Question Number 22370 by Tinkutara last updated on 16/Oct/17

Commented by mrW1 last updated on 17/Oct/17
![AC^(⌢) =((πr)/2)=(π/2)×(2/π)= 1 m BC=(3/2)−1=(1/2) m ρ=mass each meter a=acceleration T_0 =tension at point C, at θ=0 (1/2)ρg−T_0 =(1/2)ρa ⇒T_0 =(1/2)ρ(g−a) −dT+ρgdscos θ=aρds −dT=ρds(a−gcos θ) −dT=ρr(a−gcos θ)dθ −∫_T_0 ^( 0) dT=ρr∫_0 ^(π/2) (a−gcos θ)dθ T_0 =ρr[aθ−gsin θ]_0 ^(π/2) T_0 =ρr[((aπ)/2)−g] (1/2)ρ(g−a)=ρ(2/π)(((aπ)/2)−g) (g−a)=(4/π)(((aπ)/2)−g) g−a=2a−(4/π)g ⇒a=((4+π)/(3π))g ≈ 7.58 m/s^2](https://www.tinkutara.com/question/Q22383.png)
Commented by Tinkutara last updated on 16/Oct/17

Commented by Tinkutara last updated on 17/Oct/17

Answered by revenge last updated on 18/Oct/17

Commented by revenge last updated on 18/Oct/17

Commented by Tinkutara last updated on 18/Oct/17
