Question Number 138696 by mr W last updated on 16/Apr/21
Commented by mr W last updated on 16/Apr/21
$${a}\:{small}\:{mass}\:{is}\:{released}\:{on}\:{the} \\ $$$${interior}\:{surface}\:{of}\:{a}\:{hollow}\:{cone} \\ $$$${at}\:{the}\:{position}\:{as}\:{shown}.\:{find}\:{the} \\ $$$${time}\:{the}\:{mass}\:{needs}\:{to}\:{reach}\:{the}\:{tip} \\ $$$${of}\:{the}\:{cone}.\:{there}\:{is}\:{no}\:{friction}. \\ $$
Answered by mr W last updated on 18/Apr/21
Commented by ajfour last updated on 28/Apr/21
![Let a point on axis of cone be A(ρcos φ, 0, ρsin φ) eq. of plane normal to axis and through A is (r^ −iρcos φ−kρsin φ)∙(icos φ+ksin φ)=0 cos φ(x−ρcos φ)+sin φ(z−ρsin φ)=0 ⇒ xcos φ+zsin φ=ρ (x−ρcos φ)^2 +(y−ρsin φ)^2 +z^2 =r^2 x^2 +y^2 +z^2 =ρ^2 +r^2 let y=rcos θ, x=ρcos φ−rsin θsin φ z=ρsin φ+rsin θcos φ (r/ρ)=tan α=m=(R/h) a_s =((−gz)/( (√(ρ^2 +r^2 )))) a_s =((−(gcos α)z)/ρ) s=(ρ/(cos^2 α)) ds=dρ(1+m^2 ) ((v_s dv_s )/ds)=−(((gcos α)z)/ρ) v_s dv_s =−(((gcos α)(1+m^2 )zdρ)/ρ) (dx)cos φ+(dz)sin φ=dρ dx=(dρ)cos φ−(msin φ)d(ρsin θ) dρ−(dz)sin φ=(dρ)cos^2 φ −msin φcos φ[(dρ)sin θ+ρcos θdθ]](https://www.tinkutara.com/question/Q139533.png)
$${Let}\:{a}\:{point}\:{on}\:{axis}\:{of}\:{cone}\:{be} \\ $$$${A}\left(\rho\mathrm{cos}\:\phi,\:\mathrm{0},\:\rho\mathrm{sin}\:\phi\right) \\ $$$${eq}.\:{of}\:{plane}\:{normal}\:{to}\:{axis}\:{and} \\ $$$${through}\:{A}\:{is} \\ $$$$\:\left(\bar {{r}}−{i}\rho\mathrm{cos}\:\phi−{k}\rho\mathrm{sin}\:\phi\right)\centerdot\left({i}\mathrm{cos}\:\phi+{k}\mathrm{sin}\:\phi\right)=\mathrm{0} \\ $$$$\mathrm{cos}\:\phi\left({x}−\rho\mathrm{cos}\:\phi\right)+\mathrm{sin}\:\phi\left({z}−\rho\mathrm{sin}\:\phi\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{x}\mathrm{cos}\:\phi+{z}\mathrm{sin}\:\phi=\rho \\ $$$$\left({x}−\rho\mathrm{cos}\:\phi\right)^{\mathrm{2}} +\left({y}−\rho\mathrm{sin}\:\phi\right)^{\mathrm{2}} +{z}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\rho^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$${let}\:\:{y}={r}\mathrm{cos}\:\theta,\:\:{x}=\rho\mathrm{cos}\:\phi−{r}\mathrm{sin}\:\theta\mathrm{sin}\:\phi \\ $$$${z}=\rho\mathrm{sin}\:\phi+{r}\mathrm{sin}\:\theta\mathrm{cos}\:\phi \\ $$$$\frac{{r}}{\rho}=\mathrm{tan}\:\alpha={m}=\frac{{R}}{{h}} \\ $$$${a}_{{s}} =\frac{−{gz}}{\:\sqrt{\rho^{\mathrm{2}} +{r}^{\mathrm{2}} }} \\ $$$${a}_{{s}} =\frac{−\left({g}\mathrm{cos}\:\alpha\right){z}}{\rho} \\ $$$${s}=\frac{\rho}{\mathrm{cos}\:^{\mathrm{2}} \alpha} \\ $$$${ds}={d}\rho\left(\mathrm{1}+{m}^{\mathrm{2}} \right) \\ $$$$\frac{{v}_{{s}} {dv}_{{s}} }{{ds}}=−\frac{\left({g}\mathrm{cos}\:\alpha\right){z}}{\rho} \\ $$$${v}_{{s}} {dv}_{{s}} =−\frac{\left({g}\mathrm{cos}\:\alpha\right)\left(\mathrm{1}+{m}^{\mathrm{2}} \right){zd}\rho}{\rho} \\ $$$$\left({dx}\right)\mathrm{cos}\:\phi+\left({dz}\right)\mathrm{sin}\:\phi={d}\rho \\ $$$${dx}=\left({d}\rho\right)\mathrm{cos}\:\phi−\left({m}\mathrm{sin}\:\phi\right){d}\left(\rho\mathrm{sin}\:\theta\right) \\ $$$${d}\rho−\left({dz}\right)\mathrm{sin}\:\phi=\left({d}\rho\right)\mathrm{cos}\:^{\mathrm{2}} \phi \\ $$$$\:\:\:\:\:\:\:\:\:−{m}\mathrm{sin}\:\phi\mathrm{cos}\:\phi\left[\left({d}\rho\right)\mathrm{sin}\:\theta+\rho\mathrm{cos}\:\theta{d}\theta\right] \\ $$