MHT CET202525 Apr 2025Morning ShiftPhysicsRotational MotionActual
Two discs of moment of inertia ' I ₁ ' and ' I ₂ ' and angular speeds ' ₁ ^ and ' ₂ ' are rotating along the collinear axes passing through their centre of mass and perpendicular to their plane. If the two discs are made to rotate together along the same axis. The rotational kinetic energy of the system will be
Options
- AI ₁ ₁+ I ₂ ₂ 2 ( I ₁+ I ₂ )^2
- B(I₁ ₁-I₂ ₂ )^2 2 (I₁+I₂ )
- C( I ₁ ₁+ I ₂ ₂ )^2 2 ( I ₁- I ₂ )
- D( I ₁ ₁+ I ₂ ₂ )^2 2 ( I ₁+ I ₂ )
Correct answer
D. ( I ₁ ₁+ I ₂ ₂ )^2 2 ( I ₁+ I ₂ )
Step-by-step solution
Conservation of angular momentum applies since no external torques act on the system. The initial angular momentum is L_i = I₁ ₁ + I₂ ₂ . When rotating together, the combined moment of inertia becomes I_f = I₁ + I₂ with common angular speed _f . The final angular momentum is L_f = (I₁ + I₂) _f . Equating L_i = L_f gives: I₁ ₁ + I₂ ₂ = (I₁ + I₂) _f Solving for _f : _f = I₁ ₁ + I₂ ₂ I₁ + I₂ The final kinetic energy is: K_f = 1 2 (I₁ + I₂) _f^2 = 1 2 (I₁ + I₂) ( I₁ ₁ + I₂ ₂ I₁ + I₂ )^2 Simplifying: K_f = (I₁ ₁ + I₂ ₂)