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Question Number 127997 by mr W last updated on 05/Jan/21

Commented by mr W last updated on 05/Jan/21

$$aparaboloidbowlhasadepthofH$$$$andanopeningwithdiameterL.$$$$asmallblockofmassmisreleased$$$$fromrestattherimofthebowl.$$$$thefrictioncoefficientbetweenthe$$$$objectsis\mu .assumethefrictionis$$$$smallenoughsuchthattheblockcan$$$$passthroughthelowestpoint.$$$$findthemaximumheighththesmall$$$$blockreaches.$$

Answered by mr W last updated on 04/Jan/21

Commented by mr W last updated on 05/Jan/21

$$R=\frac{L}{2}=radiusofopening$$$$let\eta =\frac{H}{R},\xi =\frac{x}{R}\in [-1,1]$$$$y=H{\left(\frac{x}{R}\right)}^2=H{\xi }^2$$$$\tan \theta =y^{\prime }=2H\frac{x}{R^2}=2\eta \xi$$$$y^{″}=2\frac{H}{R^2}=\frac{2\eta }{R}$$$$radiusofcurvaturer=\frac{{[1+{\left(y^{\prime }\right)}^2]}^{\frac{3}{2}}}{\mid y^{″}\mid }$$$$\Rightarrow r=\frac{{(1+4{\eta }^2{\xi }^2)}^{\frac{3}{2}}R}{2\eta }$$$$ds=\sqrt{1+{\left(y^{\prime }\right)}^2}dx=R\sqrt{1+4{\eta }^2{\xi }^2}d\xi$$$$$$$$phase1:frompointAtopointO$$$$phase2:frompointOtopointB$$$$$$$$\mathit{p}\mathit{h}\mathit{a}\mathit{s}\mathit{e}1:$$$$N=mg\cos \theta +m\frac{v^2}{r}$$$$f=\mu N=\mu m(g\cos \theta +\frac{v^2}{r})$$$$ma=mg\sin \theta -f$$$$a=g\sin \theta -\mu (g\cos \theta +\frac{v^2}{r})$$$$a=g(\sin \theta -\mu \cos \theta )-\frac{\mu v^2}{r}$$$$a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=-v\frac{dv}{ds}$$$$-\frac{vdv}{R\sqrt{1+4{\eta }^2{\xi }^2}d\xi }=\frac{g(2\eta \xi -\mu )}{\sqrt{1+4{\eta }^2{\xi }^2}}-\frac{2\mu \eta v^2}{{(1+4{\eta }^2{\xi }^2)}^{\frac{3}{2}}R}$$$$\frac{vdv}{d\xi }=-gR(2\eta \xi -\mu )+\frac{2\mu \eta v^2}{1+4{\eta }^2{\xi }^2}...\left(i\right)$$$$\frac{2\eta vdv}{d\left(2\eta \xi \right)}=-gR(2\eta \xi -\mu )+\frac{2\mu \eta v^2}{1+4{\eta }^2{\xi }^2}$$$$\frac{vdv}{du}=-\frac{gR}{2\eta }(u-\mu )+\frac{\mu v^2}{1+u^2}$$$$letu=2\eta \xi ,w=v^2$$$$\Rightarrow \frac{dw}{du}-\frac{2\mu w}{1+u^2}=-\frac{gR}{\eta }(u-\mu )$$$$I=-2\mu \int \frac{du}{1+u^2}=-2\mu {\tan }^{-1}u$$$$\Rightarrow w=-\frac{gR}{\eta e^{-2\mu {\tan }^{-1}u}}[\int (u-\mu )e^{-2\mu {\tan }^{-1}u}du-C]$$$$\Rightarrow w=-\frac{gR}{2\eta e^{-2\mu {\tan }^{-1}u}}[(1+u^2)e^{-2\mu {\tan }^{-1}u}-C]$$$$\Rightarrow v^2=\frac{gR}{2\eta }{e}^{2\mu {\tan }^{-1}\left(2\eta \xi \right)}[C-(1+4{\eta }^2{\xi }^2)e^{-2\mu {\tan }^{-1}\left(2\eta \xi \right)}]$$$$at\xi =1,v=0:$$$$C=(1+4{\eta }^2)e^{-2\mu {\tan }^{-1}\left(2\eta \right)}$$$$\Rightarrow v^2=\frac{gR}{2\eta }{e}^{2\mu {\tan }^{-1}\left(2\eta \xi \right)}[(1+4{\eta }^2)e^{-2\mu {\tan }^{-1}\left(2\eta \right)}-(1+4{\eta }^2{\xi }^2)e^{-2\mu {\tan }^{-1}\left(2\eta \xi \right)}]$$$$\Rightarrow v^2=\frac{gR}{2\eta }\{(1+4{\eta }^2)e^{-2\mu [{\tan }^{-1}\left(2\eta \right)-{\tan }^{-1}\left(2\eta \xi \right)]}-(1+4{\eta }^2{\xi }^2)\}...\left(I\right)$$$$at\xi =0,v=v_0:$$$$v_0^2=\frac{gR}{2\eta }[(1+4{\eta }^2)e^{-2\mu {\tan }^{-1}\left(2\eta \right)}-1]$$$$\Rightarrow v_0=\sqrt{\frac{gR}{2\eta }[(1+4{\eta }^2)e^{-2\mu {\tan }^{-1}\left(2\eta \right)}-1]}$$$$suchthattheblockreachesthelowest$$$$point,v_0⩾0:$$$$(1+4{\eta }^2)e^{-2\mu {\tan }^{-1}\left(2\eta \right)}-1⩾0$$$$e^{2\mu {\tan }^{-1}\left(2\eta \right)}⩽1+4{\eta }^2$$$$\Rightarrow \mu ⩽\frac{\ln (1+4{\eta }^2)}{2{\tan }^{-1}\left(2\eta \right)}={\mu }_{max}$$$$$$$$\mathit{p}\mathit{h}\mathit{a}\mathit{s}\mathit{e}2:$$$$ma=-mg\sin \theta -f$$$$a=-g\sin \theta -\mu (g\cos \theta +\frac{v^2}{r})$$$$a=-g(\sin \theta +\mu \cos \theta )-\frac{\mu v^2}{r}$$$$a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=v\frac{dv}{ds}$$$$\frac{vdv}{R\sqrt{1+4{\eta }^2{\xi }^2}d\xi }=-\frac{g(2\eta \xi +\mu )}{\sqrt{1+4{\eta }^2{\xi }^2}}-\frac{2\mu \eta v^2}{{(1+4{\eta }^2{\xi }^2)}^{\frac{3}{2}}R}$$$$\frac{vdv}{d\xi }=-gR(2\eta \xi +\mu )-\frac{2\mu \eta v^2}{1+4{\eta }^2{\xi }^2}...\left(ii\right)$$$$\Rightarrow \frac{dw}{du}+\frac{2\mu w}{1+u^2}=-\frac{gR}{\eta }(u+\mu )$$$$similarlyaswith\left(i\right),$$$$\Rightarrow v^2=\frac{gR}{2\eta }{e}^{-2\mu {\tan }^{-1}\left(2\eta \xi \right)}[C-(1+4{\eta }^2{\xi }^2)e^{2\mu {\tan }^{-1}\left(2\eta \xi \right)}]$$$$at\xi =0,v=v_0:$$$$v_0^2=\frac{gR}{2\eta }(C-1)=\frac{gR}{2\eta }[(1+4{\eta }^2)e^{-2\mu {\tan }^{-1}\left(2\eta \right)}-1]$$$$\Rightarrow C=(1+4{\eta }^2)e^{-2\mu {\tan }^{-1}\left(2\eta \right)}$$$$\Rightarrow v^2=\frac{gR}{2\eta }{e}^{-2\mu {\tan }^{-1}\left(2\eta \xi \right)}[(1+4{\eta }^2)e^{-2\mu {\tan }^{-1}\left(2\eta \right)}-(1+4{\eta }^2{\xi }^2)e^{2\mu {\tan }^{-1}\left(2\eta \xi \right)}]$$$$\Rightarrow v^2=\frac{gR}{2\eta }\{(1+4{\eta }^2)e^{-2\mu [{\tan }^{-1}\left(2\eta \right)+{\tan }^{-1}\left(2\eta \xi \right)]}-(1+4{\eta }^2{\xi }^2)\}...\left(II\right)$$$$weseeweget\left(II\right)byreplacing\xi with$$$$-\xi in\left(I\right).thatmeans\left(II\right)and\left(I\right)are$$$$identical.$$$$$$$$atthehighestpointBthespeedis0:$$$$\frac{gR}{2\eta }\{(1+4{\eta }^2)e^{-2\mu [{\tan }^{-1}\left(2\eta \right)+{\tan }^{-1}\left(2\eta \xi \right)]}-(1+4{\eta }^2{\xi }^2)\}=0$$$$(1+4{\eta }^2)e^{-2\mu [{\tan }^{-1}\left(2\eta \right)+{\tan }^{-1}\left(2\eta \xi \right)]}=(1+4{\eta }^2{\xi }^2)$$$$e^{2\mu [{\tan }^{-1}\left(2\eta \right)+{\tan }^{-1}\left(2\eta \xi \right)]}=\frac{1+4{\eta }^2}{1+4{\eta }^2{\xi }^2}$$$$\Rightarrow {\tan }^{-1}\left(2\eta \right)+{\tan }^{-1}\left(2\eta \xi \right)=\frac{1}{2\mu }\ln \frac{1+4{\eta }^2}{1+4{\eta }^2{\xi }^2}$$$$\Rightarrow \ln (1+4{\eta }^2{\xi }^2)+2\mu {\tan }^{-1}\left(2\eta \xi \right)=\ln (1+4{\eta }^2)-2\mu {\tan }^{-1}\left(2\eta \right)...\left(III\right)$$$$fromthisequationwecandetermine$$$$\xi atwhichtheblockreachesthe$$$$highestpoint.$$$$$$$$example:\mu =0$$$$\ln (1+4{\eta }^2{\xi }^2)=\ln (1+4{\eta }^2)$$$$\Rightarrow \xi =\pm 1$$$$i.e.theblockcanreachtherimagain.$$$$$$$$example:\eta =\frac{H}{R}=1$$$$\Rightarrow {\mu }_{max}=\frac{\ln (1+4{\eta }^2)}{2{\tan }^{-1}\left(2\eta \right)}=\frac{\ln 5}{2{\tan }^{-1}2}=0.7268$$$$for\mu =\frac{1}{2}:$$$$\Rightarrow \xi \approx 0.1914\Rightarrow h_{max}\approx 0.037H$$$$for\mu =\frac{1}{4}:$$$$\Rightarrow \xi \approx 0.488\Rightarrow h_{max}\approx 0.238H$$$$$$$$example:\eta =\frac{H}{R}=4$$$$\Rightarrow {\mu }_{max}=\frac{\ln (1+4{\eta }^2)}{2{\tan }^{-1}\left(2\eta \right)}=\frac{\ln 65}{2{\tan }^{-1}8}=1.443$$$$for\mu =\frac{1}{2}:$$$$\Rightarrow \xi \approx 0.2514\Rightarrow h_{max}\approx 0.063H$$$$for\mu =\frac{1}{4}:$$$$\Rightarrow \xi \approx 0.4889\Rightarrow h_{max}\approx 0.239H$$$$◼$$$$v^2=\frac{gR}{2\eta }\{(1+4{\eta }^2)e^{-2\mu [{\tan }^{-1}\left(2\eta \right)-{\tan }^{-1}\left(2\eta \xi \right)]}-(1+4{\eta }^2{\xi }^2)\}$$$$v=-\frac{ds}{dt}=-\frac{R\sqrt{1+4{\eta }^2{\xi }^2}d\xi }{dt}=\sqrt{\frac{gR}{2\eta }}\sqrt{(1+4{\eta }^2)e^{-2\mu [{\tan }^{-1}\left(2\eta \right)-{\tan }^{-1}\left(2\eta \xi \right)]}-(1+4{\eta }^2{\xi }^2)}$$$$t=\sqrt{\frac{2H}{g}}{\int }_{\xi }^1\frac{1}{\sqrt{\frac{1+4{\eta }^2}{1+4{\eta }^2{\xi }^2}{e}^{-2\mu [{\tan }^{-1}\left(2\eta \right)-{\tan }^{-1}\left(2\eta \xi \right)]}-1}}d\xi$$$$$$$$N=m(g\cos \theta +\frac{v^2}{r})$$$$N=mg\{\frac{1}{\sqrt{1+4{\eta }^2{\xi }^2}}+\frac{(1+4{\eta }^2)e^{-2\mu [{\tan }^{-1}\left(2\eta \right)-{\tan }^{-1}\left(2\eta \xi \right)]}-(1+4{\eta }^2{\xi }^2)}{{(1+4{\eta }^2{\xi }^2)}^{\frac{3}{2}}}\}$$$$\frac{N}{mg}=\frac{1+4{\eta }^2}{{(1+4{\eta }^2{\xi }^2)}^{\frac{3}{2}}}{e}^{-2\mu [{\tan }^{-1}\left(2\eta \right)-{\tan }^{-1}\left(2\eta \xi \right)]}$$

Commented by ajfour last updated on 04/Jan/21

$$howyoumanaged,greatattempt$$$$Sir,andyesididignore\frac{v^2}{r}...$$

Commented by mr W last updated on 04/Jan/21

$$followingdiagramshowshowthe$$$$velocityoftheblockchanges.$$

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

$$N-mg\cos \theta =\frac{{mv}^2}{r}$$$$mg\sin \theta -f=\frac{mdv}{dt}$$$$\frac{f}{N}=\mu$$$$\tan \theta =2kx;\sec {}^2\theta d\theta =2kdx$$$$r=\frac{\sec {}^3\theta }{2k}$$$$\mu =\frac{g\sin \theta -\frac{vdv}{ds}}{g\cos \theta +\frac{v^2}{r}}$$$$\mu =\frac{g\tan \theta -\frac{d\left(v^2\right)}{\left(\frac{\sec {}^2\theta d\theta }{2k}\right)}}{g+v^2\left(\frac{\sec {}^2\theta }{2k}\right)}$$$$2gk\tan \theta \sec {}^2\theta -\frac{d\left(v^2\right)}{d\theta }$$$$=2\mu gk\sec {}^2\theta +\mu v^2$$$$\frac{d\left(v^2\right)}{d\theta }+\mu \left(v^2\right)=2gk(\tan \theta -\mu )\sec {}^2\theta$$$$d\left(v^2\right)+\mu \left(v^2\right)d\left({\tan }^{-1}s\right)=2gk(s-\mu )ds$$$$d\left(v^2\right)+\frac{\mu v^{2}ds}{1+s^2}=2gk(s-\mu )ds$$$$(1+s^2)d\left(v^2\right)+\mu v^{2}ds$$$$=2gk(1+s^2)(s-\mu )ds$$$$\Rightarrow$$$$\int d\left[(1+s^2)\left(v^2\right)\right]+\int (\mu -2s)\left(v^2\right)ds$$$$=2gk\int (1+s^2)(s-\mu )ds$$$$0+{\int }_{{\theta }_0}^{\theta }(\mu -2s)\left(v^2\right)ds$$$$=2gk{(\frac{s^4}{4}-\frac{\mu s^3}{3}+\frac{s^2}{2}-\mu s)}_{{\theta }_0}^{\theta }$$$$\Rightarrow (\mu s-s^2)v^2-\int (\mu s-s^2)d\left(v^2\right)$$$$=2gk{(\frac{s^4}{4}-\frac{\mu s^3}{3}+\frac{s^2}{2}-\mu s)}_{{\theta }_0}^{\theta }$$$$Anyhowhereistheplan$$$$\Rightarrow v^2=f\left(\theta \right)$$$$v^2=0\Rightarrow \theta ={\theta }_{0}and\theta =\beta$$$$\tan {\theta }_0=2kR,\tan \beta =-2k\mid x_f\mid$$$${kx}_f^2=h.$$

Commented by mr W last updated on 05/Jan/21

$$thanksforthisnewwayattemptsir!$$

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

$$Grapher$$

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