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. $4.0$

Answer 1

Answer: (A)

As no external torque is applied to the system, the
angular momentum of the system remains conserved.
$\therefore L_{i}=L_{f}$
where the subscripts represent initial and final
or $I_{i} \omega_{i}=I_{f} \omega_{f}$
Substituting the given values, we get
$\therefore 1 \times 100=2 \times 1 \times \omega_{f} $
$\text { or } \omega_{f}=50\, rad\, s ^{-1} \,\,\,\, ....(i)$
Initial kinetic energy,
$K_{i}=\frac{1}{2} I_{i} \omega_{i}^{2}=\frac{1}{2} \times 1 \times(100)^{2}=5 \times 10^{3} J$
Final kinetic energy, $K_{f}=\frac{1}{2} I_{f} \omega_{f}^{2}=\frac{1}{2} \times 2 \times 1 \times(50)^{2}$
$=2.5 \times 10^{3} J \,\,\,\,$ (Using (i))
Loss in kinetic energy, $\Delta K=K_{i}-K_{f}=5 \times 10^{3}-2.5 \times 10^{3} J$
$=2.5 \times 10^{3} J =2.5\, kJ$