A rotating disc of 1 m diameter has two eccentric masses of 0.5 k

SOLUTION

Concept:

Let θ3 and θ2 be the respective orientations of mass 3 and mass 2 w.r.t. mass 1.

Then for complete balancing forces along x-axis and y-axis must be zero.

Thus, Σmiri sin θi = 0 and Σmiri cos θi = 0

Here assume θ1 = 0°

Calculation:

Given:

m₁ = m₂ = 0.5 kg, r₁ = 50 mm, r₂ = 60 mm.

θ1 = 0°, θ2 = 150°

m₃ = m_b = 0.1 kg

Σmiri sin θi = 0

Σmiri sin θi = m1r1 sin θ1 + m2r2 sin θ2 + m3r3 sin θ₃ = 0

⇒  0.5 × 50 × sin 0 + 0.5 × 60 × sin 150 + 0.1 × r3 × sin θ3 = 0

r3 sin θ3 = -150 ___(1)

Similarly,

Σmiri cos θi = 0

Σmiri cos θi = m1r1 cos θ1 + m2r2 cos θ2 + m3r3 cos θ₃ = 0

 Σmiri cos θi =  0.5 × 50 × cos 0 + 0.5 × 60 × cos 150 + 0.1 × r3 × cos θ3 = 0

r3 cos θ3 = -9.8  ____(2)

Now squaring equation 1 and equation 2 and adding them

(r3 sin θ3)² + (r3 cos θ3)² = r3² = (-150)² + (-9.8)² 

∴ r₃ = 150.3 mm ≈ 150 mm.