Question: (III) Suppose a 65-kg person stands at the edge of a 5.5-m diameter merry-go-round turntable that is mounted on frictionless bearings and has a moment of inertia of \({\bf{1850}}\;{\bf{kg}} \cdot {{\bf{m}}^{\bf{2}}}\). The turntable is at rest initially, but when the person begins running at a speed of \({\bf{4}}{\bf{.0}}\;{{\bf{m}} \mathord{\left/{\vphantom {{\bf{m}} {\bf{s}}}} \right.} {\bf{s}}}\) (with respect to the turntable) around its edge, the turntable begins to rotate in the opposite direction. Calculate the angular velocity of the turntable.

The angular velocity of a rotating object may be defined as the rate at which the object shifts its angular position with respect to time.

It shows how fast an object rotates in a circular path around a center point.

Given data:

The mass of the person is\(m = 65\;{\rm{kg}}\).

The diameter of the merry-go-round is\(D = 5.5\;{\rm{m}}\).

The moment of inertia of the turntable is\({I_{\rm{T}}} = 1850\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

The speed of the person relative to the turntable is \(v = 4.0\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).

The radius of the merry-go-round can be calculated as:

\(\begin{aligned}{c}R &= \frac{D}{2}\\R &= \frac{{\left( {5.5\;{\rm{m}}} \right)}}{2}\\R &= 2.75\;{\rm{m}}\end{aligned}\)

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.

\(\begin{aligned}{l}{\omega _{\rm{T}}} &= - \frac{{\left( {65\;{\rm{kg}}} \right)\left( {4.0\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\left( {2.75\;{\rm{m}}} \right)}}{{\left( {1850\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right) + \left( {65\;{\rm{kg}}} \right){{\left( {2.75\;{\rm{m}}} \right)}^2}}}\\{\omega _{\rm{T}}} &= - 0.31\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)