Question

Two discs are rotating about their axes, normal to the disc and passing through the centres of the discs. Disc $D_1$ has $2\text{kg}$ mass and $0.2\text{m}$ radius and its initial angular velocity is $50\text{rad/s}$. Disc $D_2$ has $4\text{kg}$ mass, $0.1\text{m}$ radius and its initial angular velocity is $200\text{rad/s}$. The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in $\text{rad/s}$) of the system is

Answer 1

Given for disc $D_1$ :
Mass of disc $m_1=2\text{kg}$
${\omega }_1=50\text{rad/s}$
$r_1=0.2\text{m}$
$I_1=\frac{m_1{r}_1^2}{2}=\frac{2\times (0.2{)}^2}{2}$
$=0.04{\text{kg m}}^2\text{/s}$

For disc $D_2$ :
Mass of disc $m_2=4\text{kg}$
${\omega }_2=200\text{rad/s}$
$r_2=0.1\text{m}$
$I_2=\frac{m_2{r}_2^2}{2}=\frac{4\times (0.1{)}^2}{2}$
$=0.02{\text{kg m}}^2\text{/s}$

If we take disc $D_1$ and $D_2$ as a system, torque between $D_1$ & $D_2$ will be internal, so there will be no external torque. Hence, angular momentum is conserved.
$L_i=L_f$
$\Rightarrow I_1{\omega }_1+I_2{\omega }_2=(I_1+I_2)\omega$
$\Rightarrow \left(0.04\right)\times 50+\left(0.02\right)\times 200=(0.04+0.02)\omega$
$\Rightarrow \omega =100\text{rad/s}$