Question Number 216219 by mr W last updated on 30/Jan/25
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Commented by mr W last updated on 31/Jan/25
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Answered by mahdipoor last updated on 01/Feb/25

Commented by mr W last updated on 01/Feb/25
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Answered by mr W last updated on 01/Feb/25
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Commented by mr W last updated on 05/Feb/25
![let (h/L)=η ω=(dθ/dt) y=L cos θ at t=0: y=h=L cos θ_0 ⇒θ_0 =cos^(−1) η (1/2)[((mL^2 )/(12))+m((L/2))^2 ]ω^2 =((mg(h−y))/2) ((L^2 ω^2 )/3)=g(h−L cos θ) ω^2 =((3g)/L)(η−cos θ) ⇒ω(dω/dθ)=((3g)/(2L)) sin θ ⇒ω=(√(((3g)/L)(η−cos θ))) u_x =((Lω cos θ)/2) a_x =(L/2)(−ω^2 sin θ+cos θ ((ωdω)/dθ)) =(L/2)[−((3g)/L)(η−cos θ) sin θ+cos θ ((3g)/(2L)) sin θ] =((3g)/4)(−2η+3 cos θ)sin θ N=ma_x =((3mg)/4)(−2η+3 cos θ)sin θ N=0: ⇒−2η+3 cos θ=0 ⇒cos θ=((2η)/3) ⇒θ_1 =cos^(−1) ((2η)/3) dt=(√(L/(3g)))(dθ/( (√(η−cos θ)))) ∫_0 ^T_1 dt=(√(L/(3g)))∫_θ_0 ^θ_1 (dθ/( (√(η−cos θ)))) ⇒T_1 =(√(L/(3g)))∫_(cos^(−1) η) ^(cos^(−1) ((2η)/3)) (dθ/( (√(η−cos θ)))) ω_1 =(√(((3g)/L)(η−cos θ_1 ))) u_1 =u_x ∣_(t=T_1 ) =((Lω_1 cos θ_1 )/2)=((cos θ_1 )/2)(√(3gL(η−cos θ_1 ))) upon t=T_1 : u_x =u_1 =((cos θ_1 )/2)(√(3gL(η−cos θ_1 )))=(η/3)(√(ηgL)) v_y =((Lω sin θ)/2) (1/2)×((mL^2 ω^2 )/(12))+((m(u_x ^2 +v_y ^2 ))/2)=((mg(h−y))/2) (1/2)×((mL^2 ω^2 )/(12))+(m/2)×(((η^3 gL)/9)+((L^2 ω^2 sin^2 θ)/4))=((mg(h−L cos θ))/2) ⇒ω=(√((12g(η−(η^3 /9)−cos θ))/(L(1+3 sin^2 θ)))) (dθ/dt)=(√((12g(η−(η^3 /9)−cos θ))/(L(1+3 sin^2 θ)))) dt=(√((L(1+3 sin^2 θ))/(12g(η−(η^3 /9)−cos θ)))) dθ ∫_0 ^T_2 dt=(√(L/(3g)))∫_θ_1 ^(π/2) (√((1+3 sin^2 θ)/(4(η−(η^3 /9)−cos θ)))) dθ ⇒T_2 =(√(L/(3g)))∫_(cos^(−1) ((2η)/3)) ^(π/2) (√((1+3 sin^2 θ)/(4(η−(η^3 /9)−cos θ)))) dθ total time needed: T=T_1 +T_2 =λ(√(L/(3g))) with λ=∫_(cos^(−1) η) ^(cos^(−1) ((2η)/3)) (dθ/( (√(η−cos θ))))+∫_(cos^(−1) ((2η)/3)) ^(π/2) (√((1+3 sin^2 θ)/(4(η−(η^3 /9)−cos θ)))) dθ example: η=(h/L)=0.8 To≈1.73205081(√(L/g))](https://www.tinkutara.com/question/Q216238.png)
Commented by Tawa11 last updated on 01/Feb/25
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