Two circular discs of radius each 10 cm are joined at their centres by a rod of length 30 cm and mass 600 gm as shown in figure.
If the mass of each disc is 600 gm and applied torque between two discs is 43× 10⁵ dyne.cm, the angular acceleration of the discs about the given axis AB is ____ rad/s².

Solution

Physics sir · 2 steps

Find each piece's moment of inertia about AB, add them up, then divide the torque by the total.

1. Read the geometry. Axis AB is perpendicular to the rod, 10 cm from the left disc's centre and 20 cm from the right disc's centre. Each disc's face is perpendicular to the rod, so AB is parallel to a diameter of each disc. Start from 1/4MR² about a diameter and shift it to AB with the parallel axis theorem.

   Part	About its own centre (g cm²)	Shift to AB (Md²)	About AB (g cm²)
   Rod, axis 5 cm from its centre	1/12(600)(30)² = 45000	600 × 5² = 15000	60000
   Left disc, d = 10 cm	1/4(600)(10)² = 15000	600 × 10² = 60000	75000
   Right disc, d = 20 cm	1/4(600)(10)² = 15000	600 × 20² = 240000	255000
   Total	75000	315000	3.9 × 10⁵

   α = τ/I = (43 × 10⁵ dyne cm)/(3.9 × 10⁵ g cm²) ≈ 11.03 rad s⁻²

2. So. Dyne cm divided by g cm² gives s⁻², so the CGS units work out directly. The angular acceleration is about 11 rad s⁻², option (d).

The idea to keep. For a composite body, find each part's moment of inertia about its own centre, shift it to the given axis with Md², then add. First check which way the axis passes through each disc.

Answer from the NTA final answer key. Solution by PrepWiser.