The expression for the angular speed of the turntable is as follows:
\({\omega _{\rm{T}}} = \frac{{{v_{\rm{T}}}}}{R}\)
Here,\({v_{\rm{T}}}\)is the linear speed of the turntable and\(R\)is the radius.
The expression for the linear speed of the person relative to the ground is as follows:
\({v_{\rm{P}}} = v + {v_{\rm{T}}}\)
Here,\({v_{\rm{P}}}\)is the linear speed of the person relative to the ground and\({v_{\rm{T}}}\)is the linear speed of the turntable relative to the ground.
The expression for the angular speed of the person relative to the ground is as follows:
\(\begin{aligned}{c}{\omega _{\rm{P}}} &= \frac{{{v_{\rm{P}}}}}{R}\\{\omega _{\rm{P}}} &= \frac{{\left( {v + {v_{\rm{T}}}} \right)}}{R}\\{\omega _{\rm{P}}} &= \frac{v}{R} + \frac{{{v_{\rm{T}}}}}{R}\\{\omega _{\rm{P}}} &= \frac{v}{R} + {\omega _{\rm{T}}}\end{aligned}\)
Since the person and turntable are both at rest initially, the initial momentum of the system should be equal to zero. That is,
\({L_{\rm{i}}} = 0\)
The expression for the moment of inertia of the person is as follows:
\({I_{\rm{P}}} = m{R^2}\)
Now, apply the conservation of angular momentum.
\(\begin{aligned}{c}{L_{\rm{i}}} &= {L_{\rm{f}}}\\0 &= {I_{\rm{T}}}{\omega _{\rm{T}}} + {I_{\rm{P}}}{\omega _{\rm{P}}}\\0 &= {I_{\rm{T}}}{\omega _{\rm{T}}} + m{R^2}\left( {\frac{v}{R} + {\omega _{\rm{T}}}} \right)\\0 &= {I_{\rm{T}}}{\omega _{\rm{T}}} + mvR + m{R^2}{\omega _{\rm{T}}}\\{\omega _{\rm{T}}} &= - \frac{{mvR}}{{{I_{\rm{T}}} + m{R^2}}}\end{aligned}\)
Substitute the values in the above expression.
\(\begin{aligned}{l}{\omega _{\rm{T}}} &= - \frac{{\left( {65\;{\rm{kg}}} \right)\left( {4.0\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\left( {2.75\;{\rm{m}}} \right)}}{{\left( {1850\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right) + \left( {65\;{\rm{kg}}} \right){{\left( {2.75\;{\rm{m}}} \right)}^2}}}\\{\omega _{\rm{T}}} &= - 0.31\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)
