rotation

we can get answer directly from fbd and writing the equations

let tension in thread 1 be T1 and in Thread 2 be T2

and let angles with the horizontal be θ1 and θ2

then as they both are at a constant height and a constant distance from the pole

we can write the equations

T1cosθ1=mw²r1+T2cosθ2

and T1sinθ1=T2sinθ2=mg

and T2cosθ2=mw²r2

we see that T1cosθ1>T2cosθ2

but T1sinθ1=T2sinθ2
squaring the above equation

T1²(1-cos²Î¸1)=T2²(1-cos²Î¸2)
so we get

T1²-T2²=(T1cosθ1)²-(T2cosθ2)²

as T1cosθ1>T2cosθ2
T1>T2

but T1sinθ1=T2sinθ2
as T1>T2
sinθ1<sinθ2

so θ1<θ2