Question

A rigid rod of mass 5.30 kg and length of 1.50 m rotates in a vertical (x,y) plane about a frictionless pivot through its center. Particles m_1 (mass=6.30 kg) and m_2 (mass=2.20 kg) are attached at the ends of the rod. Determine the size of the angular acceleration of the system when the rod makes an angle of 42.1° with the horizontal.

Answer 1

Moment of inertia of a rod is given by -

I_rod = (1/3) m R²

Moment of inertia of two particle is given by -

I₁ = m₁ R²

I₂ = m₂ R²

The total moment of inertia will be given as -

I_total = I₁ + I₂ + I_rod

I_total = [(m₁ R²) + (m₂ R²) + (1/3) m R²]

I_total = [m₁ + m₂ + (m / 3)] R²

Then, the size of an angular acceleration of the system which will be given by -

= I_total

(m₁ - m₂) g R cos = {[m₁ + m₂ + (m / 3)] R²}

(m₁ - m₂) g cos = [m₁ + m₂ + (m / 3)] R

= (m₁ - m₂) g cos / [m₁ + m₂ + (m / 3)] R

where, R = L / 2

= [(6.30 kg) - (2.20 kg)] (9.8 m/s²) cos 42.1⁰ / {(6.30 kg) + (2.20 kg) + [(5.30 kg) / 3]} (0.75 m)

= [(4.1 kg) (9.8 m/s²) (0.7419)] / [(10.2 kg) (0.75 m)]

= [(29.809542 kg.m/s²) / (7.65 kg.m)]

= 3.89 rad/s²