Question Number 152388 by ajfour last updated on 28/Aug/21
Commented by ajfour last updated on 28/Aug/21
$${Find}\:{the}\:{maximum}\:{speed}\:{u}, \\ $$$${that}\:{can}\:{be}\:{given}\:{to}\:{the}\:{solid} \\ $$$${ball},\:{so}\:{that}\:{it}\:{goes}\:{through}\:{the} \\ $$$${paraboloid}\:{ditch}\:{y}={x}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$${and}\:{emerges}\:{out}\:{at}\:{speed}\:{u}, \\ $$$${having}\:{purely}\:{rolled}\:{all}\:{along}. \\ $$$$\left({assume}\:{upper}\:{limit}\:{of}\:{coeff}.\:\right. \\ $$$$\left.\:{of}\:{friction}\:{to}\:{be}\:\mu\:{everywhere}\right) \\ $$
Commented by mr W last updated on 28/Aug/21
$${seems}\:{to}\:{be}\:{very}\:{hard} \\ $$
Commented by ajfour last updated on 28/Aug/21
Commented by ajfour last updated on 29/Aug/21
![say at vertex ball is given a max rolling speed so that it cones over without slipping. so, consider v reversed in diagram above. I=(2/5)mr^2 +mr^2 =I_0 +A (1/2)I(ω_0 ^2 −ω^2 )=mg(b^2 +rcos θ) ⇒ (7/(10))(u^2 −v^2 )=g(b^2 +rcos θ) { v^2 =u^2 −((10g)/7)(b^2 +rcos θ) } ★ differentiating wrt θ [ ((mvdv)/(rdθ))=((5mgsin θ)/7) ] ★ mgcos θ−N=((mv^2 )/r) ⇒ N=m(gcos θ−(v^2 /r)) (normal rxn goes ↓ with θ) mgsin θ−f=m((vdv)/(rdθ)) dynamic μ=(f/N) ≤μ_0 (in Q) mgsin θ−((mdv)/(rdθ))=μm(gcos θ−(v^2 /r)) μ(u,θ)=((grsin θ−((10grsin θ)/7)+u^2 )/((grcos θ−v^2 ))) μ(u,θ)=((grsin θ−((10grsin θ)/7)+u^2 )/(grcos θ−u^2 +((10g)/7)(b^2 +rcos θ))) μ(u,θ)=((7(u^2 /gr)−3sin θ)/(17cos θ+10(b^2 /r)−7(u^2 /gr))) (∂μ/∂θ)=(((−3cos θ){17cos θ+10B−7U}−{7U−3sin θ}(−17sin θ))/([17U+10B−7U]^2 )) (∂μ/∂θ)=0 ⇒ 51cos^2 θ+30Bcos θ−21Ucos θ =119Usin θ−51sin^2 θ ⇒ U(21cos θ+119sin θ)=51+30Bcos θ .......∗∗ Now μ(θ)=((7U−3sin θ)/(17cos θ+10B−7U))=μ_0 hence 7(1+μ_0 )(((51+30Bcos θ)/(21cos θ+119sin θ))) = 3sin θ+μ_0 (17cos θ+10B) from above we get θ=δ U=((51+30Bcos δ)/(21cos δ+119sin δ)) U=(u^2 /(rg)) then for θ=0° { v^2 =u^2 −((10g)/7)(b^2 +rcos θ) } ★ gives (v^2 )_Q =rg(((51+30Bcos δ)/(21cos δ+119sin δ))) −((10g)/7)(b^2 +r) _____________________](https://www.tinkutara.com/question/Q152550.png)
$${say}\:{at}\:{vertex}\:{ball}\:{is}\:{given}\:{a} \\ $$$${max}\:{rolling}\:{speed}\:{so}\:{that}\:{it} \\ $$$${cones}\:{over}\:{without}\:{slipping}. \\ $$$${so},\:{consider}\:{v}\:{reversed}\:{in} \\ $$$${diagram}\:{above}. \\ $$$${I}=\frac{\mathrm{2}}{\mathrm{5}}{mr}^{\mathrm{2}} +{mr}^{\mathrm{2}} ={I}_{\mathrm{0}} +{A} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}\left(\omega_{\mathrm{0}} ^{\mathrm{2}} −\omega^{\mathrm{2}} \right)={mg}\left({b}^{\mathrm{2}} +{r}\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\:\:\frac{\mathrm{7}}{\mathrm{10}}\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} \right)={g}\left({b}^{\mathrm{2}} +{r}\mathrm{cos}\:\theta\right) \\ $$$$\left\{\:\:{v}^{\mathrm{2}} ={u}^{\mathrm{2}} −\frac{\mathrm{10}{g}}{\mathrm{7}}\left({b}^{\mathrm{2}} +{r}\mathrm{cos}\:\theta\right)\:\:\right\}\:\:\bigstar \\ $$$${differentiating}\:{wrt}\:\theta \\ $$$$\left[\:\:\:\:\frac{{mvdv}}{{rd}\theta}=\frac{\mathrm{5}{mg}\mathrm{sin}\:\theta}{\mathrm{7}}\:\:\:\right]\:\:\:\bigstar \\ $$$${mg}\mathrm{cos}\:\theta−{N}=\frac{{mv}^{\mathrm{2}} }{{r}} \\ $$$$\Rightarrow\:\:{N}={m}\left({g}\mathrm{cos}\:\theta−\frac{{v}^{\mathrm{2}} }{{r}}\right) \\ $$$$\left({normal}\:{rxn}\:{goes}\:\downarrow\:{with}\:\theta\right) \\ $$$${mg}\mathrm{sin}\:\theta−{f}={m}\frac{{vdv}}{{rd}\theta} \\ $$$${dynamic}\:\:\:\mu=\frac{{f}}{{N}}\:\leqslant\mu_{\mathrm{0}} \:\left({in}\:{Q}\right) \\ $$$$\:\:\:{mg}\mathrm{sin}\:\theta−\frac{{mdv}}{{rd}\theta}=\mu{m}\left({g}\mathrm{cos}\:\theta−\frac{{v}^{\mathrm{2}} }{{r}}\right) \\ $$$$ \\ $$$$\mu\left({u},\theta\right)=\frac{{gr}\mathrm{sin}\:\theta−\frac{\mathrm{10}{gr}\mathrm{sin}\:\theta}{\mathrm{7}}+{u}^{\mathrm{2}} }{\left({gr}\mathrm{cos}\:\theta−{v}^{\mathrm{2}} \right)} \\ $$$$\mu\left({u},\theta\right)=\frac{{gr}\mathrm{sin}\:\theta−\frac{\mathrm{10}{gr}\mathrm{sin}\:\theta}{\mathrm{7}}+{u}^{\mathrm{2}} }{{gr}\mathrm{cos}\:\theta−{u}^{\mathrm{2}} +\frac{\mathrm{10}{g}}{\mathrm{7}}\left({b}^{\mathrm{2}} +{r}\mathrm{cos}\:\theta\right)} \\ $$$$\mu\left({u},\theta\right)=\frac{\mathrm{7}\left({u}^{\mathrm{2}} /{gr}\right)−\mathrm{3sin}\:\theta}{\mathrm{17cos}\:\theta+\mathrm{10}\left({b}^{\mathrm{2}} /{r}\right)−\mathrm{7}\left({u}^{\mathrm{2}} /{gr}\right)} \\ $$$$\frac{\partial\mu}{\partial\theta}=\frac{\left(−\mathrm{3cos}\:\theta\right)\left\{\mathrm{17cos}\:\theta+\mathrm{10}{B}−\mathrm{7}{U}\right\}−\left\{\mathrm{7}{U}−\mathrm{3sin}\:\theta\right\}\left(−\mathrm{17sin}\:\theta\right)}{\left[\mathrm{17}{U}+\mathrm{10}{B}−\mathrm{7}{U}\right]^{\mathrm{2}} } \\ $$$$\frac{\partial\mu}{\partial\theta}=\mathrm{0}\:\:\Rightarrow \\ $$$$\mathrm{51cos}\:^{\mathrm{2}} \theta+\mathrm{30}{B}\mathrm{cos}\:\theta−\mathrm{21}{U}\mathrm{cos}\:\theta \\ $$$$\:\:=\mathrm{119}{U}\mathrm{sin}\:\theta−\mathrm{51sin}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow\:\:{U}\left(\mathrm{21cos}\:\theta+\mathrm{119sin}\:\theta\right)=\mathrm{51}+\mathrm{30}{B}\mathrm{cos}\:\theta \\ $$$$\:\:\:\:…….\ast\ast \\ $$$${Now}\: \\ $$$$\:\mu\left(\theta\right)=\frac{\mathrm{7}{U}−\mathrm{3sin}\:\theta}{\mathrm{17cos}\:\theta+\mathrm{10}{B}−\mathrm{7}{U}}=\mu_{\mathrm{0}} \\ $$$${hence} \\ $$$$\mathrm{7}\left(\mathrm{1}+\mu_{\mathrm{0}} \right)\left(\frac{\mathrm{51}+\mathrm{30}{B}\mathrm{cos}\:\theta}{\mathrm{21cos}\:\theta+\mathrm{119sin}\:\theta}\right) \\ $$$$\:\:\:=\:\mathrm{3sin}\:\theta+\mu_{\mathrm{0}} \left(\mathrm{17cos}\:\theta+\mathrm{10}{B}\right) \\ $$$${from}\:{above}\:{we}\:{get}\:\theta=\delta \\ $$$${U}=\frac{\mathrm{51}+\mathrm{30}{B}\mathrm{cos}\:\delta}{\mathrm{21cos}\:\delta+\mathrm{119sin}\:\delta}\: \\ $$$${U}=\frac{{u}^{\mathrm{2}} }{{rg}}\:\:\:\:\:{then}\:{for}\:\theta=\mathrm{0}° \\ $$$$\left\{\:\:{v}^{\mathrm{2}} ={u}^{\mathrm{2}} −\frac{\mathrm{10}{g}}{\mathrm{7}}\left({b}^{\mathrm{2}} +{r}\mathrm{cos}\:\theta\right)\:\:\right\}\:\:\bigstar \\ $$$${gives} \\ $$$$\left({v}^{\mathrm{2}} \right)_{{Q}} ={rg}\left(\frac{\mathrm{51}+\mathrm{30}{B}\mathrm{cos}\:\delta}{\mathrm{21cos}\:\delta+\mathrm{119sin}\:\delta}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{10}{g}}{\mathrm{7}}\left({b}^{\mathrm{2}} +{r}\right) \\ $$$$\:\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$
Commented by ajfour last updated on 29/Aug/21
$${have}\:{assumed}\:{r}\leqslant{R} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{2}\:\:\:\&\:\:{from} \\ $$$${x}^{\mathrm{2}} +\left({y}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}+\mathrm{2}\left({y}−{R}\right)\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=\frac{{x}}{{R}−{y}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{R}−{y}+\frac{{x}^{\mathrm{2}} }{{R}−{y}}}{\left({R}−{y}\right)^{\mathrm{2}} }=\mathrm{2} \\ $$$$\Rightarrow\:\:\:{now}\:\:{x}=\mathrm{0}\:,\:{y}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{R}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:{so}\:\:\:\:{r}\leqslant{R}=\frac{\mathrm{1}}{\mathrm{2}}\:{unit} \\ $$$${unit}\:{comes}\:{from}\:\:{unit}\:{of}\:{y} \\ $$$${of}\:{parabola}. \\ $$$${say}\:{if}\:\:{track}\:{is}\:\mathrm{1}{m}\:{deep},\:{then} \\ $$$${width}\:{is}\:\mathrm{2}{m};\:{while}\:{r}\leqslant\frac{\mathrm{1}}{\mathrm{2}}{m}. \\ $$
Commented by mr W last updated on 29/Aug/21
$${i}\:{think}\:{in}\:{mg}\mathrm{cos}\:\theta−{N}=\frac{{mv}^{\mathrm{2}} }{{r}} \\ $$$${r}\:{should}\:{be}\:{the}\:{radius}\:{of}\:{the}\:{locus} \\ $$$${of}\:{the}\:{center}\:{of}\:{the}\:{ball},{not}\:{the}\: \\ $$$${radius}\:{of}\:{the}\:{ball}. \\ $$$$ \\ $$
Commented by ajfour last updated on 29/Aug/21
$${ball}\:{is}\:{turning}\:{about}\:{the}\:{edge} \\ $$$${so}\:{center}\:{of}\:{ball}\:{goes}\:{about}\:{the} \\ $$$${edge}\:{in}\:{a}\:{circle}\:{of}\:{radius}\:{r}\:{only}. \\ $$
Commented by mr W last updated on 29/Aug/21
$${yes},\:{at}\:{that}\:{point}. \\ $$
Answered by mr W last updated on 28/Aug/21
