Question: (II) A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.0 m and moment of inertia\({\bf{820}}\;{\bf{kg}} \cdot {{\bf{m}}^{\bf{2}}}\). The platform rotates without friction with angular velocity\({\bf{0}}{\bf{.95}}\;{{{\bf{rad}}} \mathord{\left/{\vphantom {{{\bf{rad}}} {\bf{s}}}} \right.} {\bf{s}}}\). The person walks radially to the edge of the platform. (a) Calculate the angular velocity when the person reaches the edge. (b) Calculate the rotational kinetic energy of the system of platform plus person before and after the person’s walk.

The rotational kinetic energy of a rotating object is equal to half the product of its moment of inertia and the square of its angular speed.

Given data:

The initial moment of inertia is \({I_{\rm{i}}} = 820\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

The initial angular velocity is \({\omega _{\rm{i}}} = 0.95\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\).

The mass of the person is \(m = 75\;{\rm{kg}}\).

The radius of the platform is \(R = 3.0\;{\rm{m}}\).

(a)

As the person reaches the edge, the final moment of inertia of the system is equal to the initial moment of inertia and the moment of inertia of the person. Thus, you can write:

\({I_{\rm{f}}} = {I_{\rm{i}}} + m{R^2}\)

Now, apply the conservation of angular momentum to calculate the final angular velocity.

\(\begin{aligned}{c}{I_{\rm{f}}}{\omega _{\rm{f}}} &= {I_{\rm{i}}}{\omega _{\rm{i}}}\\{\omega _{\rm{f}}} &= \frac{{{I_{\rm{i}}}{\omega _{\rm{i}}}}}{{{I_{\rm{f}}}}}\\{\omega _{\rm{f}}} &= \frac{{{I_{\rm{i}}}{\omega _{\rm{i}}}}}{{\left( {{I_{\rm{i}}} + m{R^2}} \right)}}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}{\omega _{\rm{f}}} &= \frac{{\left( {820\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right)\left( {0.95\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.}{\rm{s}}}} \right)}}{{\left( {\left( {820\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right) + \left( {75\;{\rm{kg}}} \right){{\left( {3.0\;{\rm{m}}} \right)}^2}} \right)}}\\{\omega _{\rm{f}}} &= 0.52\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the angular velocity of the system when the person reaches the edge is\(0.52\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\).

(b)

The rotational kinetic energy of the system of platform and person before the person’s walk can be calculated as:

\(\begin{aligned}{c}{K_{\rm{i}}} &= \frac{1}{2}{I_{\rm{i}}}\omega _{\rm{i}}^2\\{K_{\rm{i}}} &= \frac{1}{2}\left( {820\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right){\left( {0.95\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.}{\rm{s}}}} \right)^2}\\{K_{\rm{i}}} &= 370.0\;{\rm{J}}\end{aligned}\)

Thus, the rotational kinetic energy of the system of platform and person before the person’s walk is \(370.0\;{\rm{J}}\).

The rotational kinetic energy of the system of platform and person after the person’s walk can be calculated as:

\(\begin{aligned}{c}{K_{\rm{f}}} &= \frac{1}{2}{I_{\rm{f}}}\omega _{\rm{f}}^2\\{K_{\rm{f}}} &= \frac{1}{2}\left( {{I_{\rm{i}}} + m{R^2}} \right)\omega _{\rm{f}}^2\\{K_{\rm{f}}} &= \frac{1}{2}\left( {\left( {820\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right) + \left( {75\;{\rm{kg}}} \right){{\left( {3.0\;{\rm{m}}} \right)}^2}} \right){\left( {0.52\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)^2}\\{K_{\rm{f}}} &= 2.02 \times {10^2}\;{\rm{J}}\end{aligned}\)

Thus, the rotational kinetic energy of the system of platform and person after the person’s walk is \(2.02 \times {10^2}\;{\rm{J}}\).