Question Number 225407 by mr W last updated on 25/Oct/25
Commented by mr W last updated on 25/Oct/25
$${the}\:{frustum}\:{of}\:{a}\:{cone}\:{is}\:{cut}\:{into} \\ $$$${two}\:{parts}\:{as}\:{shown}.\:{find}\:{the}\:{volume} \\ $$$${of}\:{each}\:{part}\:{and}\:{the}\:{area}\:{of}\:{the}\:{cut} \\ $$$${section}. \\ $$
Commented by ajfour last updated on 25/Oct/25
https://youtu.be/z6HoWsJmro8?si=3QF_n0sIhAdg0Dcj
Commented by fantastic last updated on 25/Oct/25
$${sir}\:{the}\:{ellips}\:{passes}\: \\ $$$${through}\:{the}\:{diameter}. \\ $$$${will}\:{not}\:{the}\:{b}\:{length}\:{will}\:{start}\:{to} \\ $$$${decrease}\:{after}\:{that}? \\ $$
Answered by mr W last updated on 25/Oct/25
Commented by mr W last updated on 25/Oct/25
![area of segment of ellipse A_S =ab[cos^(−1) (1−(h/a))−(1−(h/a))(√((h/a)(2−(h/a))))] this will be useful later.](https://www.tinkutara.com/question/Q225420.png)
$${area}\:{of}\:{segment}\:{of}\:{ellipse} \\ $$$${A}_{{S}} ={ab}\left[\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{1}−\frac{{h}}{{a}}\right)−\left(\mathrm{1}−\frac{{h}}{{a}}\right)\sqrt{\frac{{h}}{{a}}\left(\mathrm{2}−\frac{{h}}{{a}}\right)}\right] \\ $$$${this}\:{will}\:{be}\:{useful}\:{later}. \\ $$
Answered by mr W last updated on 26/Oct/25
Commented by mr W last updated on 26/Oct/25
$${some}\:{spatial}\:{ability}\:{is}\:{needed}. \\ $$$${semi}\:{minor}\:{axis}\:\left({b}\right)\:{lies}\:{in}\:{the}\:{circle} \\ $$$${with}\:{radius}\:{r}_{\mathrm{4}} . \\ $$
Commented by mr W last updated on 26/Oct/25
Commented by mr W last updated on 26/Oct/25
Commented by fantastic last updated on 26/Oct/25
$${my}\:{apology}\:{for} \\ $$$${causing}\:{problems}\:{to}\:{the} \\ $$$${other}\:{users}. \\ $$$${will}\:{only}\:{post}\:{meaningful} \\ $$$${things}\: \\ $$
Commented by mr W last updated on 26/Oct/25
$${thanks}! \\ $$
Commented by fantastic last updated on 26/Oct/25
$${ongoing}….. \\ $$
Commented by mr W last updated on 26/Oct/25
![tan φ=(h/r_1 ) ⇒cos φ=(r_1 /( (√(h^2 +r_1 ^2 )))) (h_1 /r_1 )=(h/(r_2 −r_1 )) ⇒h_1 =((r_1 h)/(r_2 −r_1 )) ⇒h_2 =((r_2 h)/(r_2 −r_1 )) ((h_3 −h_2 )/h)=(r_3 /r_1 )=((r_3 −r_2 )/(r_2 −r_1 )) ⇒r_3 =((r_1 r_2 )/(2r_1 −r_2 )) ⇒h_3 =h_2 +((r_2 h)/(2r_1 −r_2 ))=((r_1 r_2 h)/((r_2 −r_1 )(2r_1 −r_2 ))) r_4 =((r_1 +r_3 )/2)=(r_1 ^2 /(2r_1 −r_2 )) (2a)^2 =(h_3 −h_1 )^2 +(r_3 +r_1 )^2 (2a)^2 =[((r_1 r_2 h)/((r_2 −r_1 )(2r_1 −r_2 )))−((r_1 h)/(r_2 −r_1 ))]^2 +(((r_1 r_2 )/(2r_1 −r_2 ))+r_1 )^2 a^2 =((r_1 ^2 (h^2 +r_1 ^2 ))/((2r_1 −r_(1p2) )^2 )) ⇒a=((r_1 (√(h^2 +r_1 ^2 )))/(2r_1 −r_2 )) b^2 =r_4 ^2 −(r_4 −r_1 )^2 =((r_1 ^2 r_2 )/(2r_1 −r_2 )) ⇒b=r_1 (√(r_2 /(2r_1 −r_2 ))) a=(√(h^2 +r_1 ^2 )) λ=(( a)/a)=2−(r_2 /r_1 ) Area of cut section (segment of ellipse): A_s =ab[cos^(−1) (1−λ)−(1−λ)(√(λ(2−λ)))] =((r_1 ^2 (√(r_2 (h^2 +r_1 ^2 )))[cos^(−1) ((r_2 /r_1 )−1)−((r_2 /r_1 )−1)(√((r_2 /r_1 )(2−(r_2 /r_1 ))))])/((2r_1 −r_2 )^(3/2) )) h_s =h_2 cos φ=((r_1 r_2 h)/( (r_2 −r_1 )(√(h^2 +r_1 ^2 )))) V_2 =((πr_2 ^2 h_2 )/6)−((A_s h_s )/3) =((πr_2 ^3 h)/(6(r_2 −r_1 )))−((r_1 ^3 r_2 ^(3/2) h[cos^(−1) ((r_2 /r_1 )−1)−((r_2 /r_1 )−1)(√((r_2 /r_1 )(2−(r_2 /r_1 ))))])/(3(r_2 −r_1 )(2r_1 −r_2 )^(3/2) )) V_1 =((π(r_1 ^2 +r_2 ^2 +r_1 r_2 )h)/3)−V_2 example: r_1 =3 cm, r_2 =5 cm, h=9 cm A_s =135(√2)(cos^(−1) (2/3)−((2(√5))/9)) ≈65.7075 cm^2 V_2 =((375π)/4)−((405(√5)(cos^(−1) (2/3)−((2(√5))/9)))/2) ≈138.6853 cm^3 V_1 =((213π)/4)+((405(√5)(cos^(−1) (2/3)−((2(√5))/9)))/2) ≈323.1289 cm^3 Note: this is only for the case that the cut section is ellipse, i.e. r_2 <2r_1 .](https://www.tinkutara.com/question/Q225432.png)
$$\mathrm{tan}\:\phi=\frac{{h}}{{r}_{\mathrm{1}} }\:\Rightarrow\mathrm{cos}\:\phi=\frac{{r}_{\mathrm{1}} }{\:\sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }} \\ $$$$\frac{{h}_{\mathrm{1}} }{{r}_{\mathrm{1}} }=\frac{{h}}{{r}_{\mathrm{2}} −{r}_{\mathrm{1}} }\:\Rightarrow{h}_{\mathrm{1}} =\frac{{r}_{\mathrm{1}} {h}}{{r}_{\mathrm{2}} −{r}_{\mathrm{1}} } \\ $$$$\Rightarrow{h}_{\mathrm{2}} =\frac{{r}_{\mathrm{2}} {h}}{{r}_{\mathrm{2}} −{r}_{\mathrm{1}} } \\ $$$$\frac{{h}_{\mathrm{3}} −{h}_{\mathrm{2}} }{{h}}=\frac{{r}_{\mathrm{3}} }{{r}_{\mathrm{1}} }=\frac{{r}_{\mathrm{3}} −{r}_{\mathrm{2}} }{{r}_{\mathrm{2}} −{r}_{\mathrm{1}} } \\ $$$$\Rightarrow{r}_{\mathrm{3}} =\frac{{r}_{\mathrm{1}} {r}_{\mathrm{2}} }{\mathrm{2}{r}_{\mathrm{1}} −{r}_{\mathrm{2}} } \\ $$$$\Rightarrow{h}_{\mathrm{3}} ={h}_{\mathrm{2}} +\frac{{r}_{\mathrm{2}} {h}}{\mathrm{2}{r}_{\mathrm{1}} −{r}_{\mathrm{2}} }=\frac{{r}_{\mathrm{1}} {r}_{\mathrm{2}} {h}}{\left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)\left(\mathrm{2}{r}_{\mathrm{1}} −{r}_{\mathrm{2}} \right)} \\ $$$${r}_{\mathrm{4}} =\frac{{r}_{\mathrm{1}} +{r}_{\mathrm{3}} }{\mathrm{2}}=\frac{{r}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}{r}_{\mathrm{1}} −{r}_{\mathrm{2}} } \\ $$$$\left(\mathrm{2}{a}\right)^{\mathrm{2}} =\left({h}_{\mathrm{3}} −{h}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({r}_{\mathrm{3}} +{r}_{\mathrm{1}} \right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}{a}\right)^{\mathrm{2}} =\left[\frac{{r}_{\mathrm{1}} {r}_{\mathrm{2}} {h}}{\left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)\left(\mathrm{2}{r}_{\mathrm{1}} −{r}_{\mathrm{2}} \right)}−\frac{{r}_{\mathrm{1}} {h}}{{r}_{\mathrm{2}} −{r}_{\mathrm{1}} }\right]^{\mathrm{2}} +\left(\frac{{r}_{\mathrm{1}} {r}_{\mathrm{2}} }{\mathrm{2}{r}_{\mathrm{1}} −{r}_{\mathrm{2}} }+{r}_{\mathrm{1}} \right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} =\frac{{r}_{\mathrm{1}} ^{\mathrm{2}} \left({h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} \right)}{\left(\mathrm{2}{r}_{\mathrm{1}} −{r}_{\mathrm{1}{p}\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow{a}=\frac{{r}_{\mathrm{1}} \sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }}{\mathrm{2}{r}_{\mathrm{1}} −{r}_{\mathrm{2}} } \\ $$$${b}^{\mathrm{2}} ={r}_{\mathrm{4}} ^{\mathrm{2}} −\left({r}_{\mathrm{4}} −{r}_{\mathrm{1}} \right)^{\mathrm{2}} =\frac{{r}_{\mathrm{1}} ^{\mathrm{2}} {r}_{\mathrm{2}} }{\mathrm{2}{r}_{\mathrm{1}} −{r}_{\mathrm{2}} } \\ $$$$\Rightarrow{b}={r}_{\mathrm{1}} \sqrt{\frac{{r}_{\mathrm{2}} }{\mathrm{2}{r}_{\mathrm{1}} −{r}_{\mathrm{2}} }} \\ $$$$\:{a}=\sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$$\lambda=\frac{\:{a}}{{a}}=\mathrm{2}−\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} } \\ $$$${Area}\:{of}\:{cut}\:{section}\:\left({segment}\:{of}\:{ellipse}\right): \\ $$$$\left.{A}_{{s}} ={ab}\left[\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{1}−\lambda\right)−\left(\mathrm{1}−\lambda\right)\sqrt{\lambda\left(\mathrm{2}−\lambda\right.}\right)\right] \\ $$$$\:\:\:\:=\frac{{r}_{\mathrm{1}} ^{\mathrm{2}} \sqrt{{r}_{\mathrm{2}} \left({h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} \right)}\left[\mathrm{cos}^{−\mathrm{1}} \left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }−\mathrm{1}\right)−\left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }−\mathrm{1}\right)\sqrt{\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }\left(\mathrm{2}−\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }\right)}\right]}{\left(\mathrm{2}{r}_{\mathrm{1}} −{r}_{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$${h}_{{s}} ={h}_{\mathrm{2}} \:\mathrm{cos}\:\phi=\frac{{r}_{\mathrm{1}} {r}_{\mathrm{2}} {h}}{\:\left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)\sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }} \\ $$$${V}_{\mathrm{2}} =\frac{\pi{r}_{\mathrm{2}} ^{\mathrm{2}} {h}_{\mathrm{2}} }{\mathrm{6}}−\frac{{A}_{{s}} {h}_{{s}} }{\mathrm{3}} \\ $$$$\:\:\:=\frac{\pi{r}_{\mathrm{2}} ^{\mathrm{3}} {h}}{\mathrm{6}\left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)}−\frac{{r}_{\mathrm{1}} ^{\mathrm{3}} {r}_{\mathrm{2}} ^{\frac{\mathrm{3}}{\mathrm{2}}} {h}\left[\mathrm{cos}^{−\mathrm{1}} \left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }−\mathrm{1}\right)−\left(\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }−\mathrm{1}\right)\sqrt{\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }\left(\mathrm{2}−\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }\right)}\right]}{\mathrm{3}\left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)\left(\mathrm{2}{r}_{\mathrm{1}} −{r}_{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$${V}_{\mathrm{1}} =\frac{\pi\left({r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{1}} {r}_{\mathrm{2}} \right){h}}{\mathrm{3}}−{V}_{\mathrm{2}} \\ $$$$ \\ $$$${example}:\: \\ $$$${r}_{\mathrm{1}} =\mathrm{3}\:{cm},\:{r}_{\mathrm{2}} =\mathrm{5}\:{cm},\:{h}=\mathrm{9}\:{cm} \\ $$$${A}_{{s}} =\mathrm{135}\sqrt{\mathrm{2}}\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{9}}\right) \\ $$$$\:\:\:\:\:\approx\mathrm{65}.\mathrm{7075}\:{cm}^{\mathrm{2}} \\ $$$${V}_{\mathrm{2}} =\frac{\mathrm{375}\pi}{\mathrm{4}}−\frac{\mathrm{405}\sqrt{\mathrm{5}}\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{9}}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\approx\mathrm{138}.\mathrm{6853}\:{cm}^{\mathrm{3}} \\ $$$$\:{V}_{\mathrm{1}} =\frac{\mathrm{213}\pi}{\mathrm{4}}+\frac{\mathrm{405}\sqrt{\mathrm{5}}\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{9}}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\approx\mathrm{323}.\mathrm{1289}\:{cm}^{\mathrm{3}} \\ $$$$ \\ $$$${Note}: \\ $$$${this}\:{is}\:{only}\:{for}\:{the}\:{case}\:{that}\:{the} \\ $$$${cut}\:{section}\:{is}\:{ellipse},\:{i}.{e}.\:{r}_{\mathrm{2}} <\mathrm{2}{r}_{\mathrm{1}} . \\ $$
Commented by fantastic last updated on 26/Oct/25
$${or}\:\mathrm{0}<\theta\leqslant\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{9}}\right) \\ $$$$\theta\Rightarrow{angle}\:{between}\:{horizontal} \\ $$$${and}\:{the}\:{plane}\:\:{y}=\frac{\mathrm{9}−{z}}{\mathrm{3}}−\mathrm{6} \\ $$
Answered by mr W last updated on 26/Oct/25
Commented by mr W last updated on 01/Nov/25
![δ=(r_2 /r_1 ) tan φ=(r_1 /h) (h_1 /r_1 )=(h/(r_2 −r_1 )) ⇒h_1 =(h/(δ−1)) ⇒h_2 =((δh)/(δ−1)) O_2 P_1 =c=(√(h^2 +r_1 ^2 )) O_2 Q=x cos φ=((hx)/( (√(h^2 +r_1 ^2 )))) QP=x sin φ=((r_1 x)/( (√(h^2 +r_1 ^2 )))) (r/(h_2 −O_2 Q))=((r_2 −r_1 )/h) ⇒r=r_1 [δ−(((δ−1)x)/( (√(h^2 +r_1 ^2 ))))] b=2(√(r^2 −QP^2 ))=2r_1 (√([δ−(((δ−1)x)/( (√(h^2 +r_1 ^2 ))))]^2 −(x^2 /( h^2 +r_1 ^2 )))) dA_s =bdx=2r_1 (√([δ−(((δ−1)x)/( (√(h^2 +r_1 ^2 ))))]^2 −(x^2 /( h^2 +r_1 ^2 ))))dx A_s =2r_1 ∫_0 ^(√(h^2 +r_1 ^2 )) (√([δ−(((δ−1)x)/( (√(h^2 +r_1 ^2 ))))]^2 −(x^2 /( h^2 +r_1 ^2 ))))dx A_s =2r_1 (√(h^2 +r_1 ^2 ))∫_0 ^1 (√([δ−(δ−1)ξ]^2 −ξ^2 ))dξ A_s =2r_1 (√(h^2 +r_1 ^2 ))∫_0 ^1 (√(δ(δ−2)(1−ξ)((δ/(δ−2))−ξ)))dξ h_s =h_2 sin φ=((δr_1 h)/( (δ−1)(√(h^2 +r_1 ^2 )))) V_s =((A_s h_s )/3)=((2δr_1 ^2 h)/( 3(δ−1)))∫_0 ^1 (√(δ(δ−2)(1−ξ)((δ/(δ−2))−ξ)))dξ example 1: r_1 =3, r_2 =5, h=9 ⇒δ=(r_2 /r_1 )<2 A_s =18(√(10))∫_0 ^1 (√(((5/3)−(2/3)ξ)^2 −ξ^2 ))dξ ≈65.7075 cm^2 (✓) example 2: r_1 =3, r_2 =6, h=9 ⇒δ=(r_2 /r_1 )=2 A_s =18(√(10))∫_0 ^1 (√((2−ξ)^2 −ξ^2 ))dξ ≈75.8947 cm^2 example 3: r_1 =3, r_2 =9, h=9 ⇒δ=(r_2 /r_1 )>2 A_s =18(√(10))∫_0 ^1 (√((3−2ξ)^2 −ξ^2 ))dξ ≈105.8435 cm^2 example 4: r_1 =3, r_2 =3, h=9 ⇒δ=(r_2 /r_1 )=1 A_s =18(√(10))∫_0 ^1 (√(1−ξ^2 ))dξ ≈44.7056 cm^2 (=((π×3×(√(3^2 +9^2 )))/2)) ✓](https://www.tinkutara.com/question/Q225455.png)
$$\delta=\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} } \\ $$$$\mathrm{tan}\:\phi=\frac{{r}_{\mathrm{1}} }{{h}} \\ $$$$\frac{{h}_{\mathrm{1}} }{{r}_{\mathrm{1}} }=\frac{{h}}{{r}_{\mathrm{2}} −{r}_{\mathrm{1}} }\:\Rightarrow{h}_{\mathrm{1}} \:=\frac{{h}}{\delta−\mathrm{1}} \\ $$$$\Rightarrow{h}_{\mathrm{2}} =\frac{\delta{h}}{\delta−\mathrm{1}} \\ $$$${O}_{\mathrm{2}} {P}_{\mathrm{1}} ={c}=\sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$${O}_{\mathrm{2}} {Q}={x}\:\mathrm{cos}\:\phi=\frac{{hx}}{\:\sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }} \\ $$$${QP}={x}\:\mathrm{sin}\:\phi=\frac{{r}_{\mathrm{1}} {x}}{\:\sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }} \\ $$$$\frac{{r}}{{h}_{\mathrm{2}} −{O}_{\mathrm{2}} {Q}}=\frac{{r}_{\mathrm{2}} −{r}_{\mathrm{1}} }{{h}} \\ $$$$\Rightarrow{r}={r}_{\mathrm{1}} \left[\delta−\frac{\left(\delta−\mathrm{1}\right){x}}{\:\sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }}\right] \\ $$$${b}=\mathrm{2}\sqrt{{r}^{\mathrm{2}} −{QP}^{\mathrm{2}} }=\mathrm{2}{r}_{\mathrm{1}} \sqrt{\left[\delta−\frac{\left(\delta−\mathrm{1}\right){x}}{\:\sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }}\right]^{\mathrm{2}} −\frac{{x}^{\mathrm{2}} }{\:{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }} \\ $$$${dA}_{{s}} ={bdx}=\mathrm{2}{r}_{\mathrm{1}} \sqrt{\left[\delta−\frac{\left(\delta−\mathrm{1}\right){x}}{\:\sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }}\right]^{\mathrm{2}} −\frac{{x}^{\mathrm{2}} }{\:{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }}{dx} \\ $$$${A}_{{s}} =\mathrm{2}{r}_{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }} \sqrt{\left[\delta−\frac{\left(\delta−\mathrm{1}\right){x}}{\:\sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }}\right]^{\mathrm{2}} −\frac{{x}^{\mathrm{2}} }{\:{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }}{dx} \\ $$$${A}_{{s}} =\mathrm{2}{r}_{\mathrm{1}} \sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\left[\delta−\left(\delta−\mathrm{1}\right)\xi\right]^{\mathrm{2}} −\xi^{\mathrm{2}} }{d}\xi \\ $$$${A}_{{s}} =\mathrm{2}{r}_{\mathrm{1}} \sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\delta\left(\delta−\mathrm{2}\right)\left(\mathrm{1}−\xi\right)\left(\frac{\delta}{\delta−\mathrm{2}}−\xi\right)}{d}\xi \\ $$$${h}_{{s}} ={h}_{\mathrm{2}} \:\mathrm{sin}\:\phi=\frac{\delta{r}_{\mathrm{1}} {h}}{\:\left(\delta−\mathrm{1}\right)\sqrt{{h}^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }} \\ $$$${V}_{{s}} =\frac{{A}_{{s}} {h}_{{s}} }{\mathrm{3}}=\frac{\mathrm{2}\delta{r}_{\mathrm{1}} ^{\mathrm{2}} {h}}{\:\mathrm{3}\left(\delta−\mathrm{1}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\delta\left(\delta−\mathrm{2}\right)\left(\mathrm{1}−\xi\right)\left(\frac{\delta}{\delta−\mathrm{2}}−\xi\right)}{d}\xi \\ $$$$ \\ $$$${example}\:\mathrm{1}: \\ $$$${r}_{\mathrm{1}} =\mathrm{3},\:{r}_{\mathrm{2}} =\mathrm{5},\:{h}=\mathrm{9}\:\Rightarrow\delta=\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }<\mathrm{2} \\ $$$${A}_{{s}} =\mathrm{18}\sqrt{\mathrm{10}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\left(\frac{\mathrm{5}}{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{3}}\xi\right)^{\mathrm{2}} −\xi^{\mathrm{2}} }{d}\xi \\ $$$$\:\:\:\:\approx\mathrm{65}.\mathrm{7075}\:{cm}^{\mathrm{2}} \:\:\:\left(\checkmark\right) \\ $$$${example}\:\mathrm{2}: \\ $$$${r}_{\mathrm{1}} =\mathrm{3},\:{r}_{\mathrm{2}} =\mathrm{6},\:{h}=\mathrm{9}\:\Rightarrow\delta=\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }=\mathrm{2} \\ $$$${A}_{{s}} =\mathrm{18}\sqrt{\mathrm{10}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\left(\mathrm{2}−\xi\right)^{\mathrm{2}} −\xi^{\mathrm{2}} }{d}\xi \\ $$$$\:\:\:\:\:\approx\mathrm{75}.\mathrm{8947}\:{cm}^{\mathrm{2}} \\ $$$${example}\:\mathrm{3}: \\ $$$${r}_{\mathrm{1}} =\mathrm{3},\:{r}_{\mathrm{2}} =\mathrm{9},\:{h}=\mathrm{9}\:\Rightarrow\delta=\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }>\mathrm{2} \\ $$$${A}_{{s}} =\mathrm{18}\sqrt{\mathrm{10}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\left(\mathrm{3}−\mathrm{2}\xi\right)^{\mathrm{2}} −\xi^{\mathrm{2}} }{d}\xi \\ $$$$\:\:\:\:\:\approx\mathrm{105}.\mathrm{8435}\:{cm}^{\mathrm{2}} \\ $$$${example}\:\mathrm{4}: \\ $$$${r}_{\mathrm{1}} =\mathrm{3},\:{r}_{\mathrm{2}} =\mathrm{3},\:{h}=\mathrm{9}\:\Rightarrow\delta=\frac{{r}_{\mathrm{2}} }{{r}_{\mathrm{1}} }=\mathrm{1} \\ $$$${A}_{{s}} =\mathrm{18}\sqrt{\mathrm{10}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−\xi^{\mathrm{2}} }{d}\xi \\ $$$$\:\:\:\:\:\approx\mathrm{44}.\mathrm{7056}\:{cm}^{\mathrm{2}} \:\:\left(=\frac{\pi×\mathrm{3}×\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} }}{\mathrm{2}}\right)\:\checkmark \\ $$
Commented by mr W last updated on 26/Oct/25
$${this}\:{approach}\:{can}\:{be}\:{applied}\:{for}\:{any} \\ $$$${case}\:{with}\:{r}_{\mathrm{2}} \geqslant{r}_{\mathrm{1}} . \\ $$
Answered by ajfour last updated on 26/Oct/25
Commented by ajfour last updated on 27/Oct/25
Commented by ajfour last updated on 26/Oct/25
Commented by mr W last updated on 26/Oct/25
$${thanks}\:{sir}! \\ $$
Commented by mr W last updated on 26/Oct/25
$${here}\:{a}\:{alternative}\:{solution}\:{to}\:{the} \\ $$$${question}\:{from}\:{your}\:{video}\:{channel}. \\ $$
Commented by mr W last updated on 26/Oct/25
Commented by mr W last updated on 26/Oct/25
$${at}\:{the}\:{instant}\:{as}\:{the}\:{rod}\:{is}\:{just}\:{about} \\ $$$${to}\:{reach}\:{the}\:{ground},\:{it}\:{rotates}\:{about} \\ $$$${the}\:{point}\:{O}. \\ $$$${OA}=\frac{{L}}{\mathrm{tan}\:\alpha},\:{OB}=\frac{{L}}{\mathrm{sin}\:\alpha} \\ $$$${OC}^{\mathrm{2}} =\left(\frac{{L}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{L}}{\mathrm{tan}\:\alpha}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\alpha}\right){L}^{\mathrm{2}} \\ $$$${I}_{{O}} =\frac{{ML}^{\mathrm{2}} }{\mathrm{12}}+{M}\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\alpha}\right){L}^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\alpha}\right){ML}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}_{{O}} \omega^{\mathrm{2}} ={Mg}×\frac{{L}\:\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\alpha}\right){ML}^{\mathrm{2}} \omega^{\mathrm{2}} ={Mg}×\frac{{L}\:\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$$\Rightarrow\omega=\sqrt{\frac{{g}\:\mathrm{sin}\:\alpha}{{L}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\alpha}\right)}} \\ $$$${u}=\omega×\frac{{L}}{\mathrm{tan}\:\alpha}=\sqrt{\frac{\mathrm{3}{gL}\:\mathrm{sin}\:\alpha}{\mathrm{3}+\mathrm{tan}^{\mathrm{2}} \:\alpha}} \\ $$$${v}=\omega×\frac{{L}}{\mathrm{sin}\:\alpha}=\frac{\mathrm{1}}{\mathrm{cos}\:\alpha}\sqrt{\frac{\mathrm{3}{gL}\:\mathrm{sin}\:\alpha}{\mathrm{3}+\mathrm{tan}^{\mathrm{2}} \:\alpha}} \\ $$
Commented by mr W last updated on 27/Oct/25
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Commented by ajfour last updated on 26/Oct/25
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Commented by mr W last updated on 29/Oct/25
$${ses}\:{Q}\mathrm{225474}. \\ $$$${both}\:{chatgpt}\:{and}\:{gemini}\:{deliver} \\ $$$${different}\:{answer}\:{as}\:{i}.\:{they}\:{say} \\ $$$${T}=\sqrt{\frac{{L}}{\mathrm{3}{g}}}\int_{\mathrm{0}} ^{\theta} \frac{{d}\phi}{\:\sqrt{\mathrm{sin}\:\theta−\mathrm{sin}\:\phi}} \\ $$$${what}\:{do}\:{you}\:{think}? \\ $$
Commented by mr W last updated on 29/Oct/25
