Question

Two spheres A and B of negligible dimensions and masses 1 kg and √3 kg respectively, are supported on the smooth circular surface, fixed to the ground with a centre O and radius of 0.1m. The spheres are joined by the cord shown in length π/20 m; determine the angles α and β corresponding to the position of equilibrium of the spheres with respect to the vertical passing through O.

Answer 1

Answer:

α = π/3

β = π/6

Explanation:

Use arc length equation to find the sum of the angles.

s = rθ

π/20 m = (0.1 m) (α + β)

π/2 = α + β

Draw a free body diagram for each sphere.  Both spheres have three forces acting on them:

Weight force mg pulling down,

Normal force N pushing perpendicular to the surface,

and tension force T pulling tangential to the surface.

Sum of forces on A in the tangential direction:

∑F = ma

T − m₁g sin α = 0

T = m₁g sin α

Sum of forces on B in the tangential direction:

∑F = ma

T − m₂g sin β = 0

T = m₂g sin β

Substituting:

m₁g sin α = m₂g sin β

m₁ sin α = m₂ sin β

(1 kg) sin α = (√3 kg) sin (π/2 − α)

1 sin α = √3 cos α

tan α = √3

α = π/3

β = π/6