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Question Number 52409 by ajfour last updated on 07/Jan/19
Commented by ajfour last updated on 07/Jan/19
Find θ (assuming no friction) specially if R=2r and M=m.
$$\:\:\:\:{Find}\:\theta\:\left({assuming}\:{no}\:{friction}\right) \\ $$$${specially}\:{if}\:{R}=\mathrm{2}{r}\:\:{and}\:{M}={m}. \\ $$
Answered by mr W last updated on 07/Jan/19
Commented by Tawa1 last updated on 07/Jan/19
Sir, please help me look at question 52412. God bless you sir
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{look}\:\mathrm{at}\:\mathrm{question}\:\:\mathrm{52412}.\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 07/Jan/19
parabola: y=cx^2 y′=2cx P(−a,ca^2 ) y′=−2ca=−(1/(tan α)) ⇒a=(1/(2c tan α)) Q(b,cb^2 ) y′=2cb=(1/(tan β)) ⇒b=(1/(2c tan β)) r cos α+R cos β+(R+r) cos θ=a+b r cos α+R cos β+(R+r) cos θ=(1/(2c tan α))+(1/(2c tan β)) ⇒2cr cos α−(1/(tan α))+2cR cos β−(1/(tan β)) +2c(R+r) cos θ=0 ...(i) r sin α−R sin β+(R+r) sin θ=cb^2 −ca^2 r sin α−R sin β+(R+r) sin θ=(1/(4c tan^2 β))−(1/(4c tan^2 α)) ⇒4cr sin α+(1/(tan^2 α))−4cR sin β−(1/(tan^2 β))+4c(R+r) sin θ=0 ...(ii) ((M(R+r) sin (α−θ))/((M+m) cos α))=((m(R+r) sin (β+θ))/((M+m) cos β)) with μ=(m/M) ((sin (α−θ))/(cos α))=((μ sin (β+θ))/(cos β)) ⇒tan θ=((tan α−μ tan β)/(1+μ)) ...(iii) 3 unknowns α,β,θ from (iii): sin θ=((tan α−μ tan β)/( (√((1+μ)^2 +(tan α−μ tan β)^2 )))) cos θ=((1−μ)/( (√((1+μ)^2 +(tan α−μ tan β)^2 )))) put this into (i) and (ii): ⇒2cr cos α−(1/(tan α))+2cR cos β−(1/(tan β)) +((2c(R+r)(1−μ))/( (√((1+μ)^2 +(tan α−μ tan β)^2 ))))=0 ⇒4cr sin α+(1/(tan^2 α))−4cR sin β−(1/(tan^2 β))+((4c(R+r)(tan α−μ tan β))/( (√((1+μ)^2 +(tan α−μ tan β)^2 ))))=0 or ⇒((2cr)/( (√(1+tan^2 α))))−(1/(tan α))+((2cR)/( (√(1+tan^2 β))))−(1/(tan β)) +((2c(R+r)(1−μ))/( (√((1+μ)^2 +(tan α−μ tan β)^2 ))))=0 ⇒((4cr tan α)/( (√(1+tan^2 α))))+(1/(tan^2 α))−((4cR tan β)/( (√(1+tan^2 β))))−(1/(tan^2 β))+((4c(R+r)(tan α−μ tan β))/( (√((1+μ)^2 +(tan α−μ tan β)^2 ))))=0 or with s=tan α, t=tan β ⇒((2cr)/( (√(1+s^2 ))))−(1/s)+((2cR)/( (√(1+t^2 ))))−(1/t) +((2c(R+r)(1−μ))/( (√((1+μ)^2 +(s−μt)^2 ))))=0 ...(1) ⇒((4crs)/( (√(1+s^2 ))))+(1/s^2 )−((4cRt)/( (√(1+t^2 ))))−(1/t^2 )+((4c(R+r)(s−μt))/( (√((1+μ)^2 +(s−μt)^2 ))))=0 ...(2) ⇒solve for s and t....
$${parabola}: \\ $$$${y}={cx}^{\mathrm{2}} \\ $$$${y}'=\mathrm{2}{cx} \\ $$$${P}\left(−{a},{ca}^{\mathrm{2}} \right) \\ $$$${y}'=−\mathrm{2}{ca}=−\frac{\mathrm{1}}{\mathrm{tan}\:\alpha} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}{c}\:\mathrm{tan}\:\alpha} \\ $$$${Q}\left({b},{cb}^{\mathrm{2}} \right) \\ $$$${y}'=\mathrm{2}{cb}=\frac{\mathrm{1}}{\mathrm{tan}\:\beta} \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{2}{c}\:\mathrm{tan}\:\beta} \\ $$$${r}\:\mathrm{cos}\:\alpha+{R}\:\mathrm{cos}\:\beta+\left({R}+{r}\right)\:\mathrm{cos}\:\theta={a}+{b} \\ $$$${r}\:\mathrm{cos}\:\alpha+{R}\:\mathrm{cos}\:\beta+\left({R}+{r}\right)\:\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}{c}\:\mathrm{tan}\:\alpha}+\frac{\mathrm{1}}{\mathrm{2}{c}\:\mathrm{tan}\:\beta} \\ $$$$\Rightarrow\mathrm{2}{cr}\:\mathrm{cos}\:\alpha−\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}+\mathrm{2}{cR}\:\mathrm{cos}\:\beta−\frac{\mathrm{1}}{\mathrm{tan}\:\beta}\:+\mathrm{2}{c}\left({R}+{r}\right)\:\mathrm{cos}\:\theta=\mathrm{0}\:\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${r}\:\mathrm{sin}\:\alpha−{R}\:\mathrm{sin}\:\beta+\left({R}+{r}\right)\:\mathrm{sin}\:\theta={cb}^{\mathrm{2}} −{ca}^{\mathrm{2}} \\ $$$${r}\:\mathrm{sin}\:\alpha−{R}\:\mathrm{sin}\:\beta+\left({R}+{r}\right)\:\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{4}{c}\:\mathrm{tan}^{\mathrm{2}} \:\beta}−\frac{\mathrm{1}}{\mathrm{4}{c}\:\mathrm{tan}^{\mathrm{2}} \:\alpha} \\ $$$$\Rightarrow\mathrm{4}{cr}\:\mathrm{sin}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\alpha}−\mathrm{4}{cR}\:\mathrm{sin}\:\beta−\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\beta}+\mathrm{4}{c}\left({R}+{r}\right)\:\mathrm{sin}\:\theta=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$$\frac{{M}\left({R}+{r}\right)\:\mathrm{sin}\:\left(\alpha−\theta\right)}{\left({M}+{m}\right)\:\mathrm{cos}\:\alpha}=\frac{{m}\left({R}+{r}\right)\:\mathrm{sin}\:\left(\beta+\theta\right)}{\left({M}+{m}\right)\:\mathrm{cos}\:\beta} \\ $$$${with}\:\mu=\frac{{m}}{{M}} \\ $$$$\frac{\mathrm{sin}\:\left(\alpha−\theta\right)}{\mathrm{cos}\:\alpha}=\frac{\mu\:\mathrm{sin}\:\left(\beta+\theta\right)}{\mathrm{cos}\:\beta} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{tan}\:\alpha−\mu\:\mathrm{tan}\:\beta}{\mathrm{1}+\mu}\:\:\:…\left({iii}\right) \\ $$$$\mathrm{3}\:{unknowns}\:\alpha,\beta,\theta \\ $$$${from}\:\left({iii}\right): \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{tan}\:\alpha−\mu\:\mathrm{tan}\:\beta}{\:\sqrt{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} +\left(\mathrm{tan}\:\alpha−\mu\:\mathrm{tan}\:\beta\right)^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{1}−\mu}{\:\sqrt{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} +\left(\mathrm{tan}\:\alpha−\mu\:\mathrm{tan}\:\beta\right)^{\mathrm{2}} }} \\ $$$${put}\:{this}\:{into}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\Rightarrow\mathrm{2}{cr}\:\mathrm{cos}\:\alpha−\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}+\mathrm{2}{cR}\:\mathrm{cos}\:\beta−\frac{\mathrm{1}}{\mathrm{tan}\:\beta}\:+\frac{\mathrm{2}{c}\left({R}+{r}\right)\left(\mathrm{1}−\mu\right)}{\:\sqrt{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} +\left(\mathrm{tan}\:\alpha−\mu\:\mathrm{tan}\:\beta\right)^{\mathrm{2}} }}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}{cr}\:\mathrm{sin}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\alpha}−\mathrm{4}{cR}\:\mathrm{sin}\:\beta−\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\beta}+\frac{\mathrm{4}{c}\left({R}+{r}\right)\left(\mathrm{tan}\:\alpha−\mu\:\mathrm{tan}\:\beta\right)}{\:\sqrt{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} +\left(\mathrm{tan}\:\alpha−\mu\:\mathrm{tan}\:\beta\right)^{\mathrm{2}} }}=\mathrm{0} \\ $$$${or} \\ $$$$\Rightarrow\frac{\mathrm{2}{cr}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\alpha}}−\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}+\frac{\mathrm{2}{cR}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\beta}}−\frac{\mathrm{1}}{\mathrm{tan}\:\beta}\:+\frac{\mathrm{2}{c}\left({R}+{r}\right)\left(\mathrm{1}−\mu\right)}{\:\sqrt{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} +\left(\mathrm{tan}\:\alpha−\mu\:\mathrm{tan}\:\beta\right)^{\mathrm{2}} }}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{4}{cr}\:\mathrm{tan}\:\alpha}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\alpha}}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\alpha}−\frac{\mathrm{4}{cR}\:\mathrm{tan}\:\beta}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\beta}}−\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\beta}+\frac{\mathrm{4}{c}\left({R}+{r}\right)\left(\mathrm{tan}\:\alpha−\mu\:\mathrm{tan}\:\beta\right)}{\:\sqrt{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} +\left(\mathrm{tan}\:\alpha−\mu\:\mathrm{tan}\:\beta\right)^{\mathrm{2}} }}=\mathrm{0} \\ $$$${or}\:{with}\:{s}=\mathrm{tan}\:\alpha,\:{t}=\mathrm{tan}\:\beta \\ $$$$\Rightarrow\frac{\mathrm{2}{cr}}{\:\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}−\frac{\mathrm{1}}{{s}}+\frac{\mathrm{2}{cR}}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}−\frac{\mathrm{1}}{{t}}\:+\frac{\mathrm{2}{c}\left({R}+{r}\right)\left(\mathrm{1}−\mu\right)}{\:\sqrt{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} +\left({s}−\mu{t}\right)^{\mathrm{2}} }}=\mathrm{0}\:\:\:…\left(\mathrm{1}\right) \\ $$$$\Rightarrow\frac{\mathrm{4}{crs}}{\:\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}+\frac{\mathrm{1}}{{s}^{\mathrm{2}} }−\frac{\mathrm{4}{cRt}}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }+\frac{\mathrm{4}{c}\left({R}+{r}\right)\left({s}−\mu{t}\right)}{\:\sqrt{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} +\left({s}−\mu{t}\right)^{\mathrm{2}} }}=\mathrm{0}\:\:\:…\left(\mathrm{2}\right) \\ $$$$\Rightarrow{solve}\:{for}\:{s}\:{and}\:{t}…. \\ $$
Commented by ajfour last updated on 07/Jan/19
hereforth i will avoid posting such physics questions Sir, that dont fetch concrete answers. Thanks, but please solve latest geometry question i created with some thinking.
$${hereforth}\:{i}\:{will}\:{avoid}\:{posting} \\ $$$${such}\:{physics}\:{questions}\:{Sir},\:{that} \\ $$$${dont}\:{fetch}\:{concrete}\:{answers}. \\ $$$${Thanks},\:{but}\:{please}\:{solve}\:{latest} \\ $$$${geometry}\:{question}\:{i}\:{created}\:{with} \\ $$$${some}\:{thinking}. \\ $$
Commented by ajfour last updated on 07/Jan/19
what a beauty and symmetry in the equations, Sir. I pray for all happiness & health for you and your family.
$${what}\:{a}\:{beauty}\:{and}\:{symmetry}\:{in}\:{the} \\ $$$${equations},\:{Sir}.\:{I}\:{pray}\:{for}\:{all} \\ $$$${happiness}\:\&\:{health}\:{for}\:{you}\:{and}\:{your} \\ $$$${family}. \\ $$
Commented by mr W last updated on 07/Jan/19
thank you sir! the same to you and your family!
$${thank}\:{you}\:{sir}! \\ $$$${the}\:{same}\:{to}\:{you}\:{and}\:{your}\:{family}! \\ $$

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