Question Number 29418 by ajfour last updated on 08/Feb/18
Commented by ajfour last updated on 08/Feb/18
$${If}\:{sphere}\:\left({blue}\right)\:{starts}\:{rolling} \\ $$$${down}\:{from}\:{top}\:{of}\:{a}\:{semi}\:{cylindrical} \\ $$$${surface}\:\left({motion}\:{just}\:{initiated}\:{here}\right), \\ $$$${and}\:{coefficient}\:{of}\:{friction}\:{between} \\ $$$${them}\:{is}\:\boldsymbol{\mu},\:{find}\:{at}\:{what}\:{angle}\:\boldsymbol{\theta} \\ $$$${does}\:{slipping}\:{begin}\:? \\ $$
Answered by mrW2 last updated on 09/Feb/18
![N=normal force f=friction force CASE 1: the sphere keeps the contact. rotation of sphere ϕ=((Rθ)/r) ω_1 =(dθ/dt) α_1 =(dω_1 /dt) ω_2 =(dϕ/dt)=(R/r)ω_1 α_2 =(dω_2 /dt)=(R/r)α_1 m(R+r)α_1 =mg sin θ−f m(R+r)ω_1 ^2 =mg cos θ−N ⇒N=mg cos θ−m(R+r)ω_1 ^2 Iα_2 =fr ⇒((2mr^2 )/5)×(R/r)×α_1 =fr ⇒((2mR)/5)×α_1 =f ⇒α_1 =((5f)/(2mR)) m(R+r)×((5f)/(2mR))=mg sin θ−f ⇒[((7R+5r)/(2R))]f=mg sin θ ⇒f=((2mg sin θ R)/(7R+5r)) mg(R+r)(1−cos θ)=((mv^2 )/2)+((Iω_2 ^2 )/2) ⇒mg(R+r)(1−cos θ)=((m(R+r)^2 ω_1 ^2 )/2)+(1/2)×((2mr^2 )/5)×((R^2 ω_1 ^2 )/r^2 ) ⇒g(1−cos θ)=(((R+r)ω_1 ^2 )/2)+((R^2 ω_1 ^2 )/(5(R+r))) ⇒g(1−cos θ)=[((7R^2 +10Rr+5r^2 )/(10(R+r)))]ω_1 ^2 ⇒ω_1 ^2 =((10g(1−cos θ)(R+r))/(7R^2 +10Rr+5r^2 )) ⇒N=mg[cos θ−((10(1−cos θ)(R+r)^2 )/(7R^2 +10Rr+5r^2 ))] ⇒N=mg[(((7R^2 +10Rr+5r^2 )cos θ−10(1−cos θ)(R+r)^2 )/(7R^2 +10Rr+5r^2 ))] ⇒N=mg[(((17R^2 +30Rr+15r^2 )cos θ−10(R+r)^2 )/(7R^2 +10Rr+5r^2 ))] ∣(f/N)∣≤μ ((2mg sin θ R)/(7R+5r))×((7R^2 +10Rr+5r^2 )/(mg[(17R^2 +30Rr+15r^2 )cos θ−10(R+r)^2 ]))=μ ((2 sin θ R)/(7R+5r))×((7R^2 +10Rr+5r^2 )/((17R^2 +30Rr+15r^2 )cos θ−10(R+r)^2 ))=μ 2R(7R^2 +10Rr+5r^2 )sin θ=μ(7R+5r)(17R^2 +30Rr+15r^2 )cos θ−10μ(7R+5r)(R+r)^2 μ(7R+5r)(17R^2 +30Rr+15r^2 )cos θ−2R(7R^2 +10Rr+5r^2 )sin θ=10μ(7R+5r)(R+r)^2 with ψ=tan^(−1) [((μ(7R+5r)(17R^2 +30Rr+15r^2 ))/(2R(7R^2 +10Rr+5r^2 )))] sin ψ cos θ−cos ψ sin θ=((10μ(7R+5r)(R+r)^2 )/( (√([μ(7R+5r)(17R^2 +30Rr+15r^2 )]^2 +[2R(7R^2 +10Rr+5r^2 )]^2 )))) sin (ψ−θ)=((10μ(7R+5r)(R+r)^2 )/( (√(μ^2 (7R+5r)^2 (17R^2 +30Rr+15r^2 )^2 +4R^2 (7R^2 +10Rr+5r^2 )^2 )))) ψ−θ=sin^(−1) {((10μ(7R+5r)(R+r)^2 )/( (√(μ^2 (7R+5r)^2 (17R^2 +30Rr+15r^2 )^2 +4R^2 (7R^2 +10Rr+5r^2 )^2 ))))} θ=sin^(−1) {((10μ(7R+5r)(R+r)^2 )/( (√(μ^2 (7R+5r)^2 (17R^2 +30Rr+15r^2 )^2 +4R^2 (7R^2 +10Rr+5r^2 )^2 ))))}−ψ ⇒θ_(case 1) =sin^(−1) [((10μ(7R+5r)(R+r)^2 )/( (√(μ^2 (7R+5r)^2 (17R^2 +30Rr+15r^2 )^2 +4R^2 (7R^2 +10Rr+5r^2 )^2 ))))]−tan^(−1) [((μ(7R+5r)(17R^2 +30Rr+15r^2 ))/(2R(7R^2 +10Rr+5r^2 )))] with λ=(r/R) ⇒θ_(case 1) =sin^(−1) [((10μ(7+5λ)(1+λ)^2 )/( (√(μ^2 (7+5λ)^2 (17+30λ+15λ^2 )^2 +4(7+10λ+5λ^2 )^2 ))))]−tan^(−1) [((μ(7+5λ)(17+30λ+15λ^2 ))/(2(7+10λ+5λ^2 )))]](https://www.tinkutara.com/question/Q29426.png)
$${N}={normal}\:{force} \\ $$$${f}={friction}\:{force} \\ $$$${CASE}\:\mathrm{1}:\: \\ $$$${the}\:{sphere}\:{keeps}\:{the}\:{contact}. \\ $$$${rotation}\:{of}\:{sphere}\:\varphi=\frac{{R}\theta}{{r}} \\ $$$$\omega_{\mathrm{1}} =\frac{{d}\theta}{{dt}} \\ $$$$\alpha_{\mathrm{1}} =\frac{{d}\omega_{\mathrm{1}} }{{dt}} \\ $$$$\omega_{\mathrm{2}} =\frac{{d}\varphi}{{dt}}=\frac{{R}}{{r}}\omega_{\mathrm{1}} \\ $$$$\alpha_{\mathrm{2}} =\frac{{d}\omega_{\mathrm{2}} }{{dt}}=\frac{{R}}{{r}}\alpha_{\mathrm{1}} \\ $$$${m}\left({R}+{r}\right)\alpha_{\mathrm{1}} ={mg}\:\mathrm{sin}\:\theta−{f} \\ $$$${m}\left({R}+{r}\right)\omega_{\mathrm{1}} ^{\mathrm{2}} ={mg}\:\mathrm{cos}\:\theta−{N} \\ $$$$\Rightarrow{N}={mg}\:\mathrm{cos}\:\theta−{m}\left({R}+{r}\right)\omega_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${I}\alpha_{\mathrm{2}} ={fr} \\ $$$$\Rightarrow\frac{\mathrm{2}{mr}^{\mathrm{2}} }{\mathrm{5}}×\frac{{R}}{{r}}×\alpha_{\mathrm{1}} ={fr} \\ $$$$\Rightarrow\frac{\mathrm{2}{mR}}{\mathrm{5}}×\alpha_{\mathrm{1}} ={f} \\ $$$$\Rightarrow\alpha_{\mathrm{1}} =\frac{\mathrm{5}{f}}{\mathrm{2}{mR}} \\ $$$${m}\left({R}+{r}\right)×\frac{\mathrm{5}{f}}{\mathrm{2}{mR}}={mg}\:\mathrm{sin}\:\theta−{f} \\ $$$$\Rightarrow\left[\frac{\mathrm{7}{R}+\mathrm{5}{r}}{\mathrm{2}{R}}\right]{f}={mg}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow{f}=\frac{\mathrm{2}{mg}\:\mathrm{sin}\:\theta\:{R}}{\mathrm{7}{R}+\mathrm{5}{r}} \\ $$$${mg}\left({R}+{r}\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right)=\frac{{mv}^{\mathrm{2}} }{\mathrm{2}}+\frac{{I}\omega_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{mg}\left({R}+{r}\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right)=\frac{{m}\left({R}+{r}\right)^{\mathrm{2}} \omega_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}{mr}^{\mathrm{2}} }{\mathrm{5}}×\frac{{R}^{\mathrm{2}} \omega_{\mathrm{1}} ^{\mathrm{2}} }{{r}^{\mathrm{2}} } \\ $$$$\Rightarrow{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)=\frac{\left({R}+{r}\right)\omega_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}}+\frac{{R}^{\mathrm{2}} \omega_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{5}\left({R}+{r}\right)} \\ $$$$\Rightarrow{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)=\left[\frac{\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} }{\mathrm{10}\left({R}+{r}\right)}\right]\omega_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Rightarrow\omega_{\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{10}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left({R}+{r}\right)}{\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} } \\ $$$$\Rightarrow{N}={mg}\left[\mathrm{cos}\:\theta−\frac{\mathrm{10}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left({R}+{r}\right)^{\mathrm{2}} }{\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} }\right] \\ $$$$\Rightarrow{N}={mg}\left[\frac{\left(\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} \right)\mathrm{cos}\:\theta−\mathrm{10}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left({R}+{r}\right)^{\mathrm{2}} }{\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} }\right] \\ $$$$\Rightarrow{N}={mg}\left[\frac{\left(\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} \right)\mathrm{cos}\:\theta−\mathrm{10}\left({R}+{r}\right)^{\mathrm{2}} }{\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} }\right] \\ $$$$\mid\frac{{f}}{{N}}\mid\leqslant\mu \\ $$$$\frac{\mathrm{2}{mg}\:\mathrm{sin}\:\theta\:{R}}{\mathrm{7}{R}+\mathrm{5}{r}}×\frac{\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} }{{mg}\left[\left(\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} \right)\mathrm{cos}\:\theta−\mathrm{10}\left({R}+{r}\right)^{\mathrm{2}} \right]}=\mu \\ $$$$\frac{\mathrm{2}\:\mathrm{sin}\:\theta\:{R}}{\mathrm{7}{R}+\mathrm{5}{r}}×\frac{\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} }{\left(\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} \right)\mathrm{cos}\:\theta−\mathrm{10}\left({R}+{r}\right)^{\mathrm{2}} }=\mu \\ $$$$\mathrm{2}{R}\left(\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} \right)\mathrm{sin}\:\theta=\mu\left(\mathrm{7}{R}+\mathrm{5}{r}\right)\left(\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} \right)\mathrm{cos}\:\theta−\mathrm{10}\mu\left(\mathrm{7}{R}+\mathrm{5}{r}\right)\left({R}+{r}\right)^{\mathrm{2}} \\ $$$$\mu\left(\mathrm{7}{R}+\mathrm{5}{r}\right)\left(\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} \right)\mathrm{cos}\:\theta−\mathrm{2}{R}\left(\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} \right)\mathrm{sin}\:\theta=\mathrm{10}\mu\left(\mathrm{7}{R}+\mathrm{5}{r}\right)\left({R}+{r}\right)^{\mathrm{2}} \\ $$$${with}\:\psi=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mu\left(\mathrm{7}{R}+\mathrm{5}{r}\right)\left(\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} \right)}{\mathrm{2}{R}\left(\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} \right)}\right] \\ $$$$\mathrm{sin}\:\psi\:\mathrm{cos}\:\theta−\mathrm{cos}\:\psi\:\mathrm{sin}\:\theta=\frac{\mathrm{10}\mu\left(\mathrm{7}{R}+\mathrm{5}{r}\right)\left({R}+{r}\right)^{\mathrm{2}} }{\:\sqrt{\left[\mu\left(\mathrm{7}{R}+\mathrm{5}{r}\right)\left(\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} \right)\right]^{\mathrm{2}} +\left[\mathrm{2}{R}\left(\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} \right)\right]^{\mathrm{2}} }} \\ $$$$\mathrm{sin}\:\left(\psi−\theta\right)=\frac{\mathrm{10}\mu\left(\mathrm{7}{R}+\mathrm{5}{r}\right)\left({R}+{r}\right)^{\mathrm{2}} }{\:\sqrt{\mu^{\mathrm{2}} \left(\mathrm{7}{R}+\mathrm{5}{r}\right)^{\mathrm{2}} \left(\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{2}} \left(\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$\psi−\theta=\mathrm{sin}^{−\mathrm{1}} \left\{\frac{\mathrm{10}\mu\left(\mathrm{7}{R}+\mathrm{5}{r}\right)\left({R}+{r}\right)^{\mathrm{2}} }{\:\sqrt{\mu^{\mathrm{2}} \left(\mathrm{7}{R}+\mathrm{5}{r}\right)^{\mathrm{2}} \left(\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{2}} \left(\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} \right)^{\mathrm{2}} }}\right\} \\ $$$$\theta=\mathrm{sin}^{−\mathrm{1}} \left\{\frac{\mathrm{10}\mu\left(\mathrm{7}{R}+\mathrm{5}{r}\right)\left({R}+{r}\right)^{\mathrm{2}} }{\:\sqrt{\mu^{\mathrm{2}} \left(\mathrm{7}{R}+\mathrm{5}{r}\right)^{\mathrm{2}} \left(\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{2}} \left(\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} \right)^{\mathrm{2}} }}\right\}−\psi \\ $$$$\Rightarrow\theta_{{case}\:\mathrm{1}} =\mathrm{sin}^{−\mathrm{1}} \left[\frac{\mathrm{10}\mu\left(\mathrm{7}{R}+\mathrm{5}{r}\right)\left({R}+{r}\right)^{\mathrm{2}} }{\:\sqrt{\mu^{\mathrm{2}} \left(\mathrm{7}{R}+\mathrm{5}{r}\right)^{\mathrm{2}} \left(\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{2}} \left(\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} \right)^{\mathrm{2}} }}\right]−\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mu\left(\mathrm{7}{R}+\mathrm{5}{r}\right)\left(\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} \right)}{\mathrm{2}{R}\left(\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} \right)}\right] \\ $$$${with}\:\lambda=\frac{{r}}{{R}} \\ $$$$\Rightarrow\theta_{{case}\:\mathrm{1}} =\mathrm{sin}^{−\mathrm{1}} \left[\frac{\mathrm{10}\mu\left(\mathrm{7}+\mathrm{5}\lambda\right)\left(\mathrm{1}+\lambda\right)^{\mathrm{2}} }{\:\sqrt{\mu^{\mathrm{2}} \left(\mathrm{7}+\mathrm{5}\lambda\right)^{\mathrm{2}} \left(\mathrm{17}+\mathrm{30}\lambda+\mathrm{15}\lambda^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{7}+\mathrm{10}\lambda+\mathrm{5}\lambda^{\mathrm{2}} \right)^{\mathrm{2}} }}\right]−\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mu\left(\mathrm{7}+\mathrm{5}\lambda\right)\left(\mathrm{17}+\mathrm{30}\lambda+\mathrm{15}\lambda^{\mathrm{2}} \right)}{\mathrm{2}\left(\mathrm{7}+\mathrm{10}\lambda+\mathrm{5}\lambda^{\mathrm{2}} \right)}\right] \\ $$
Commented by ajfour last updated on 09/Feb/18
$${Thanks}\:{Sir}.\:\mathbb{INCREDIBLE}\:! \\ $$
Commented by mrW2 last updated on 09/Feb/18
![CASE 2: at some angle θ the sphere losses contact m(R+r)ω_1 ^2 =mg cos θ−N with N=0 ω_1 ^2 =((10g(1−cos θ)(R+r))/(7R^2 +10Rr+5r^2 )) ((10mg(1−cos θ)(R+r)^2 )/(7R^2 +10Rr+5r^2 ))=mgcos θ [1+((7R^2 +10Rr+5r^2 )/(10(R+r)^2 ))]cos θ=1 cos θ=((10(R+r)^2 )/(17R^2 +30Rr+15r^2 )) ⇒θ_(case 2) =cos^(−1) [((10(R+r)^2 )/(17R^2 +30Rr+15r^2 ))] ⇒θ_(case 2) =cos^(−1) [((10(1+λ)^2 )/(17+30λ+15λ^2 ))] ⇒θ=min(θ_(case 1) , θ_(case 2) )](https://www.tinkutara.com/question/Q29474.png)
$${CASE}\:\mathrm{2}:\: \\ $$$${at}\:{some}\:{angle}\:\theta\:{the}\:{sphere}\:{losses}\:{contact} \\ $$$${m}\left({R}+{r}\right)\omega_{\mathrm{1}} ^{\mathrm{2}} ={mg}\:\mathrm{cos}\:\theta−{N}\:{with}\:{N}=\mathrm{0} \\ $$$$\omega_{\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{10}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left({R}+{r}\right)}{\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{10}{mg}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left({R}+{r}\right)^{\mathrm{2}} }{\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} }={mg}\mathrm{cos}\:\theta \\ $$$$\left[\mathrm{1}+\frac{\mathrm{7}{R}^{\mathrm{2}} +\mathrm{10}{Rr}+\mathrm{5}{r}^{\mathrm{2}} }{\mathrm{10}\left({R}+{r}\right)^{\mathrm{2}} }\right]\mathrm{cos}\:\theta=\mathrm{1} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{10}\left({R}+{r}\right)^{\mathrm{2}} }{\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\theta_{{case}\:\mathrm{2}} =\mathrm{cos}^{−\mathrm{1}} \left[\frac{\mathrm{10}\left({R}+{r}\right)^{\mathrm{2}} }{\mathrm{17}{R}^{\mathrm{2}} +\mathrm{30}{Rr}+\mathrm{15}{r}^{\mathrm{2}} }\right] \\ $$$$\Rightarrow\theta_{{case}\:\mathrm{2}} =\mathrm{cos}^{−\mathrm{1}} \left[\frac{\mathrm{10}\left(\mathrm{1}+\lambda\right)^{\mathrm{2}} }{\mathrm{17}+\mathrm{30}\lambda+\mathrm{15}\lambda^{\mathrm{2}} }\right] \\ $$$$ \\ $$$$\Rightarrow\theta={min}\left(\theta_{{case}\:\mathrm{1}} ,\:\theta_{{case}\:\mathrm{2}} \right) \\ $$