Question and Answers Forum
Question Number 159085 by ajfour last updated on 12/Nov/21
Answered by mr W last updated on 15/Nov/21
Commented by mr W last updated on 15/Nov/21
![2Rsin θ=R+Rsin ϕ sin ϕ=2 sin θ−1 (dϕ/dθ)=((2 cos θ)/(cos ϕ))=((cos θ)/( (√(sin θ(1−sin θ))))) x_A =x_E =4R sin θ x_B =x_A +2R cos θ+R cos ϕ x_B =4R sin θ+2R cos θ+R cos ϕ ω=(dθ/dt) v_B =(dx_B /dt)=4Rcos θ ω−2R sin θ ω−R sin ϕ ω (dϕ/dθ) v_B =2R(2cos θ−sin θ−cos θ tan ϕ)ω (dv_B /dθ)=2R(2cos θ−sin θ−cos θ tan ϕ)(dω/dθ) +2Rω(−2sin θ−cos θ+sin θ tan ϕ−((2cos^2 θ)/(cos^3 ϕ))) a_B =(dv_B /dt)=ω(dv_B /dθ) =2Rω(2cos θ−sin θ−cos θ tan ϕ)(dω/dθ) +2Rω^2 (−2sin θ−cos θ+sin θ tan ϕ−((2cos^2 θ)/(cos^3 ϕ))) Ncos ϕ=ma_B N=0 ⇒a_B =0 (2cos θ−sin θ−cos θ tan ϕ)(dω/dθ) −ω(2sin θ+cos θ−sin θ tan ϕ+((2cos^2 θ)/(cos^3 ϕ)))=0 ..(I) x_G −x_A =(−2sin θ+cos θ)R y_G =2R cos θ+R sin θ y_E =4R cos θ I_E =I+m(x_E −x_G )^2 +m(y_E −y_G )^2 =((m(4R^2 +16R^2 ))/(12))+m(−2 sin θ+cos θ)^2 R^2 +m(−2 cos θ+sin θ)^2 R^2 =(((20−12 sin 2θ)mR^2 )/3) ((I_E ω^2 )/2)+((mv_B ^2 )/2)=mg((√5)R−y_G ) ((I_E ω^2 )/2)+((mv_B ^2 )/2)=mgR((√5)−2 cos θ−sin θ) (((20−12 sin 2θ)mω^2 R^2 )/6)+((4mR^2 (2cos θ−sin θ−cos θ tan ϕ)^2 ω^2 )/2)=mgR((√5)−2 cos θ−sin θ) [10−6 sin 2θ+6(2cos θ−sin θ−cos θ tan ϕ)^2 ]ω^2 =((3g)/R)((√5)−2 cos θ−sin θ) ω^2 =((3g)/R)×(((√5)−2 cos θ−sin θ)/(10−6 sin 2θ+6(2cos θ−sin θ−cos θ tan ϕ)^2 )) ω=(√((3g)/R))×f(θ) with f(θ)=(√(((√5)−2 cos θ−sin θ)/(10−6 sin 2θ+6(2cos θ−sin θ−cos θ tan ϕ)^2 ))) we get from (I) θ≈0.7296≈41.8° at which N=0 all above is valid for tan^(−1) (1/2)≤θ≤θ_(max) ϕ+θ_(max) =90° sin (90°−θ_(max) )=2 sin θ_(max) −1 2 sin θ_(max) −cos θ_(max) =1 sin (θ_(max) −tan^(−1) (1/2))=(1/( (√5))) θ_(max) =sin^(−1) (1/( (√5)))+tan^(−1) (1/2)≈53.1° N=0 occours at θ≈41.8°, before θ_(max) is reached.](Q159224.png)
Commented by mr W last updated on 14/Nov/21
Commented by mr W last updated on 18/Nov/21
Commented by mr W last updated on 14/Nov/21
