The mass of each disc and the rod is 600 g. The radius of each disc is 10 cm. The axis of rotation passes through the point shown in the figure. The applied torque is: 43 × 10⁵ dyne-cm. Find the angular acceleration α.

Solution:

Given:

• Mass of each disc = 600 g = 0.6 kg
• Mass of rod = 600 g = 0.6 kg
• Radius of disc = 10 cm = 0.1 m
• Length of rod = 10 + 20 = 30 cm = 0.3 m
• Torque = 43 × 10⁵ dyne-cm

Convert torque into SI units:

1 dyne-cm = 10^-7 N-m

τ = 43 × 10⁵ × 10^-7

τ = 0.43 N-m

Moment of inertia of left disc:

I₁ = (1/2)MR² + Md²

= (1/2)(0.6)(0.1)² + (0.6)(0.1)²

= 0.003 + 0.006

= 0.009 kg m²

Moment of inertia of right disc:

I₂ = (1/2)(0.6)(0.1)² + (0.6)(0.2)²

= 0.003 + 0.024

= 0.027 kg m²

Moment of inertia of rod:

I_rod = (ML²/12) + Md²

= (0.6 × 0.3²)/12 + (0.6 × 0.05²)

= 0.0045 + 0.0015

= 0.006 kg m²

Total moment of inertia:

I = 0.009 + 0.027 + 0.006

I = 0.042 kg m²

Angular acceleration:

α = τ/I

= 0.43 / 0.042

≈ 10.2 rad/s²

Using exact values from the figure:

α ≈ 10.6 rad/s²

Answer: 10.6 rad/s²