Question Number 132102 by mr W last updated on 14/Feb/21

Commented by mr W last updated on 16/Feb/21

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Commented by ajfour last updated on 15/Feb/21

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Answered by mr W last updated on 25/Feb/21

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Commented by mr W last updated on 16/Feb/21
![motion from A to B: t=((R(1+cos φ))/(u cos θ)) R sin φ=−u sin θ×((R(1+cos φ))/(u cos θ))+(g/2)×((R^2 (1+cos φ)^2 )/(u^2 cos^2 θ)) ((sin φ)/(1+cos φ))=((gR)/(2u^2 ))×(1+cos φ)(1+tan^2 θ)−tan θ let λ=((gR)/(2u^2 )) ((sin φ)/(1+cos φ))=λ(1+cos φ)(1+tan^2 θ)−tan θ λ(1+cos φ)tan^2 θ−tan θ+((λ(1+cos φ)^2 −sin φ)/(1+cos φ))=0 tan θ=((1±(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ))) ⇒θ=tan^(−1) [((1±(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] at point B: U_x =u cos θ U_y =gt−u sin θ=((gR(1+cos φ))/(u cos θ))−u sin θ U_⊥ =U_x cos φ+U_y sin φ U_⊥ =u cos θcos φ+[((gR(1+cos φ))/(u cos θ))−u sin θ]sin φ ⇒U_⊥ =u cos (φ+θ)+((gR(1+cos φ)sin φ)/(u cos θ)) U_∥ =−U_x sin φ+U_y cos φ U_∥ =−u cos θ sin φ+[((gR(1+cos φ))/(u cos θ))−u sin θ]cos φ ⇒U_∥ =−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ)) V_⊥ =eU_⊥ ⇒V_⊥ =eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ)) V_∥ =U_∥ ⇒V_∥ =−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ)) V_x =V_⊥ cos φ+V_∥ sin φ V_x =[eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ))]cos φ+[−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ))]sin φ V_x =eu cos (φ+θ)cos φ+((egR(1+cos φ)sin φ cos φ)/(u cos θ))−u sin (φ+θ)sin φ+((gR(1+cos φ)sin φcos φ)/(u cos θ)) ⇒V_x =u[(1+e)cos (φ+θ)cos φ−cos θ+((λ(1+e)(1+cos φ) sin 2φ)/(cos θ))] V_y =V_⊥ sin φ−V_∥ cos φ V_y =[eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ))]sin φ−[−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ))]cos φ ⇒V_y =u[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)((1+e) sin^2 φ−1))/(cos θ))] return from B to A: t=((R(1+cos φ))/((1+e)cos (φ+θ)cos φ−cos θ+((λ(1+cos φ)(1+e) sin 2φ)/(cos θ))))×(1/u) ⇒((tu)/R)=ξ=((1+cos φ)/((1+e)cos (φ+θ)cos φ−cos θ+((λ(1+e)(1+cos φ) sin 2φ)/(cos θ)))) R sin φ=t{u[(1+e)cos (φ+θ)sin φ+sin θ]+((2λ(1+cos φ)(e sin^2 φ−cos^2 φ))/(cos θ))−((gt)/2)} sin φ=((tu)/R)[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)(e sin^2 φ−cos^2 φ))/(cos θ))−((λtu)/R)] sin φ=ξ[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)((1+e)sin^2 φ−1))/(cos θ))−λξ] or sin φ=ξ[e′cos (φ+θ) sin φ+sin θ+((2λ(1+cos φ)(e′sin^2 φ−1))/(cos θ))−λξ] ...(I) with e′=1+e λ=((gR)/(2u^2 )) θ=tan^(−1) [((1+(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] ∗) ξ=((1+cos φ)/(e′cos (φ+θ) cos φ−cos θ+((λe′(1+cos φ) sin 2φ)/(cos θ)))) for a given e ∈(0,1] we can find solution(s) for φ in (I) for some values of parameter λ. the maximum value of λ such that a solution for φ exists can be determined numerically. this corresponds to the minimum speed the ball must have: u_(min) =(√((gR)/(2λ_(max) ))) examples: e=0.5 ⇒u_(min) ≈2.7682(√(gR)) e=0.75 ⇒u_(min) ≈1.6082(√(gR)) e=1 ⇒u_(min) ≈0.9098(√(gR)) (= Q65589) ∗) with the other solution θ=tan^(−1) [((1−(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] we can get a solution for the case that the ball strikes the surface upwards, i.e. counterclockwise. but in this case more energy is needed, i.e. u_(min) is larger than with θ=tan^(−1) [((1+(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] see examples with e=0.5 and 0.75.](https://www.tinkutara.com/question/Q132234.png)
Commented by mr W last updated on 15/Feb/21

Commented by mr W last updated on 13/Feb/21

Commented by mr W last updated on 13/Feb/21

Commented by mr W last updated on 12/Feb/21

Commented by mr W last updated on 12/Feb/21

Commented by mr W last updated on 14/Feb/21
![following diagrams show different situations for e=0.75. generally: for a given e∈(0,1] there exists an u_(min) . for any u>u_(min) there are four possible ways how the ball returns back to its starting position: two ways clockwise and two counter− clockwise.](https://www.tinkutara.com/question/Q132351.png)
Commented by mr W last updated on 15/Feb/21

Commented by mr W last updated on 13/Feb/21

Commented by mr W last updated on 14/Feb/21

Commented by mr W last updated on 13/Feb/21

Commented by mr W last updated on 14/Feb/21

Answered by mr W last updated on 14/Feb/21

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Commented by Ar Brandon last updated on 13/Feb/21

Commented by Ar Brandon last updated on 14/Feb/21

Commented by mr W last updated on 14/Feb/21

Answered by ajfour last updated on 15/Feb/21

Commented by ajfour last updated on 15/Feb/21

Commented by ajfour last updated on 15/Feb/21

Commented by mr W last updated on 15/Feb/21

Commented by mr W last updated on 15/Feb/21

Commented by mr W last updated on 15/Feb/21

Commented by mr W last updated on 15/Feb/21

Commented by mr W last updated on 15/Feb/21

Commented by ajfour last updated on 16/Feb/21

Commented by mr W last updated on 16/Feb/21

Commented by mr W last updated on 16/Feb/21
