Question

Two discs are mounted on frictionless bearings on a common shaft. The first disc has rotational inertia I and is spinning with angular velocity $\omega$. The second disc has rotational inertia 21 and is spinning in opposite direction with angular velocity $3\omega$, as shown in figure. The two discs are slowly forced towards each other along the shaft until they couple and have a final common angular velocity of

• A. $\frac{-5\omega }{3}$
• B. $\frac{7\omega }{3}$
• C. $2\omega$
• D. $1.5\omega$

Answer 1

The correct option is A $\frac{-5\omega }{3}$

$\begin{array}{c}\text{Initial angular momentum of disc}1\to \\ L_1=I\omega \\ \text{For disc}2\to \\ L_2=-6I\omega \\ L_i=I\omega -6I\omega =-5I\omega \end{array}$

$\begin{array}{c}\text{By conservation of angular momentum-}\\ \text{Final angular momentum-}\\ L_f=L_i=-5\text{Iw}D\\ \text{Final moment of Inertia of the combined disc will be}\\ I^{\prime }=I+2I=3I\\ \Rightarrow L_f=3I{\omega }_{+}-\left(2\right)\\ \text{From eq}D\text{and}\left(2\right)\text{we get}{\omega }_f=-\frac{5}{3}{\omega }_0\end{array}$