The moment of inertia of a uniform disc about an axis passing through its centre and perpendicular to its plane is 1 kg m 2. It is rotating with an angular velocity 100 rad s -1. Another identical disc is gently placed on it so that their centres coincide. Now these two discs together continue to rotate about the same axis. Then the loss in kinetic energy in kilo joules is

Question

The moment of inertia of a uniform disc about an axis passing through its centre and perpendicular to its plane is 1 kg m 2. It is rotating with an angular velocity 100 rad s -1. Another identical disc is gently placed on it so that their centres coincide. Now these two discs together continue to rotate about the same axis. Then the loss in kinetic energy in kilo joules is (A) 2.5 (B) 3.0 (C) 3.5 (D) 4.0

Tardigrade Admin · Accepted Answer

As no external torque is applied to the system, the
angular momentum of the system remains conserved.

∴ L_{i} = L_{f}

where the subscripts represent initial and final
or

I_{i} ω_{i} = I_{f} ω_{f}

Substituting the given values, we get

∴ 1 \times 100 = 2 \times 1 \times ω_{f}

or ω_{f} = 50 r a d s^{- 1} .... (i)

Initial kinetic energy,

K_{i} = \frac{1}{2} I_{i} ω_{i}^{2} = \frac{1}{2} \times 1 \times (100)^{2} = 5 \times 1 0^{3} J

Final kinetic energy,

K_{f} = \frac{1}{2} I_{f} ω_{f}^{2} = \frac{1}{2} \times 2 \times 1 \times (50)^{2}

= 2.5 \times 1 0^{3} J

(Using (i))
Loss in kinetic energy,

Δ K = K_{i} - K_{f} = 5 \times 1 0^{3} - 2.5 \times 1 0^{3} J

= 2.5 \times 1 0^{3} J = 2.5 k J

$2.5$

$3.0$

$3.5$

$4.0$