Question

A wheel is rotating freely with an angular speed $\omega$ on a shaft. The moment of inertia of the wheel is $I$ and the moment ot inertia of the shaft is negligible. Another wheel of moment of inertia $3I$ initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is

• A. $\frac{5}{6}$
• B. $\frac{1}{4}$
• C. $0$
• D. $\frac{3}{4}$

Answer 1

The correct option is D $\frac{3}{4}$

By angular momentum conservation, $\Rightarrow L_c=L_f$
$\Rightarrow \omega I+3I\times 0=4I{\omega }^{\prime }\Rightarrow {\omega }^{\prime }=\frac{\omega }{4}$
Initial Kinetic energy of system is
$(KE{)}_i=\frac{1}{2}I{\omega }^2$
Final Kinetic energy of system is
$(KE{)}_f=\frac{1}{2}(3I+I){\omega }^{\prime 2}=\frac{1}{2}\times (4I)\times {\left(\frac{\omega }{4}\right)}^2=\frac{I{\omega }^2}{8}$
Loss of Kinetic energy is
$\Delta KE=\frac{1}{2}I{\omega }^2-\frac{1}{8}I{\omega }^2=\frac{3}{8}I{\omega }^2$
$\therefore$ Fractional loss in K.E. $\frac{\Delta KE}{K{E}_{I_i}}=\frac{\frac{3}{8}I{\omega }^2}{\frac{1}{2}I{\omega }^2}=\frac{3}{4}$