91Ó°ÊÓ

$\begin{array}{c}{\mathrm{Ó\neg }}_{\mathrm{i}}=4.7\text{rad}/\text{s}\\ {\mathrm{I}}_{\mathrm{i}}=5×{10}^{−4}{\text{kg.m}}^{\text{2}}\\ \mathrm{M}=0.0\text{2kg}\\ \mathrm{R}=0.10\text{m}\end{array}$

Find the initial angular momentum of the record before dropping the putty on it. As angular momentum is conserved when no external torque acts on it, find the final angular velocity of the record after dropping the putty on it.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula is as follow:

$\mathbf{Initial}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}\mathbf{}\mathbf{=}\mathbf{}\mathbf{Final}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}$

According to law of conservation of momentum:

$\text{Initialangularmomentum}=\text{Finalangularmomentum}$

$\begin{array}{c}{\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}}\\ {\mathrm{I}}_{\mathrm{i}}{\mathrm{Ó\neg }}_{\mathrm{i}}={\mathrm{I}}_{\mathrm{f}}{\mathrm{Ó\neg }}_{\mathrm{f}}\\ {\mathrm{I}}_{\mathrm{f}}={\mathrm{I}}_{\mathrm{i}}+{\mathrm{mR}}^2=5×{10}^{−4}+(0.02×{0.1}^2)=7×{10}^{−4}{\text{kg.m}}^{\text{2}}\\ 5×{10}^{−4}×4.7=(7×{10}^{−4})×{\mathrm{Ó\neg }}_{\mathrm{f}}\end{array}$

${\mathrm{Ó\neg }}_{\mathrm{f}}=3.4\text{rad/s}$

Hence,the angular speed of record immediately after dropping the putty on the record is$\mathrm{Ó\neg }=3.\text{4rad/s.}$

Therefore, by applying law of conservation of angular momentum, the final angular speed can be calculated.