Question

A round disc of moment of inertia ${{I}_{2}}$ about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia ${{I}_{1}}$ rotating with angular velocity $\omega$ long the same axis. The final angular velocity of the combination of the discs is:-
$\begin{aligned} & a)\omega \\\ & b)\dfrac{{{I}_{1}}\omega }{{{I}_{1}}+{{I}_{2}}} \\\ & c)\dfrac{({{I}_{1}}+{{I}_{2}})\omega }{{{I}_{1}}} \\\ & d)\dfrac{{{I}_{2}}}{\omega }{{I}_{1}}+{{I}_{2}} \\\ \end{aligned}$

Answer 1

The above problem can be solved by using the law of conservation of angular momentum. When the disc is placed over another disc with some angular velocity, both of them will move together with common angular velocity. Hence we can find the angular velocity by equating the final angular momentum with the initial angular momentum of a round disc. The angular momentum of body with radius r and linear momentum i.e. p=’mv’ (where m is the mass of the body and v is the tangential velocity to the circular path) is given by $L=r\times p$

Complete answer:
To begin with let us first express the above equation of angular momentum in terms of Moment of inertia and angular velocity.
Let us consider a circular disc of radius r and the mass of the disc as M. Let linear velocity or the tangential velocity be v and its angular velocity be $\omega$. The angular momentum of the disc is given by,
$L=r\times \text{p where p is linear momentum the disc}\text{.}$ Let us say a particle of mass m moves along the circumference of the disc with linear velocity v. Hence its angular momentum is given by,

$\begin{aligned} & L=r\times \text{p } \\\ & L=rmv.\text{ } \\\ \end{aligned}$

Multiplying the above equation by r and dividing by r we get,

$\begin{aligned} & L=rmv \\\ & \implies L=\dfrac{{{r}^{2}}mv}{r}\text{, since }\dfrac{v}{r}=\omega \\\ & \implies L={{r}^{2}}m\omega ,\text{Now let us sum this equation for entire disc,} \\\ & \implies L=\sum{{{r}^{2}}m\omega } \\\ & \implies L=\omega \sum{{{r}^{2}}m} \\\ & \implies L=\omega M{{r}^{2}},\text{now since}M{{r}^{2}}=I(\text{moment of inertia of the disc}) \\\ & \implies L=I\omega \\\ \end{aligned}$

By law of conservation of momentum,
$\begin{aligned} & {{L}_{INITIAL}}={{L}_{FINAL}} \\\ & \implies {{I}_{1}}\omega ={{I}_{1}}\alpha +{{I}_{2}}\alpha \text{, where }\,\,\alpha\,\,\text{ is the common angular velocity of both the discs} \\\ & {{I}_{1}}\omega =\alpha \left( {{I}_{1}}+{{I}_{2}} \right) \\\ & \implies \alpha =\dfrac{{{I}_{1}}\omega }{\left( {{I}_{1}}+{{I}_{2}} \right)} \\\ \end{aligned}$

So, the correct answer is “Option B”.

Note:
The axle along which both the discs are rotating is considered as massless . Hence the law of conservation of momentum is applicable. If the axle had some mass some of the momentum would have got lost to the axle in rotating that as well.