Question

The moment of inertia of a uniform disc about an axis passing through its centre and perpendicular to its plane is $1kg{m}^2$. It is rotating with an angular velocity $100rad/s$. 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 $kJ$ is:

• A. $2.5$
• B. $3.0$
• C. $3.5$
• D. $4.0$

Answer 1

The correct option is A $2.5$

$I_1{\omega }_1+I_2{\omega }_2=(I_1+I_2)\omega$, by conservation of momentum

$\Rightarrow \omega =\frac{{\omega }_1}{2}$

$\therefore lossinkineticenergy=\frac{1}{2}{I}_1{\omega }_1^2-\frac{1}{2}(I_1+I_2){\omega }_2$

$=\frac{1}{2}\times 1\times (100{)}^2-\frac{1}{2}(1+1){\left(\frac{100}{2}\right)}^2$

$=\frac{1}{2}(10000-5000)$

$=2500=2.5kJ$