Two circular discs of radius each 10 cm are joined at their centres by a rod of length 30 cm and mas

Question

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 $43 	imes 10^5$ dyne.cm, the angular acceleration of the discs about the given axis $A B$ is $\_\_\_\_$ $\mathrm{rad} / \mathrm{s}^2$.

Professor PC Thomas Classes · Answer

The total moment of inertia of the system about the central perpendicular axis AB assumes the planes of the identical discs lie parallel to the central rod. For the rod: $I_{rod} = \frac{mL^2}{12} = \frac{600 	imes 30^2}{12} = 45000$ g cm$^2$. For the two discs (using the parallel axis theorem): $I_{discs} = 2 	imes [I_{cm} + M(L/2)^2] = 2 	imes [\frac{600 	imes 10^2}{2} + 600 	imes 15^2] = 2 	imes [30000 + 135000] = 330000$ g cm$^2$. The total inertia is $I = 45000 + 330000 = 375000$ g cm$^2$. Using $	au = I\alpha$, the angular acceleration is $\alpha = \frac{43 	imes 10^5}{375000} \approx 11.46$ rad/s$^2$, which closely matches 11.