Question Number 226217 by mr W last updated on 22/Nov/25
Commented by mr W last updated on 22/Nov/25
Commented by fantastic2 last updated on 22/Nov/25
$${But}\:{if}\:{M}\geqslant\geqslant{m}\:{we}\:{can}\:{say}\:{the}\:{hemisphere}\:{will} \\ $$$${not}\:{move}\:{and}\:\theta\:{will}\:{be}\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$
Answered by mr W last updated on 23/Nov/25
Commented by mr W last updated on 23/Nov/25
![say the particle moves relatively to hemisphere with velocity v. let ω=(dθ/dt) v=Rω momentum conservation in x−direction: MV=m(v cos θ−V) (M+m)V=mv cos θ=mRω cos θ let λ=(m/(M+m)) ⇒V=λRω cos θ A=(dV/dt)=ω(dV/dθ)=λR(cos θ ((ωdω)/dθ)−ω^2 sin θ) energy conservation: ((MV^2 )/2)+(m/2)[v^2 sin^2 θ+(v cos θ−V)^2 ]=mgR(1−cos θ) ((M/m)+1)V^2 +v^2 −2Vv cos θ=2gR(1−cos θ) λR^2 ω^2 cos^2 θ+R^2 ω^2 −2λR^2 ω^2 cos^2 θ=2gR(1−cos θ) (1−λ cos^2 θ)ω^2 =((2g)/R)(1−cos θ) ⇒ω^2 =((2g(1−cos θ))/(R(1−λ cos^2 θ))) 2λ cos θ sin θ ω^2 +2(1−λ cos^2 θ)ω(dω/dθ)=((2g)/R) sin θ λ cos θ sin θ ω^2 +(1−λ cos^2 θ)ω(dω/dθ)=(g/R) sin θ λ cos θ sin θ ((2g(1−cos θ))/(R(1−λ cos^2 θ))) +(1−λ cos^2 θ)ω(dω/dθ)=(g/R) sin θ ⇒ω(dω/dθ)= ((g sin θ(λ cos^2 θ−2λ cos θ+1))/(R(1−λ cos^2 θ)^2 )) N sin θ=MA N sin θ=MλR(cos θ ((ωdω)/dθ)−ω^2 sin θ) N=0: ⇒cos θ ((ωdω)/dθ)−ω^2 sin θ=0 ⇒ω(dω/dθ)=((ω^2 sin θ)/(cos θ)) ((λ cos^2 θ−2λ cos θ+1)/(1−λ cos^2 θ))=((2(1−cos θ))/(cos θ)) λ cos^3 θ−3 cos θ+2=0 ⇒cos θ=(2/( (√λ))) sin ((1/3)sin^(−1) (√λ)) ⇒θ=cos^(−1) [(2/( (√λ))) sin ((1/3)sin^(−1) (√λ))] if M≫m, λ→0 θ=lim_(λ→0) cos^(−1) [(2/( (√λ))) sin ((1/3)sin^(−1) (√λ))] =lim_(λ→0) cos^(−1) [(2/( 3))×((sin ((1/3)sin^(−1) (√λ)))/((1/3)(√λ)))] =cos^(−1) (2/( 3)) (≈53.5541°) this is the case when hemisphere is fixed on the ground. N=MλR(cos θ ω(dω/dθ)−ω^2 sin θ)(1/(sin θ)) =MλR(((cos θ)/(sin θ))×((ωdω)/dθ)−ω^2 ) =MλR(((cos θ)/(sin θ))×((g sin θ(λ cos^2 θ−2λ cos θ+1))/(R(1−λ cos^2 θ)^2 ))−((2g(1−cos θ))/(R(1−λ cos^2 θ)))) =Mλg×((cos θ(λ cos^2 θ−2λ cos θ+1)−2(1−cos θ)(1−λ cos^2 θ))/((1−λ cos^2 θ)^2 )) =((λMg(−2+3cos θ−λ cos^3 θ))/((1−λ cos^2 θ)^2 )) A=((N sin θ)/M) =((λg(−2+3cos θ−λ cos^3 θ) sin θ)/((1−λ cos^2 θ)^2 )) ω=(dθ/dt)=(√((2g(1−cos θ))/(R(1−λ cos^2 θ)))) dt=(√((R(1−λ cos^2 θ))/(2g(1−cos θ))))dθ ⇒t=(√(R/(2g)))∫_θ_0 ^θ (√((1−λ cos^2 θ)/(1−cos θ)))dθ example: λ=(m/(M+m))=(1/4) θ_m =cos^(−1) [4 sin ((1/3)sin^(−1) (1/2))] =cos^(−1) [4 sin ((π/(18)))]≈0.8029≈46.005°](https://www.tinkutara.com/question/Q226236.png)
$${say}\:{the}\:{particle}\:{moves}\:{relatively}\:{to} \\ $$$${hemisphere}\:{with}\:{velocity}\:{v}. \\ $$$${let}\:\omega=\frac{{d}\theta}{{dt}} \\ $$$${v}={R}\omega \\ $$$${momentum}\:{conservation}\:{in}\:{x}−{direction}: \\ $$$${MV}={m}\left({v}\:\mathrm{cos}\:\theta−{V}\right) \\ $$$$\left({M}+{m}\right){V}={mv}\:\mathrm{cos}\:\theta={mR}\omega\:\mathrm{cos}\:\theta \\ $$$${let}\:\lambda=\frac{{m}}{{M}+{m}} \\ $$$$\Rightarrow{V}=\lambda{R}\omega\:\mathrm{cos}\:\theta \\ $$$${A}=\frac{{dV}}{{dt}}=\omega\frac{{dV}}{{d}\theta}=\lambda{R}\left(\mathrm{cos}\:\theta\:\frac{\omega{d}\omega}{{d}\theta}−\omega^{\mathrm{2}} \:\mathrm{sin}\:\theta\right) \\ $$$${energy}\:{conservation}: \\ $$$$\frac{{MV}^{\mathrm{2}} }{\mathrm{2}}+\frac{{m}}{\mathrm{2}}\left[{v}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta+\left({v}\:\mathrm{cos}\:\theta−{V}\right)^{\mathrm{2}} \right]={mgR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\left(\frac{{M}}{{m}}+\mathrm{1}\right){V}^{\mathrm{2}} +{v}^{\mathrm{2}} −\mathrm{2}{Vv}\:\mathrm{cos}\:\theta=\mathrm{2}{gR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\lambda{R}^{\mathrm{2}} \omega^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta+{R}^{\mathrm{2}} \omega^{\mathrm{2}} −\mathrm{2}\lambda{R}^{\mathrm{2}} \omega^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta=\mathrm{2}{gR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)\omega^{\mathrm{2}} =\frac{\mathrm{2}{g}}{{R}}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{2}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)} \\ $$$$\mathrm{2}\lambda\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\:\omega^{\mathrm{2}} \:+\mathrm{2}\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)\omega\frac{{d}\omega}{{d}\theta}=\frac{\mathrm{2}{g}}{{R}}\:\mathrm{sin}\:\theta \\ $$$$\lambda\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\:\omega^{\mathrm{2}} \:+\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)\omega\frac{{d}\omega}{{d}\theta}=\frac{{g}}{{R}}\:\mathrm{sin}\:\theta \\ $$$$\lambda\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\:\frac{\mathrm{2}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)}\:+\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)\omega\frac{{d}\omega}{{d}\theta}=\frac{{g}}{{R}}\:\mathrm{sin}\:\theta \\ $$$$\:\Rightarrow\omega\frac{{d}\omega}{{d}\theta}=\:\frac{{g}\:\mathrm{sin}\:\theta\left(\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{2}\lambda\:\mathrm{cos}\:\theta+\mathrm{1}\right)}{{R}\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} } \\ $$$${N}\:\mathrm{sin}\:\theta={MA} \\ $$$${N}\:\mathrm{sin}\:\theta={M}\lambda{R}\left(\mathrm{cos}\:\theta\:\frac{\omega{d}\omega}{{d}\theta}−\omega^{\mathrm{2}} \:\mathrm{sin}\:\theta\right) \\ $$$${N}=\mathrm{0}: \\ $$$$\Rightarrow\mathrm{cos}\:\theta\:\frac{\omega{d}\omega}{{d}\theta}−\omega^{\mathrm{2}} \:\mathrm{sin}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\omega\frac{{d}\omega}{{d}\theta}=\frac{\omega^{\mathrm{2}} \:\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta} \\ $$$$\frac{\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{2}\lambda\:\mathrm{cos}\:\theta+\mathrm{1}}{\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta}=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{cos}\:\theta} \\ $$$$\lambda\:\mathrm{cos}^{\mathrm{3}} \:\theta−\mathrm{3}\:\mathrm{cos}\:\theta+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{2}}{\:\sqrt{\lambda}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \sqrt{\lambda}\right) \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \left[\frac{\mathrm{2}}{\:\sqrt{\lambda}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \sqrt{\lambda}\right)\right] \\ $$$$ \\ $$$${if}\:{M}\gg{m},\:\lambda\rightarrow\mathrm{0} \\ $$$$\theta=\underset{\lambda\rightarrow\mathrm{0}} {\mathrm{lim}cos}^{−\mathrm{1}} \left[\frac{\mathrm{2}}{\:\sqrt{\lambda}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \sqrt{\lambda}\right)\right] \\ $$$$\:=\underset{\lambda\rightarrow\mathrm{0}} {\mathrm{lim}cos}^{−\mathrm{1}} \left[\frac{\mathrm{2}}{\:\mathrm{3}}×\frac{\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \sqrt{\lambda}\right)}{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\lambda}}\right] \\ $$$$\:=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\:\mathrm{3}}\:\:\:\left(\approx\mathrm{53}.\mathrm{5541}°\right) \\ $$$${this}\:{is}\:{the}\:{case}\:{when}\:{hemisphere}\:{is} \\ $$$${fixed}\:{on}\:{the}\:{ground}. \\ $$$$ \\ $$$${N}={M}\lambda{R}\left(\mathrm{cos}\:\theta\:\omega\frac{{d}\omega}{{d}\theta}−\omega^{\mathrm{2}} \:\mathrm{sin}\:\theta\right)\frac{\mathrm{1}}{\mathrm{sin}\:\theta} \\ $$$$\:\:\:={M}\lambda{R}\left(\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}×\frac{\omega{d}\omega}{{d}\theta}−\omega^{\mathrm{2}} \right) \\ $$$$\:\:\:={M}\lambda{R}\left(\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}×\frac{{g}\:\mathrm{sin}\:\theta\left(\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{2}\lambda\:\mathrm{cos}\:\theta+\mathrm{1}\right)}{{R}\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} }−\frac{\mathrm{2}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)}\right) \\ $$$$\:\:\:={M}\lambda{g}×\frac{\mathrm{cos}\:\theta\left(\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{2}\lambda\:\mathrm{cos}\:\theta+\mathrm{1}\right)−\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)}{\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\lambda{Mg}\left(−\mathrm{2}+\mathrm{3cos}\:\theta−\lambda\:\mathrm{cos}^{\mathrm{3}} \:\theta\right)}{\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${A}=\frac{{N}\:\mathrm{sin}\:\theta}{{M}} \\ $$$$\:\:\:=\frac{\lambda{g}\left(−\mathrm{2}+\mathrm{3cos}\:\theta−\lambda\:\mathrm{cos}^{\mathrm{3}} \:\theta\right)\:\mathrm{sin}\:\theta}{\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\omega=\frac{{d}\theta}{{dt}}=\sqrt{\frac{\mathrm{2}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{R}\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)}} \\ $$$${dt}=\sqrt{\frac{{R}\left(\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta\right)}{\mathrm{2}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}}{d}\theta \\ $$$$\Rightarrow{t}=\sqrt{\frac{{R}}{\mathrm{2}{g}}}\int_{\theta_{\mathrm{0}} } ^{\theta} \sqrt{\frac{\mathrm{1}−\lambda\:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{1}−\mathrm{cos}\:\theta}}{d}\theta \\ $$$$ \\ $$$${example}:\:\lambda=\frac{{m}}{{M}+{m}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\theta_{{m}} =\mathrm{cos}^{−\mathrm{1}} \left[\mathrm{4}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right)\right] \\ $$$$\:\:\:\:\:=\mathrm{cos}^{−\mathrm{1}} \left[\mathrm{4}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{18}}\right)\right]\approx\mathrm{0}.\mathrm{8029}\approx\mathrm{46}.\mathrm{005}° \\ $$
Commented by fantastic2 last updated on 23/Nov/25
$${wow}\:{sir}\:{really}\:{good} \\ $$
Answered by ajfour last updated on 23/Nov/25
$$\phi\:{is}\:{angle}\:{from}\:{vertical} \\ $$$$\left({M}+{m}\right){V}={mu}\mathrm{cos}\:\phi \\ $$$${gR}\mathrm{cos}\:\phi={u}^{\mathrm{2}} \\ $$$$\mathrm{2}{gR}\left(\mathrm{1}−\mathrm{cos}\:\phi\right)={u}^{\mathrm{2}} +\left(\frac{{M}+{m}}{{m}}\right){V}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{uV}\mathrm{cos}\:\phi \\ $$$${so} \\ $$$$\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\phi\right)=\mathrm{cos}\:\phi−\left(\frac{{m}}{{M}+{m}}\right)\mathrm{cos}\:^{\mathrm{3}} \phi \\ $$$${so} \\ $$$$\left(\frac{{m}}{{M}+{m}}\right)\mathrm{cos}\:^{\mathrm{3}} \phi−\mathrm{3cos}\:\phi+\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by mr W last updated on 23/Nov/25
$${sir}:\:{please}\:{try}\:{Q}\mathrm{226245}.\:{i}'{m}\:{unsure}, \\ $$$${because}\:{the}\:{youtube}\:{channel},\:{from} \\ $$$${which}\:{the}\:{question}\:{came},\:{has} \\ $$$${presented}\:{a}\:{different}\:{solution}\:{which} \\ $$$${i}\:{don}'{t}\:{agree}\:{with}. \\ $$
Commented by mr W last updated on 23/Nov/25
$${perfect}! \\ $$$${you}\:{treat}\:{directly}\:{the}\:{special}\:{case} \\ $$$${N}=\mathrm{0}.\:{i}\:{usually}\:{prefer}\:{to}\:{treat}\:{the} \\ $$$${general}\:{case}\:{in}\:{order}\:{to}\:{get} \\ $$$${N}\left(\theta\right),\:{V}\left(\theta\right),\:{A}\left(\theta\right),\:{v}\left(\theta\right)\:{etc}.\:{and}\:{then}\: \\ $$$${set}\:{N}\left(\theta\right)=\mathrm{0}\:{for}\:{the}\:{special}\:{case}. \\ $$