91Ó°ÊÓ

If the value of the net torque on a body is zero such that\(\Delta L/\Delta t = 0\), then\(L = {\rm{constant}}\). This is the law of conservation of angular momentum for a rotating body.

Given data:

The initial angular frequency of the uniform disk is \({\omega _1} = 3.3\;{{{\rm{rev}}} \mathord{\left/{\vphantom {{{\rm{rev}}} {\rm{s}}}} \right.} {\rm{s}}}\).

The expression for the moment of inertia of the uniform disk is as follows:

\({I_1} = \frac{1}{2}M{R^2}\)

Here,\(M\)is the mass of the uniform disk and\(R\)is the radius of the uniform disk.

The distance between the mass at each end and the rotation axis is equal to half its length. That is:

\(\begin{aligned}{c}R &= \frac{l}{2}\\l &= 2R\end{aligned}\)

The expression for the moment of inertia of the system of uniform rod and the disk is as follows:

\(\begin{aligned}{c}{I_2} &= {I_{{\rm{rod}}}} + {I_{{\rm{disk}}}}\\{I_2} &= \frac{1}{2}M{l^2} + \frac{1}{2}M{R^2}\\{I_2} &= \frac{1}{{12}}M{\left( {2R} \right)^2} + \frac{1}{2}M{R^2}\\{I_2} &= \frac{5}{6}M{R^2}\end{aligned}\)

Apply the conservation of angular momentum to calculate the final angular frequency of the combination.

\(\begin{aligned}{c}{I_1}{\omega _1} &= {I_2}{\omega _2}\\{\omega _2} &= \frac{{{I_1}{\omega _1}}}{{{I_2}}}\\{\omega _2} &= \frac{{\left( {\frac{1}{2}M{R^2}} \right)}}{{\left( {\frac{5}{6}M{R^2}} \right)}}\left( {3.3\;{{{\rm{rev}}} \mathord{\left/{\vphantom {{{\rm{rev}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\\{\omega _2} &= 1.98\;{{{\rm{rev}}} \mathord{\left/{\vphantom {{{\rm{rev}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the final angular frequency of the combination is \(1.98\;{{{\rm{rev}}} \mathord{\left/{\vphantom {{{\rm{rev}}} {\rm{s}}}} \right.} {\rm{s}}}\).