Question Number 100928 by ajfour last updated on 29/Jun/20

Commented byajfour last updated on 29/Jun/20

Find r in terms of R.

Commented byajfour last updated on 30/Jun/20

Its physics plus maths, Sir.

Answered by bramlex last updated on 29/Jun/20

Commented bybramlex last updated on 29/Jun/20

x^2 +(R−r)^2 =(R+r)^2   x^2 −2rR = 2rR →x = 2(√(rR))  (1)(x−R)^2 +(y−R)^2 =R^2   →x^2 +y^2 −2Rx−2Ry+R^2 =0  (2)(x−(2(√(rR))+r)^2 +(y−r)^2 =r^2   →x^2 −2(r+2(√(rR)))x+(r+2(√(rR)))^2 +y^2 −2ry=0  ...

Answered by mr W last updated on 29/Jun/20

Commented bymr W last updated on 29/Jun/20

at position θ:  (1/2)mu^2 =R(1−cos θ)mg  ⇒u=(√(2R(1−cos θ)g))  mg cos θ−N=((mu^2 )/R)  N=0 when contact gets lost  mg cos θ=((m×2R(1−cos θ)g)/R)  ⇒cos θ=(2/3)  ⇒u=(√((2Rg)/3))  track of mass m after point B:  x=R sin θ+u cos θ t  ⇒x=(((√5)R)/3)+(2/3)(√((2Rg)/3))t  y=R(1+cos θ)−u sin θ t−(1/2)gt^2   ⇒y=((5R)/3)−((√5)/3)(√((2Rg)/3)) t−(1/2)gt^2   D((√((R+r)^2 −(R−r)^2 )), r)≡D(2(√(Rr)),r)  let Φ=DC^2   Φ=(x−2(√(Rr)))^2 +(y−r)^2   Φ=((((√5)R)/3)+(2/3)(√((2Rg)/3))t−2(√(Rr)))^2 +(((5R)/3)−((√5)/3)(√((2Rg)/3)) t−(1/2)gt^2 −r)^2   (dΦ/dt)=2((((√5)R)/3)+(2/3)(√((2Rg)/3))t−2(√(Rr)))((2/3)(√((2Rg)/3)))+2(((5R)/3)−((√5)/3)(√((2Rg)/3)) t−(1/2)gt^2 −r)(−((√5)/3)(√((2Rg)/3))−gt)=0  ⇒12((√(30))+4(√(g/R))t−6(√6)(√(r/R)))=(30−2(√(30))(√(g/R)) t−9((√(g/R))t)^2 −18(r/R))((√(30))+9(√(g/R))t)  Φ_(min) =r^2   ⇒(((√5)/3)+((2(√6))/9)(√(g/R))t−2(√(r/R)))^2 +((5/3)−((√(30))/9)(√(g/R)) t−(1/2)((√(g/R))t)^2 −(r/R))^2 =((r/R))^2   let δ=(√(g/R))t, λ=(√(r/R))  ⇒12((√(30))+4δ−6(√6)λ)=(30−2(√(30))δ−9δ^2 −18λ^2 )((√(30))+9δ)   ...(i)  ⇒4(3(√5)+2(√6)δ−18λ)^2 +(30−2(√(30))δ−9δ^2 −18λ^2 )^2 =324λ^4    ...(ii)  we get from (i) and (ii):  λ≈0.53716  ⇒r=λ^2 R≈0.2885R

Commented bymr W last updated on 29/Jun/20

Commented byajfour last updated on 29/Jun/20

haven′t been through all of it sir, but  really looks superb solution, i want  to try on my own for a little while...

Commented byajfour last updated on 29/Jun/20

can you help me sir, with the equation  of parabola with shown axes..?

Commented bymr W last updated on 30/Jun/20

Commented byajfour last updated on 02/Jul/20

x=Rsin θ+(ucos θ)t  y=R(1+cos θ)−(usin θ)t−((gt^2 )/2)  As  cos θ=(2/3) , sin θ=((√5)/3) , u^2 =((2Rg)/3)  y=((5R)/3)−(((u(√5))/3))(((x−((R(√5))/3)))/((((2u)/3))))−(g/2)(((x−((R(√5))/3))^2 )/((((2u)/3))^2 ))  y=((5R)/3)−((√5)/2)(x−((R(√5))/3))−((27)/(16R))(x−((R(√5))/3))^2   (dy/dx)=−((√5)/2)−((27)/(8R))(x−((R(√5))/3))=0  ⇒   x_0 =((R(√5))/3)−((8R)/(27))(((√5)/2)) = ((5(√5)R)/(27))  y_0 =((5R)/3)−((√5)/2)(((5(√5)R)/(27))−((9R(√5))/(27)))                    −((27)/(16R))(((80R^2 )/(27×27)))  ⇒ y_0 =((5R)/3)+((10R)/(27))−((5R)/(27))=((50R)/(27))  < 2R  Thus eq. of parabola is    y=((50R)/(27))−((27)/(16R))(x−((5(√5)R)/(27)))^2     (dy/dx)=−((27)/(8R))(x−((5(√5)R)/(27)))  say this parabola touches the smaller  circle at P (h,k)  and let center of  smaller circle is C(2(√(Rr)), r);  then  h=2(√(Rr))+rcos 𝛃  k=r+rsin 𝛃  cot β=((27)/(8R))(h−((5(√5)R)/(27)))=((h−2(√(Rr)))/(k−r))    ⇒     ((27)/8)z=(((z+((5(√5))/(27)))−2(√s))/(((50)/(27))−((27)/(16))z^2 −s))       ......(i)      k=((50R)/(27))−((27)/(16R))(h−((5(√5)R)/(27)))^2        = r+(r/((1+[((27)/(8R))(h−((5(√5)R)/(27)))]^2 ))^(1/) )    ⇒  ((50)/(27))−((27)/(16))z^2 =s+(s/((1+(((27)/(16))z)^2 ))^(1/) )  .....(ii)  ⇒  s=(((((50)/(27))−((27)/(16))z^2 ))/(1+(1/((1+(((27z)/(16)))^2 ))^(1/) )))      ;   Now  ⇒     ((27)/8)z=(((z+((5(√5))/(27)))−2{(((((50)/(27))−((27)/(16))z^2 ))/(1+(1/((1+(((27z)/(16)))^2 ))^(1/) )))}^(1/2) )/(((50)/(27))−((27)/(16))z^2 −((((((50)/(27))−((27)/(16))z^2 ))/(1+(1/((1+(((27z)/(16)))^2 ))^(1/) ))))))       ......(A)  z is obtained from above eq.  Then   s=(r/R) = (((((50)/(27))−((27)/(16))z^2 ))/(1+(1/((1+(((27z)/(16)))^2 ))^(1/) )))  ■

Commented byajfour last updated on 02/Jul/20

Commented bymr W last updated on 02/Jul/20

beautifully solved!