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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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