Given data
The mass of the first solid is m1 = 14 kg
The mass of the second solid is m2 = 6 kg
The radius of the first solid is r1 = 0.6 m
The radius of the second solid is r2 = 0.4 m
The angular speed of the first solid is w1 = 50 rad/s
The angular speed of the second solid is w2 = 0 rad/s
The expression for the final common angular speed of the system from the conservation of angular momentum is given as:
[tex]\begin{gathered} I_1\omega_1+I_2\omega_2=(I_1+I_2)\omega_f \\ \omega_f=\frac{I_1\omega_1+I_2\omega_2}{(I_1+I_2)} \\ \omega_f=\frac{\frac{m_1(r_1)^2}{2}\omega_1+\frac{m_2(r_2)^2}{2}_{}\omega_2}{(\frac{m_1(r_1)^2}{2}+\frac{m_2(r_2)^2}{2})} \end{gathered}[/tex]
Substitute the value in the above equation.
[tex]\begin{gathered} \omega_f=\frac{\frac{14\text{ kg}\times(0.6m)^2}{2}\times50\text{ rad/s +}\frac{6\text{ kg}\times(0.4m)^2}{2}\times0\text{ rad/s}}{\frac{14\text{ kg}\times(0.6m)^2}{2}+\frac{6\text{ kg}\times(0.4m)^2}{2}} \\ \omega_f=42\text{ rad/s} \end{gathered}[/tex]
Thus, the final common angular speed of the system is 42 rad/s.
(b)
The expression for the energy lost to friction in the collision is given as:
[tex]\begin{gathered} \Delta E=E_i-E_f \\ \Delta E=\frac{1}{2}I_1(\omega_1)^2-\frac{1}{2}(I_1+I_2)(\omega_f)^2 \end{gathered}[/tex]
Substitute the value in the above equation.
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