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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.

Findθ(assumingnofriction)speciallyifR=2randM=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

Sir,pleasehelpmelookatquestion52412.Godblessyousir

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=cx2y′=2cxP(−a,ca2)y′=−2ca=−1tanα⇒a=12ctanαQ(b,cb2)y′=2cb=1tanβ⇒b=12ctanβrcosα+Rcosβ+(R+r)cosθ=a+brcosα+Rcosβ+(R+r)cosθ=12ctanα+12ctanβ⇒2crcosα−1tanα+2cRcosβ−1tanβ+2c(R+r)cosθ=0...(i)rsinα−Rsinβ+(R+r)sinθ=cb2−ca2rsinα−Rsinβ+(R+r)sinθ=14ctan2β−14ctan2α⇒4crsinα+1tan2α−4cRsinβ−1tan2β+4c(R+r)sinθ=0...(ii)M(R+r)sin(α−θ)(M+m)cosα=m(R+r)sin(β+θ)(M+m)cosβwithμ=mMsin(α−θ)cosα=μsin(β+θ)cosβ⇒tanθ=tanα−μtanβ1+μ...(iii)3unknownsα,β,θfrom(iii):sinθ=tanα−μtanβ(1+μ)2+(tanα−μtanβ)2cosθ=1−μ(1+μ)2+(tanα−μtanβ)2putthisinto(i)and(ii):⇒2crcosα−1tanα+2cRcosβ−1tanβ+2c(R+r)(1−μ)(1+μ)2+(tanα−μtanβ)2=0⇒4crsinα+1tan2α−4cRsinβ−1tan2β+4c(R+r)(tanα−μtanβ)(1+μ)2+(tanα−μtanβ)2=0or⇒2cr1+tan2α−1tanα+2cR1+tan2β−1tanβ+2c(R+r)(1−μ)(1+μ)2+(tanα−μtanβ)2=0⇒4crtanα1+tan2α+1tan2α−4cRtanβ1+tan2β−1tan2β+4c(R+r)(tanα−μtanβ)(1+μ)2+(tanα−μtanβ)2=0orwiths=tanα,t=tanβ⇒2cr1+s2−1s+2cR1+t2−1t+2c(R+r)(1−μ)(1+μ)2+(s−μt)2=0...(1)⇒4crs1+s2+1s2−4cRt1+t2−1t2+4c(R+r)(s−μt)(1+μ)2+(s−μt)2=0...(2)⇒solveforsandt....

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.

hereforthiwillavoidpostingsuchphysicsquestionsSir,thatdontfetchconcreteanswers.Thanks,butpleasesolvelatestgeometryquestionicreatedwithsomethinking.

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.

whatabeautyandsymmetryintheequations,Sir.Iprayforallhappiness&healthforyouandyourfamily.

Commented by mr W last updated on 07/Jan/19

thank you sir! the same to you and your family!

thankyousir!thesametoyouandyourfamily!

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