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Question Number 70423 by ajfour last updated on 04/Oct/19
Commented by ajfour last updated on 04/Oct/19
Within a hollow sphere, mass M_0 , radius R, rests two smaller spheres masses, M and m, radii b, a , respectively. Find equilibrium angles α, β.
$${Within}\:{a}\:{hollow}\:{sphere},\:{mass} \\ $$$${M}_{\mathrm{0}} \:,\:{radius}\:{R},\:{rests}\:{two}\:{smaller} \\ $$$${spheres}\:{masses},\:{M}\:{and}\:{m},\:{radii} \\ $$$${b},\:{a}\:,\:{respectively}.\:{Find}\:{equilibrium} \\ $$$${angles}\:\alpha,\:\beta. \\ $$
Answered by ajfour last updated on 04/Oct/19
Commented by ajfour last updated on 04/Oct/19
Let the normal reaction between the two smaller spheres be F. Rsin γ=(R−b)cos α−(R−a)cos β ....(i) Rcos γ=(R−b)sin α+(R−a)sin β ....(ii) From concurrency of forces F = ((Mgsin α)/(cos (γ+α)))=((mgsin β)/(cos (β−γ))) ⇒ M((1/(tan β))+ tan γ) = m((1/(tan α))−tan γ) ⇒ tan γ = ((mcot α−Mcot β)/(m+M)) ..(I) Also from (i) & (ii) tan γ = (((R−b)cos α−(R−a)cos β)/((R−b)sin α+(R−a)sin β)) ......(II) let R−a=p , R−b=q , M=μm, r=a+b from cosine rule in △ACB cos (α+β)=((p^2 +q^2 −r^2 )/(2pq)) = k ..(1) And from (I) & (II) ((cot α−μcot β)/(1+μ))=((qcos α−pcos β)/(qsin α+psin β)) ⇒ (qsin α+psin β)(sin βcos α−μcos βsin α) =(1+μ)(qcos α−pcos β)(sin αsin β) ⇒ qsin αcos αsin β−μqsin^2 αcos β +pcos αsin^2 β−μpsin αsin βcos β =(1+μ)qsin αcos αsin β −(1+μ)psin αsin βcos β ⇒ μqsin αcos αsin β−psin αsin βcos β =pcos αsin^2 β−μqsin^2 αcos β ⇒ (μqsin α)sin (α+β) =psin β sin (α+β) ⇒ ((sin α)/(sin β)) = (p/(μq)) (has to be a short way to this) Now using (1) cos α(√(1−sin^2 β))−sin αsin β =k ⇒ cos α(√(1−((μ^2 q^2 )/p^2 )sin^2 α)) −((μq)/p) sin^2 α = k (1−sin^2 α)(1−((μ^2 q^2 )/p^2 )sin^2 α) = (k+((μq)/p)sin^2 α)^2 lets call sin α=x , ((μq)/p)=c ⇒ 1−(1+c^2 )x^2 =k^2 +2ckx^2 ⇒ x^2 =((1−k^2 )/((c^2 +2ck+1))) y^2 = c^2 x^2 = ((c^2 (1−k^2 ))/((c^2 +2ck+1))) __________________________ ⇒ 𝛂=sin^(−1) (√((1−k^2 )/(c^2 +2ck+1))) 𝛃=sin^(−1) (√((c^2 (1−k^2 ))/(c^2 +2ck+1))) c = ((M(R−b))/(m(R−a))) k=(((R−a)^2 +(R−b)^2 −(a+b)^2 )/(2(R−a)(R−b))) _________________________■ If as an example μ=8 , b=2a = (R/2) (same densities) c=((16)/3) , k=(((9/(16))+(4/(16))−(9/(16)))/((12)/(16)))= (1/3) sin α=(√((1−(1/9))/(((256)/9)+2((1/3))(((16)/3))+1))) α = sin^(−1) (((2(√2))/( (√(297)))))≈ 9.446° β = sin^(−1) ((√((2048)/(2673)))) ≈ 61.082° .
$${Let}\:{the}\:{normal}\:{reaction}\:{between} \\ $$$${the}\:{two}\:{smaller}\:{spheres}\:{be}\:{F}. \\ $$$${R}\mathrm{sin}\:\gamma=\left({R}−{b}\right)\mathrm{cos}\:\alpha−\left({R}−{a}\right)\mathrm{cos}\:\beta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({i}\right) \\ $$$${R}\mathrm{cos}\:\gamma=\left({R}−{b}\right)\mathrm{sin}\:\alpha+\left({R}−{a}\right)\mathrm{sin}\:\beta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({ii}\right) \\ $$$$\:{From}\:{concurrency}\:{of}\:{forces} \\ $$$$\:{F}\:=\:\frac{{Mg}\mathrm{sin}\:\alpha}{\mathrm{cos}\:\left(\gamma+\alpha\right)}=\frac{{mg}\mathrm{sin}\:\beta}{\mathrm{cos}\:\left(\beta−\gamma\right)} \\ $$$$\Rightarrow\:{M}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\beta}+\:\mathrm{tan}\:\gamma\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:{m}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}−\mathrm{tan}\:\gamma\right) \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\gamma\:=\:\frac{{m}\mathrm{cot}\:\alpha−{M}\mathrm{cot}\:\beta}{{m}+{M}}\:\:..\left({I}\right) \\ $$$${Also}\:{from}\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$$\mathrm{tan}\:\gamma\:=\:\frac{\left({R}−{b}\right)\mathrm{cos}\:\alpha−\left({R}−{a}\right)\mathrm{cos}\:\beta}{\left({R}−{b}\right)\mathrm{sin}\:\alpha+\left({R}−{a}\right)\mathrm{sin}\:\beta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……\left({II}\right) \\ $$$${let}\:{R}−{a}={p}\:,\:{R}−{b}={q}\:,\:{M}=\mu{m}, \\ $$$$\:\:\:\:\:\:{r}={a}+{b} \\ $$$$\:\:\:\:{from}\:{cosine}\:{rule}\:{in}\:\bigtriangleup{ACB} \\ $$$$\:\:\:\mathrm{cos}\:\left(\alpha+\beta\right)=\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{pq}}\:=\:\boldsymbol{{k}}\:\:\:..\left(\mathrm{1}\right) \\ $$$${And}\:{from}\:\left({I}\right)\:\&\:\left({II}\right) \\ $$$$\frac{\mathrm{cot}\:\alpha−\mu\mathrm{cot}\:\beta}{\mathrm{1}+\mu}=\frac{{q}\mathrm{cos}\:\alpha−{p}\mathrm{cos}\:\beta}{{q}\mathrm{sin}\:\alpha+{p}\mathrm{sin}\:\beta} \\ $$$$\Rightarrow\:\:\left({q}\mathrm{sin}\:\alpha+{p}\mathrm{sin}\:\beta\right)\left(\mathrm{sin}\:\beta\mathrm{cos}\:\alpha−\mu\mathrm{cos}\:\beta\mathrm{sin}\:\alpha\right) \\ $$$$\:\:\:=\left(\mathrm{1}+\mu\right)\left({q}\mathrm{cos}\:\alpha−{p}\mathrm{cos}\:\beta\right)\left(\mathrm{sin}\:\alpha\mathrm{sin}\:\beta\right) \\ $$$$\Rightarrow \\ $$$$\:{q}\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha\mathrm{sin}\:\beta−\mu{q}\mathrm{sin}\:^{\mathrm{2}} \alpha\mathrm{cos}\:\beta \\ $$$$+{p}\mathrm{cos}\:\alpha\mathrm{sin}\:^{\mathrm{2}} \beta−\mu{p}\mathrm{sin}\:\alpha\mathrm{sin}\:\beta\mathrm{cos}\:\beta \\ $$$$=\left(\mathrm{1}+\mu\right){q}\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha\mathrm{sin}\:\beta \\ $$$$\:\:\:\:\:\:\:\:−\left(\mathrm{1}+\mu\right){p}\mathrm{sin}\:\alpha\mathrm{sin}\:\beta\mathrm{cos}\:\beta \\ $$$$\Rightarrow \\ $$$$\mu{q}\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha\mathrm{sin}\:\beta−{p}\mathrm{sin}\:\alpha\mathrm{sin}\:\beta\mathrm{cos}\:\beta \\ $$$$\:\:\:\:={p}\mathrm{cos}\:\alpha\mathrm{sin}\:^{\mathrm{2}} \beta−\mu{q}\mathrm{sin}\:^{\mathrm{2}} \alpha\mathrm{cos}\:\beta \\ $$$$\Rightarrow \\ $$$$\:\:\left(\mu{q}\mathrm{sin}\:\alpha\right)\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$$$\:\:\:\:\:\:\:\:={p}\mathrm{sin}\:\beta\:\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$$$\Rightarrow\:\:\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\beta}\:=\:\frac{{p}}{\mu{q}}\:\:\:\:\left({has}\:{to}\:{be}\:{a}\:{short}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{way}\:{to}\:{this}\right) \\ $$$${Now}\:{using}\:\left(\mathrm{1}\right) \\ $$$$\:\:\mathrm{cos}\:\alpha\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \beta}−\mathrm{sin}\:\alpha\mathrm{sin}\:\beta\:={k} \\ $$$$\Rightarrow \\ $$$$\:\:\:\mathrm{cos}\:\alpha\sqrt{\mathrm{1}−\frac{\mu^{\mathrm{2}} {q}^{\mathrm{2}} }{{p}^{\mathrm{2}} }\mathrm{sin}\:^{\mathrm{2}} \alpha} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mu{q}}{{p}}\:\mathrm{sin}\:^{\mathrm{2}} \alpha\:=\:{k} \\ $$$$\:\:\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \alpha\right)\left(\mathrm{1}−\frac{\mu^{\mathrm{2}} {q}^{\mathrm{2}} }{{p}^{\mathrm{2}} }\mathrm{sin}\:^{\mathrm{2}} \alpha\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\left({k}+\frac{\mu{q}}{{p}}\mathrm{sin}\:^{\mathrm{2}} \alpha\right)^{\mathrm{2}} \\ $$$$\:{lets}\:{call}\:\:\mathrm{sin}\:\alpha={x}\:,\:\:\frac{\mu{q}}{{p}}={c} \\ $$$$\Rightarrow\:\:\mathrm{1}−\left(\mathrm{1}+{c}^{\mathrm{2}} \right){x}^{\mathrm{2}} ={k}^{\mathrm{2}} +\mathrm{2}{ckx}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:{x}^{\mathrm{2}} =\frac{\mathrm{1}−{k}^{\mathrm{2}} }{\left({c}^{\mathrm{2}} +\mathrm{2}{ck}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} =\:{c}^{\mathrm{2}} {x}^{\mathrm{2}} \:=\:\frac{{c}^{\mathrm{2}} \left(\mathrm{1}−{k}^{\mathrm{2}} \right)}{\left({c}^{\mathrm{2}} +\mathrm{2}{ck}+\mathrm{1}\right)} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Rightarrow\:\:\boldsymbol{\alpha}=\mathrm{sin}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}−{k}^{\mathrm{2}} }{{c}^{\mathrm{2}} +\mathrm{2}{ck}+\mathrm{1}}} \\ $$$$\:\:\:\:\:\:\boldsymbol{\beta}=\mathrm{sin}^{−\mathrm{1}} \sqrt{\frac{{c}^{\mathrm{2}} \left(\mathrm{1}−{k}^{\mathrm{2}} \right)}{{c}^{\mathrm{2}} +\mathrm{2}{ck}+\mathrm{1}}} \\ $$$$\:\:\:\:\:{c}\:=\:\frac{{M}\left({R}−{b}\right)}{{m}\left({R}−{a}\right)}\: \\ $$$$\:\:\:\:\:{k}=\frac{\left({R}−{a}\right)^{\mathrm{2}} +\left({R}−{b}\right)^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{2}\left({R}−{a}\right)\left({R}−{b}\right)}\:\:\:\: \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\blacksquare \\ $$$${If}\:\:{as}\:{an}\:{example} \\ $$$$\:\:\:\mu=\mathrm{8}\:,\:{b}=\mathrm{2}{a}\:=\:\frac{{R}}{\mathrm{2}}\:\:\:\:\left({same}\:{densities}\right) \\ $$$$\:\:\:{c}=\frac{\mathrm{16}}{\mathrm{3}}\:,\:{k}=\frac{\frac{\mathrm{9}}{\mathrm{16}}+\frac{\mathrm{4}}{\mathrm{16}}−\frac{\mathrm{9}}{\mathrm{16}}}{\frac{\mathrm{12}}{\mathrm{16}}}=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\mathrm{sin}\:\alpha=\sqrt{\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}}{\frac{\mathrm{256}}{\mathrm{9}}+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\frac{\mathrm{16}}{\mathrm{3}}\right)+\mathrm{1}}} \\ $$$$\:\:\alpha\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{297}}}\right)\approx\:\mathrm{9}.\mathrm{446}° \\ $$$$\:\:\beta\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{2048}}{\mathrm{2673}}}\right)\:\approx\:\mathrm{61}.\mathrm{082}°\:. \\ $$
Answered by mr W last updated on 04/Oct/19
Commented by mr W last updated on 04/Oct/19
p=R−a q=R−b r=a+b M=μm cos θ=((q^2 +r^2 −p^2 )/(2qr))=((q^2 +(r^2 /((1+μ)^2 ))−h^2 )/(2q(r/(1+μ)))) ((q^2 +r^2 −p^2 )/(1+μ))=q^2 +(r^2 /((1+μ)^2 ))−h^2 ⇒h^2 =((μq^2 )/(1+μ))+(p^2 /(1+μ))−((μr^2 )/((1+μ)^2 )) ⇒h^2 =(1/((1+μ)^2 ))[μ(1+μ)q^2 +(1+μ)p^2 −μr^2 ] cos α=((q^2 +h^2 −(r^2 /((1+μ)^2 )))/(2qh)) =(((1+2μ)q^2 +p^2 −r^2 )/(2q(√(μ(1+μ)q^2 +(1+μ)p^2 −μr^2 )))) ⇒α=cos^(−1) (((1+2μ)q^2 +p^2 −r^2 )/(2q(√(μ(1+μ)q^2 +(1+μ)p^2 −μr^2 )))) cos β=((p^2 +h^2 −((μ^2 r^2 )/((1+μ)^2 )))/(2ph)) =((μq^2 +(2+μ)p^2 −μr^2 )/(2p(√(μ(1+μ)q^2 +(1+μ)p^2 −μr^2 )))) ⇒β=cos^(−1) ((μq^2 +(2+μ)p^2 −μr^2 )/(2p(√(μ(1+μ)q^2 +(1+μ)p^2 −μr^2 )))) example: μ=8, a=(R/4), b=(R/2) p=R−a=((3R)/4), q=R−b=((2R)/4), r=((3R)/4) cos α=((17)/(3(√(33)))) ⇒sin α=((2(√2))/(3(√(33)))) ⇒9.446° cos β=((25)/(9(√(33)))) ⇒sin β=((32(√2))/(9(√(33)))) ⇒61.083°
$${p}={R}−{a} \\ $$$${q}={R}−{b} \\ $$$${r}={a}+{b} \\ $$$${M}=\mu{m} \\ $$$$\mathrm{cos}\:\theta=\frac{{q}^{\mathrm{2}} +{r}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{2}{qr}}=\frac{{q}^{\mathrm{2}} +\frac{{r}^{\mathrm{2}} }{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} }−{h}^{\mathrm{2}} }{\mathrm{2}{q}\frac{{r}}{\mathrm{1}+\mu}} \\ $$$$\frac{{q}^{\mathrm{2}} +{r}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{1}+\mu}={q}^{\mathrm{2}} +\frac{{r}^{\mathrm{2}} }{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} }−{h}^{\mathrm{2}} \\ $$$$\Rightarrow{h}^{\mathrm{2}} =\frac{\mu{q}^{\mathrm{2}} }{\mathrm{1}+\mu}+\frac{{p}^{\mathrm{2}} }{\mathrm{1}+\mu}−\frac{\mu{r}^{\mathrm{2}} }{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{h}^{\mathrm{2}} =\frac{\mathrm{1}}{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} }\left[\mu\left(\mathrm{1}+\mu\right){q}^{\mathrm{2}} +\left(\mathrm{1}+\mu\right){p}^{\mathrm{2}} −\mu{r}^{\mathrm{2}} \right] \\ $$$$\mathrm{cos}\:\alpha=\frac{{q}^{\mathrm{2}} +{h}^{\mathrm{2}} −\frac{{r}^{\mathrm{2}} }{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} }}{\mathrm{2}{qh}} \\ $$$$=\frac{\left(\mathrm{1}+\mathrm{2}\mu\right){q}^{\mathrm{2}} +{p}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{q}\sqrt{\mu\left(\mathrm{1}+\mu\right){q}^{\mathrm{2}} +\left(\mathrm{1}+\mu\right){p}^{\mathrm{2}} −\mu{r}^{\mathrm{2}} }} \\ $$$$\Rightarrow\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{\left(\mathrm{1}+\mathrm{2}\mu\right){q}^{\mathrm{2}} +{p}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{q}\sqrt{\mu\left(\mathrm{1}+\mu\right){q}^{\mathrm{2}} +\left(\mathrm{1}+\mu\right){p}^{\mathrm{2}} −\mu{r}^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\beta=\frac{{p}^{\mathrm{2}} +{h}^{\mathrm{2}} −\frac{\mu^{\mathrm{2}} {r}^{\mathrm{2}} }{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} }}{\mathrm{2}{ph}} \\ $$$$=\frac{\mu{q}^{\mathrm{2}} +\left(\mathrm{2}+\mu\right){p}^{\mathrm{2}} −\mu{r}^{\mathrm{2}} }{\mathrm{2}{p}\sqrt{\mu\left(\mathrm{1}+\mu\right){q}^{\mathrm{2}} +\left(\mathrm{1}+\mu\right){p}^{\mathrm{2}} −\mu{r}^{\mathrm{2}} }} \\ $$$$\Rightarrow\beta=\mathrm{cos}^{−\mathrm{1}} \frac{\mu{q}^{\mathrm{2}} +\left(\mathrm{2}+\mu\right){p}^{\mathrm{2}} −\mu{r}^{\mathrm{2}} }{\mathrm{2}{p}\sqrt{\mu\left(\mathrm{1}+\mu\right){q}^{\mathrm{2}} +\left(\mathrm{1}+\mu\right){p}^{\mathrm{2}} −\mu{r}^{\mathrm{2}} }} \\ $$$$ \\ $$$${example}: \\ $$$$\mu=\mathrm{8},\:{a}=\frac{{R}}{\mathrm{4}},\:{b}=\frac{{R}}{\mathrm{2}} \\ $$$${p}={R}−{a}=\frac{\mathrm{3}{R}}{\mathrm{4}},\:{q}={R}−{b}=\frac{\mathrm{2}{R}}{\mathrm{4}},\:{r}=\frac{\mathrm{3}{R}}{\mathrm{4}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{17}}{\mathrm{3}\sqrt{\mathrm{33}}}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}\sqrt{\mathrm{33}}}\:\Rightarrow\mathrm{9}.\mathrm{446}° \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{25}}{\mathrm{9}\sqrt{\mathrm{33}}}\:\Rightarrow\mathrm{sin}\:\beta=\frac{\mathrm{32}\sqrt{\mathrm{2}}}{\mathrm{9}\sqrt{\mathrm{33}}}\:\Rightarrow\mathrm{61}.\mathrm{083}° \\ $$
Commented by ajfour last updated on 04/Oct/19
sir please consider the example if have chosen and bring out values for α, β .
$${sir}\:{please}\:{consider}\:{the}\:{example} \\ $$$${if}\:{have}\:{chosen}\:{and}\:{bring}\:{out} \\ $$$$\:{values}\:{for}\:\alpha,\:\beta\:. \\ $$
Commented by mr W last updated on 04/Oct/19
the results are the same sir.
$${the}\:{results}\:{are}\:{the}\:{same}\:{sir}. \\ $$
Commented by ajfour last updated on 04/Oct/19
Great, i had done blunders, edited, corrected. Your solution is shorter. Thank you Sir.
$${Great},\:{i}\:{had}\:{done}\:{blunders}, \\ $$$${edited},\:{corrected}.\:{Your}\:{solution} \\ $$$${is}\:{shorter}.\:{Thank}\:{you}\:{Sir}. \\ $$
Commented by mr W last updated on 04/Oct/19
i treated the two balls as a single object whose center of mass must be vertically beneath the center of hollow sphere when in equilibrium.
$${i}\:{treated}\:{the}\:{two}\:{balls}\:{as}\:{a}\:{single}\: \\ $$$${object}\:{whose}\:{center}\:{of}\:{mass}\:{must} \\ $$$${be}\:{vertically}\:{beneath}\:{the}\:{center}\:{of} \\ $$$${hollow}\:{sphere}\:{when}\:{in}\:{equilibrium}. \\ $$

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