) (2η)) −1])) such that the block reaches the lowest point, v_0 ≥0: (1+4η^2 )e^(−2μ tan^(−1) (2η)) −1≥0 e^(2μ tan^(−1) (2η)) ≤1+4η^2 ⇒μ≤((ln (1+4η^2 ))/(2 tan^(−1) (2η)))=μ_(max) phase 2: ma=−mg sin θ−f a=−g sin θ−μ(g cos θ+(v^2 /r)) a=−g(sin θ+μ cos θ)−((μv^2 )/r) a=(dv/dt)=(dv/ds)×(ds/dt)=v(dv/ds) ((vdv)/(R(√(1+4η^2 ξ^2 ))dξ))=−((g(2ηξ+μ))/( (√(1+4η^2 ξ^2 ))))−((2μηv^2 )/((1+4η^2 ξ^2 )^(3/2) R)) ((vdv)/( dξ))=−gR(2ηξ+μ)−((2μηv^2 )/(1+4η^2 ξ^2 )) ...(ii) ⇒(dw/( du))+((2μw)/(1+u^2 ))=−((gR)/η)(u+μ) similarly as with (i), ⇒v^2 =((gR)/(2η))e^(−2μ tan^(−1) (2ηξ)) [C−(1+4η^2 ξ^2 )e^(2μ tan^(−1) (2ηξ)) ] at ξ=0, v=v_0 : v_0 ^2 =((gR)/(2η))(C−1)=((gR)/(2η))[(1+4η^2 )e^(−2μ tan^(−1) (2η)) −1] ⇒C=(1+4η^2 )e^(−2μ tan^(−1) (2η)) ⇒v^2 =((gR)/(2η))e^(−2μ tan^(−1) (2ηξ)) [(1+4η^2 )e^(−2μ tan^(−1) (2η)) −(1+4η^2 ξ^2 )e^(2μ tan^(−1) (2ηξ)) ] ⇒v^2 =((gR)/(2η)){(1+4η^2 )e^(−2μ[tan^(−1) (2η)+tan^(−1) (2ηξ)]) −(1+4η^2 ξ^2 )} ...(II) we see we get (II) by replacing ξ with −ξ in (I). that means (II) and (I) are identical. at the highest point B the speed is 0: ((gR)/(2η)){(1+4η^2 )e^(−2μ[tan^(−1) (2η)+tan^(−1) (2ηξ)]) −(1+4η^2 ξ^2 )}=0 (1+4η^2 )e^(−2μ[tan^(−1) (2η)+tan^(−1) (2ηξ)]) =(1+4η^2 ξ^2 ) e^(2μ[tan^(−1) (2η)+tan^(−1) (2ηξ)]) =((1+4η^2 )/(1+4η^2 ξ^2 )) ⇒tan^(−1) (2η)+tan^(−1) (2ηξ)=(1/(2μ))ln ((1+4η^2 )/(1+4η^2 ξ^2 )) ⇒ln (1+4η^2 ξ^2 )+2μ tan^(−1) (2ηξ)=ln (1+4η^2 )−2μ tan^(−1) (2η) ...(III) from this equation we can determine ξ at which the block reaches the highest point. example: μ=0 ln (1+4η^2 ξ^2 )=ln (1+4η^2 ) ⇒ξ=±1 i.e. the block can reach the rim again. example: η=(H/R)=1 ⇒μ_(max) =((ln (1+4η^2 ))/(2 tan^(−1) (2η)))=((ln 5)/(2 tan^(−1) 2))=0.7268 for μ=(1/2): ⇒ξ≈0.1914 ⇒h_(max) ≈0.037H for μ=(1/4): ⇒ξ≈0.488 ⇒h_(max) ≈0.238H example: η=(H/R)=4 ⇒μ_(max) =((ln (1+4η^2 ))/(2 tan^(−1) (2η)))=((ln 65)/(2 tan^(−1) 8))=1.443 for μ=(1/2): ⇒ξ≈0.2514 ⇒h_(max) ≈0.063H for μ=(1/4): ⇒ξ≈0.4889 ⇒h_(max) ≈0.239H ■ v^2 =((gR)/(2η)){(1+4η^2 )e^(−2μ[tan^(−1) (2η)−tan^(−1) (2ηξ)]) −(1+4η^2 ξ^2 )} v=−(ds/dt)=−((R(√(1+4η^2 ξ^2 ))dξ)/dt)=(√((gR)/(2η)))(√((1+4η^2 )e^(−2μ[tan^(−1) (2η)−tan^(−1) (2ηξ)]) −(1+4η^2 ξ^2 ))) t=(√((2H)/g))∫_ξ ^1 (1/( (√(((1+4η^2 )/(1+4η^2 ξ^2 )) e^(−2μ[tan^(−1) (2η)−tan^(−1) (2ηξ)]) −1))))dξ N=m(g cos θ+(v^2 /r)) N=mg{(1/( (√(1+4η^2 ξ^2 ))))+(((1+4η^2 )e^(−2μ[tan^(−1) (2η)−tan^(−1) (2ηξ)]) −(1+4η^2 ξ^2 ))/((1+4η^2 ξ^2 )^(3/2) ))} (N/(mg))=((1+4η^2 )/((1+4η^2 ξ^2 )^(3/2) )) e^(−2μ[tan^(−1) (2η)−tan^(−1) (2ηξ)])](https://www.tinkutara.com/question/Q128042.png)
$${R}=\frac{{L}}{\mathrm{2}}={radius}\:{of}\:{opening} \\ $$$${let}\:\eta=\frac{{H}}{{R}},\:\xi=\frac{{x}}{{R}}\in\left[−\mathrm{1},\mathrm{1}\right] \\ $$$${y}={H}\left(\frac{{x}}{{R}}\right)^{\mathrm{2}} ={H}\xi^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta={y}'=\mathrm{2}{H}\frac{{x}}{{R}^{\mathrm{2}} }=\mathrm{2}\eta\xi \\ $$$${y}''=\mathrm{2}\frac{{H}}{{R}^{\mathrm{2}} }=\frac{\mathrm{2}\eta}{{R}} \\ $$$${radius}\:{of}\:{curvature}\:{r}=\frac{\left[\mathrm{1}+\left({y}'\right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mid{y}''\mid} \\ $$$$\Rightarrow{r}=\frac{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} {R}}{\mathrm{2}\eta} \\ $$$${ds}=\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }{dx}={R}\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{d}\xi \\ $$$$ \\ $$$${phase}\:\mathrm{1}:\:{from}\:{point}\:{A}\:{to}\:{point}\:{O} \\ $$$${phase}\:\mathrm{2}:\:{from}\:{point}\:{O}\:{to}\:{point}\:{B} \\ $$$$ \\ $$$$\boldsymbol{{phase}}\:\mathrm{1}: \\ $$$${N}={mg}\:\mathrm{cos}\:\theta+{m}\frac{{v}^{\mathrm{2}} }{{r}} \\ $$$${f}=\mu{N}=\mu{m}\left({g}\:\mathrm{cos}\:\theta+\frac{{v}^{\mathrm{2}} }{{r}}\right) \\ $$$${ma}={mg}\:\mathrm{sin}\:\theta−{f} \\ $$$${a}={g}\:\mathrm{sin}\:\theta−\mu\left({g}\:\mathrm{cos}\:\theta+\frac{{v}^{\mathrm{2}} }{{r}}\right) \\ $$$${a}={g}\left(\mathrm{sin}\:\theta−\mu\:\mathrm{cos}\:\theta\right)−\frac{\mu{v}^{\mathrm{2}} }{{r}} \\ $$$${a}=\frac{{dv}}{{dt}}=\frac{{dv}}{{ds}}×\frac{{ds}}{{dt}}=−{v}\frac{{dv}}{{ds}} \\ $$$$−\frac{{vdv}}{{R}\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{d}\xi}=\frac{{g}\left(\mathrm{2}\eta\xi−\mu\right)}{\:\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }}−\frac{\mathrm{2}\mu\eta{v}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} {R}} \\ $$$$\frac{{vdv}}{\:{d}\xi}=−{gR}\left(\mathrm{2}\eta\xi−\mu\right)+\frac{\mathrm{2}\mu\eta{v}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{2}\eta{vdv}}{\:{d}\left(\mathrm{2}\eta\xi\right)}=−{gR}\left(\mathrm{2}\eta\xi−\mu\right)+\frac{\mathrm{2}\mu\eta{v}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} } \\ $$$$\frac{{vdv}}{\:{du}}=−\frac{{gR}}{\mathrm{2}\eta}\left({u}−\mu\right)+\frac{\mu{v}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\: \\ $$$${let}\:{u}=\mathrm{2}\eta\xi,\:{w}={v}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{dw}}{\:{du}}−\frac{\mathrm{2}\mu{w}}{\mathrm{1}+{u}^{\mathrm{2}} }=−\frac{{gR}}{\eta}\left({u}−\mu\right) \\ $$$${I}=−\mathrm{2}\mu\int\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }=−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} {u} \\ $$$$\Rightarrow{w}=−\frac{{gR}}{\eta{e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} {u}} }\left[\int\left({u}−\mu\right){e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} {u}} {du}−{C}\right] \\ $$$$\Rightarrow{w}=−\frac{{gR}}{\mathrm{2}\eta{e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} {u}} }\left[\left(\mathrm{1}+{u}^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} {u}} −{C}\right] \\ $$$$\Rightarrow{v}^{\mathrm{2}} =\frac{{gR}}{\mathrm{2}\eta}{e}^{\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)} \left[{C}−\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)} \right] \\ $$$${at}\:\xi=\mathrm{1},\:{v}=\mathrm{0}: \\ $$$${C}=\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)} \\ $$$$\Rightarrow{v}^{\mathrm{2}} =\frac{{gR}}{\mathrm{2}\eta}{e}^{\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)} \left[\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)} −\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)} \right] \\ $$$$\Rightarrow{v}^{\mathrm{2}} =\frac{{gR}}{\mathrm{2}\eta}\left\{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)\right]} −\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)\right\}\:\:\:…\left({I}\right) \\ $$$${at}\:\xi=\mathrm{0},\:{v}={v}_{\mathrm{0}} : \\ $$$${v}_{\mathrm{0}} ^{\mathrm{2}} =\frac{{gR}}{\mathrm{2}\eta}\left[\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)} −\mathrm{1}\right] \\ $$$$\Rightarrow{v}_{\mathrm{0}} =\sqrt{\frac{{gR}}{\mathrm{2}\eta}\left[\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)} −\mathrm{1}\right]} \\ $$$${such}\:{that}\:{the}\:{block}\:{reaches}\:{the}\:{lowest} \\ $$$${point},\:{v}_{\mathrm{0}} \geqslant\mathrm{0}: \\ $$$$\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)} −\mathrm{1}\geqslant\mathrm{0} \\ $$$${e}^{\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)} \leqslant\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \\ $$$$\Rightarrow\mu\leqslant\frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right)}{\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)}=\mu_{{max}} \\ $$$$ \\ $$$$\boldsymbol{{phase}}\:\mathrm{2}: \\ $$$${ma}=−{mg}\:\mathrm{sin}\:\theta−{f} \\ $$$${a}=−{g}\:\mathrm{sin}\:\theta−\mu\left({g}\:\mathrm{cos}\:\theta+\frac{{v}^{\mathrm{2}} }{{r}}\right) \\ $$$${a}=−{g}\left(\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta\right)−\frac{\mu{v}^{\mathrm{2}} }{{r}} \\ $$$${a}=\frac{{dv}}{{dt}}=\frac{{dv}}{{ds}}×\frac{{ds}}{{dt}}={v}\frac{{dv}}{{ds}} \\ $$$$\frac{{vdv}}{{R}\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{d}\xi}=−\frac{{g}\left(\mathrm{2}\eta\xi+\mu\right)}{\:\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }}−\frac{\mathrm{2}\mu\eta{v}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} {R}} \\ $$$$\frac{{vdv}}{\:{d}\xi}=−{gR}\left(\mathrm{2}\eta\xi+\mu\right)−\frac{\mathrm{2}\mu\eta{v}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\frac{{dw}}{\:{du}}+\frac{\mathrm{2}\mu{w}}{\mathrm{1}+{u}^{\mathrm{2}} }=−\frac{{gR}}{\eta}\left({u}+\mu\right)\: \\ $$$${similarly}\:{as}\:{with}\:\left({i}\right), \\ $$$$\Rightarrow{v}^{\mathrm{2}} =\frac{{gR}}{\mathrm{2}\eta}{e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)} \left[{C}−\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right){e}^{\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)} \right] \\ $$$${at}\:\xi=\mathrm{0},\:{v}={v}_{\mathrm{0}} : \\ $$$${v}_{\mathrm{0}} ^{\mathrm{2}} =\frac{{gR}}{\mathrm{2}\eta}\left({C}−\mathrm{1}\right)=\frac{{gR}}{\mathrm{2}\eta}\left[\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)} −\mathrm{1}\right] \\ $$$$\Rightarrow{C}=\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)} \\ $$$$\Rightarrow{v}^{\mathrm{2}} =\frac{{gR}}{\mathrm{2}\eta}{e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)} \left[\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)} −\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right){e}^{\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)} \right] \\ $$$$\Rightarrow{v}^{\mathrm{2}} =\frac{{gR}}{\mathrm{2}\eta}\left\{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)\right]} −\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)\right\}\:\:\:…\left({II}\right) \\ $$$${we}\:{see}\:{we}\:{get}\:\left({II}\right)\:{by}\:{replacing}\:\xi\:{with} \\ $$$$−\xi\:{in}\:\left({I}\right).\:{that}\:{means}\:\left({II}\right)\:{and}\:\left({I}\right)\:{are} \\ $$$${identical}. \\ $$$$ \\ $$$${at}\:{the}\:{highest}\:{point}\:{B}\:{the}\:{speed}\:{is}\:\mathrm{0}: \\ $$$$\frac{{gR}}{\mathrm{2}\eta}\left\{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)\right]} −\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)\right\}=\mathrm{0} \\ $$$$\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)\right]} =\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right) \\ $$$${e}^{\mathrm{2}\mu\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)\right]} =\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)=\frac{\mathrm{1}}{\mathrm{2}\mu}\mathrm{ln}\:\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{ln}\:\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)+\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)=\mathrm{ln}\:\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right)−\mathrm{2}\mu\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)\:\:\:…\left({III}\right) \\ $$$${from}\:{this}\:{equation}\:{we}\:{can}\:{determine} \\ $$$$\xi\:{at}\:{which}\:{the}\:{block}\:{reaches}\:{the} \\ $$$${highest}\:{point}. \\ $$$$ \\ $$$${example}:\:\mu=\mathrm{0} \\ $$$$\mathrm{ln}\:\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)=\mathrm{ln}\:\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right) \\ $$$$\Rightarrow\xi=\pm\mathrm{1} \\ $$$${i}.{e}.\:{the}\:{block}\:{can}\:{reach}\:{the}\:{rim}\:{again}. \\ $$$$ \\ $$$${example}:\:\eta=\frac{{H}}{{R}}=\mathrm{1} \\ $$$$\Rightarrow\mu_{{max}} =\frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right)}{\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)}=\frac{\mathrm{ln}\:\mathrm{5}}{\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \mathrm{2}}=\mathrm{0}.\mathrm{7268} \\ $$$${for}\:\mu=\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$$\Rightarrow\xi\approx\mathrm{0}.\mathrm{1914}\:\Rightarrow{h}_{{max}} \approx\mathrm{0}.\mathrm{037}{H} \\ $$$${for}\:\mu=\frac{\mathrm{1}}{\mathrm{4}}: \\ $$$$\Rightarrow\xi\approx\mathrm{0}.\mathrm{488}\:\Rightarrow{h}_{{max}} \approx\mathrm{0}.\mathrm{238}{H} \\ $$$$ \\ $$$${example}:\:\eta=\frac{{H}}{{R}}=\mathrm{4} \\ $$$$\Rightarrow\mu_{{max}} =\frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right)}{\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)}=\frac{\mathrm{ln}\:\mathrm{65}}{\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \mathrm{8}}=\mathrm{1}.\mathrm{443} \\ $$$${for}\:\mu=\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$$\Rightarrow\xi\approx\mathrm{0}.\mathrm{2514}\:\Rightarrow{h}_{{max}} \approx\mathrm{0}.\mathrm{063}{H} \\ $$$${for}\:\mu=\frac{\mathrm{1}}{\mathrm{4}}: \\ $$$$\Rightarrow\xi\approx\mathrm{0}.\mathrm{4889}\:\Rightarrow{h}_{{max}} \approx\mathrm{0}.\mathrm{239}{H} \\ $$$$\blacksquare \\ $$$${v}^{\mathrm{2}} =\frac{{gR}}{\mathrm{2}\eta}\left\{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)\right]} −\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)\right\} \\ $$$${v}=−\frac{{ds}}{{dt}}=−\frac{{R}\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{d}\xi}{{dt}}=\sqrt{\frac{{gR}}{\mathrm{2}\eta}}\sqrt{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)\right]} −\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)} \\ $$$${t}=\sqrt{\frac{\mathrm{2}{H}}{{g}}}\int_{\xi} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }\:{e}^{−\mathrm{2}\mu\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)\right]} −\mathrm{1}}}{d}\xi \\ $$$$ \\ $$$${N}={m}\left({g}\:\mathrm{cos}\:\theta+\frac{{v}^{\mathrm{2}} }{{r}}\right) \\ $$$${N}={mg}\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }}+\frac{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){e}^{−\mathrm{2}\mu\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)\right]} −\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\right\} \\ $$$$\frac{{N}}{{mg}}=\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{e}^{−\mathrm{2}\mu\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\eta\xi\right)\right]} \\ $$
Commented by ajfour last updated on 04/Jan/21
$${how}\:{you}\:{managed},\:{great}\:{attempt} \\ $$$${Sir},\:\:{and}\:{yes}\:{i}\:{did}\:{ignore}\:\frac{{v}^{\mathrm{2}} }{{r}}\:\:… \\ $$
Commented by mr W last updated on 04/Jan/21
$${following}\:{diagram}\:{shows}\:{how}\:{the} \\ $$$${velocity}\:{of}\:{the}\:{block}\:{changes}. \\ $$
Commented by mr W last updated on 04/Jan/21
Commented by mr W last updated on 04/Jan/21
Commented by mr W last updated on 05/Jan/21
Commented by ajfour last updated on 05/Jan/21
![N−mgcos θ=((mv^2 )/r) mgsin θ−f=((mdv)/dt) (f/N)=μ tan θ=2kx ; sec^2 θdθ=2kdx r=((sec^3 θ)/(2k)) μ=((gsin θ−((vdv)/ds))/(gcos θ+(v^2 /r))) μ=((gtan θ−((d(v^2 ))/((((sec^2 θdθ)/(2k))))))/(g+v^2 (((sec^2 θ)/(2k))))) 2gktan θsec^2 θ−((d(v^2 ))/dθ) = 2μgksec^2 θ+μv^2 ((d(v^2 ))/dθ)+μ(v^2 )=2gk(tanθ−μ)sec^2 θ d(v^2 )+μ(v^2 )d(tan^(−1) s)=2gk(s−μ)ds d(v^2 )+((μv^2 ds)/(1+s^2 ))=2gk(s−μ)ds (1+s^2 )d(v^2 )+μv^2 ds = 2gk(1+s^2 )(s−μ)ds ⇒ ∫d[(1+s^2 )(v^2 )]+∫(μ−2s)(v^2 )ds =2gk∫(1+s^2 )(s−μ)ds 0+∫_θ_0 ^( θ) (μ−2s)(v^2 )ds =2gk((s^4 /4)−((μs^3 )/3)+(s^2 /2)−μs)_θ_0 ^θ ⇒ (μs−s^2 )v^2 −∫(μs−s^2 )d(v^2 ) =2gk((s^4 /4)−((μs^3 )/3)+(s^2 /2)−μs)_θ_0 ^θ Anyhow here is the plan ⇒ v^2 =f(θ) v^2 =0 ⇒ θ=θ_0 and θ=β tan θ_0 =2kR , tan β=−2k∣x_f ∣ kx_f ^2 =h.](https://www.tinkutara.com/question/Q128174.png)
$$\:\:\:\:{N}−{mg}\mathrm{cos}\:\theta=\frac{{mv}^{\mathrm{2}} }{{r}} \\ $$$$\:\:\:\:{mg}\mathrm{sin}\:\theta−{f}=\frac{{mdv}}{{dt}} \\ $$$$\:\:\:\:\frac{{f}}{{N}}=\mu \\ $$$$\:\:\mathrm{tan}\:\theta=\mathrm{2}{kx}\:\:\:;\:\:\mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta=\mathrm{2}{kdx} \\ $$$$\:{r}=\frac{\mathrm{sec}\:^{\mathrm{3}} \theta}{\mathrm{2}{k}}\:\:\:\:\:\: \\ $$$$\:\:\mu=\frac{{g}\mathrm{sin}\:\theta−\frac{{vdv}}{{ds}}}{{g}\mathrm{cos}\:\theta+\frac{{v}^{\mathrm{2}} }{{r}}} \\ $$$$\:\:\mu=\frac{{g}\mathrm{tan}\:\theta−\frac{{d}\left({v}^{\mathrm{2}} \right)}{\left(\frac{\mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta}{\mathrm{2}{k}}\right)}}{{g}+{v}^{\mathrm{2}} \left(\frac{\mathrm{sec}\:^{\mathrm{2}} \theta}{\mathrm{2}{k}}\right)} \\ $$$$\:\:\:\mathrm{2}{gk}\mathrm{tan}\:\theta\mathrm{sec}\:^{\mathrm{2}} \theta−\frac{{d}\left({v}^{\mathrm{2}} \right)}{{d}\theta} \\ $$$$\:\:\:=\:\mathrm{2}\mu{gk}\mathrm{sec}\:^{\mathrm{2}} \theta+\mu{v}^{\mathrm{2}} \\ $$$$\frac{{d}\left({v}^{\mathrm{2}} \right)}{{d}\theta}+\mu\left({v}^{\mathrm{2}} \right)=\mathrm{2}{gk}\left(\mathrm{tan}\theta−\mu\right)\mathrm{sec}\:^{\mathrm{2}} \theta \\ $$$${d}\left({v}^{\mathrm{2}} \right)+\mu\left({v}^{\mathrm{2}} \right){d}\left(\mathrm{tan}^{−\mathrm{1}} {s}\right)=\mathrm{2}{gk}\left({s}−\mu\right){ds} \\ $$$${d}\left({v}^{\mathrm{2}} \right)+\frac{\mu{v}^{\mathrm{2}} {ds}}{\mathrm{1}+{s}^{\mathrm{2}} }=\mathrm{2}{gk}\left({s}−\mu\right){ds} \\ $$$$\left(\mathrm{1}+{s}^{\mathrm{2}} \right){d}\left({v}^{\mathrm{2}} \right)+\mu{v}^{\mathrm{2}} {ds} \\ $$$$\:\:\:\:\:\:=\:\mathrm{2}{gk}\left(\mathrm{1}+{s}^{\mathrm{2}} \right)\left({s}−\mu\right){ds} \\ $$$$\Rightarrow \\ $$$$\:\:\int{d}\left[\left(\mathrm{1}+{s}^{\mathrm{2}} \right)\left({v}^{\mathrm{2}} \right)\right]+\int\left(\mu−\mathrm{2}{s}\right)\left({v}^{\mathrm{2}} \right){ds} \\ $$$$\:\:\:\:\:\:=\mathrm{2}{gk}\int\left(\mathrm{1}+{s}^{\mathrm{2}} \right)\left({s}−\mu\right){ds} \\ $$$$\mathrm{0}+\int_{\theta_{\mathrm{0}} } ^{\:\theta} \left(\mu−\mathrm{2}{s}\right)\left({v}^{\mathrm{2}} \right){ds} \\ $$$$\:\:\:\:=\mathrm{2}{gk}\left(\frac{{s}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mu{s}^{\mathrm{3}} }{\mathrm{3}}+\frac{{s}^{\mathrm{2}} }{\mathrm{2}}−\mu{s}\right)_{\theta_{\mathrm{0}} } ^{\theta} \\ $$$$\Rightarrow\:\:\left(\mu{s}−{s}^{\mathrm{2}} \right){v}^{\mathrm{2}} −\int\left(\mu{s}−{s}^{\mathrm{2}} \right){d}\left({v}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{gk}\left(\frac{{s}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mu{s}^{\mathrm{3}} }{\mathrm{3}}+\frac{{s}^{\mathrm{2}} }{\mathrm{2}}−\mu{s}\right)_{\theta_{\mathrm{0}} } ^{\theta} \\ $$$${Anyhow}\:\:{here}\:\:{is}\:{the}\:{plan} \\ $$$$\Rightarrow\:\:{v}^{\mathrm{2}} ={f}\left(\theta\right) \\ $$$${v}^{\mathrm{2}} =\mathrm{0}\:\:\:\Rightarrow\:\:\theta=\theta_{\mathrm{0}} \:\:{and}\:\:\theta=\beta \\ $$$$\mathrm{tan}\:\theta_{\mathrm{0}} =\mathrm{2}{kR}\:\:,\:\:\mathrm{tan}\:\beta=−\mathrm{2}{k}\mid{x}_{{f}} \mid \\ $$$${kx}_{{f}} ^{\mathrm{2}} ={h}. \\ $$
Commented by mr W last updated on 05/Jan/21
$${thanks}\:{for}\:{this}\:{new}\:{way}\:{attempt}\:{sir}! \\ $$
Commented by Ahmed1hamouda last updated on 05/Jan/21
$$ \\ $$What is the name of the application you are using in drawing functions?
Commented by mr W last updated on 05/Jan/21
$${Grapher} \\ $$