Question

Part A Max (15 kg) and Maya (12 kg) are riding on a merry-go-round that rotates at a constant speed. Max is sitting on the edge of the merry-go-round, 2.4 m from the center, and Maya is 1.2 m from the center. Considering Max and Maya to be one system of masses, what is their moment of inertia measured with respect to the center of the merry-go-round? O 1500 kg.m2 0 17 kg-m2 O50 kg.m2 O 104 kg-m2 86 kg-m2 Submit My Answers Give Up

Answer 1

Concepts and reason

The concepts that are to be sued to solve the given problem are moment of inertia, rotational kinetic energy, and angular velocity.

Use the concept of moment of inertia of the object of the disk to find the total moment of inertia of the Max and Maya. Then, use the concept of moment of inertia of the object and the moment of inertia of the disk to find the total moment of inertia of the system. Finally, use the rotational kinetic energy expression to find the distance to Maya’s location from the axis of rotation.

Fundamentals

The expression for the moment of inertia I of the object rotating about the axis is,

$I = m{r^2}$

Here, m is the mass of the object and r radius of rotation.

The rotational kinetic energy ${E_r}$ expression is,

${E_r} = \frac{1}{2}I{\omega ^2}$

Here, $I$ is the rotational inertia (moment of inertia) and $\omega$ is the angular velocity.

(A)

The moment of inertia of Max ${I_{\max }}$ is,

${I_{\max }} = M{R^2}$

Here, M is the mass and R is the distance of Max from the axis.

The moment of inertia of Maya ${I_{{\rm{maya}}}}$ is,

${I_{{\rm{maya}}}} = m{r^2}$

Here, m is the mass and r is the distance of Maya from the axis.

Total moment of inertia is,

$I = {I_{\max }} + {I_{{\rm{maya}}}}$

Replace ${I_{\max }}$ with $M{R^2}$ and ${I_{{\rm{maya}}}}$ with $m{r^2}$ .

$I = M{R^2} + m{r^2}$

Substitute 15 kg for M, 12 kg for m, 2.4 m for R, and 1.2 m for r.

$\begin{array}{c}\\I = \left( {15{\rm{ kg}}} \right){\left( {2.4{\rm{ m}}} \right)^2} + \left( {12{\rm{ kg}}} \right){\left( {1.2{\rm{ m}}} \right)^2}\\\\ = 104{\rm{ kg}} \cdot {{\rm{m}}^2}\\\end{array}$

(B)

The moment of inertia of marry-go-round ${I_M}$ is,

${I_M} = \frac{1}{2}{M_0}{R^2}$

Here, ${M_0}$ is the mass of marry-go-round and R is the radius of marry-go-round.

Total moment of inertia of the system is,

$I = {I_{\max }} + {I_{{\rm{maya}}}} + {I_M}$

Replace ${I_{\max }}$ with $M{R^2}$ , ${I_M}$ with $\frac{1}{2}{M_0}{R^2}$ and ${I_{{\rm{maya}}}}$ with $m{r^2}$ .

$I = M{R^2} + m{r^2} + \frac{1}{2}{M_0}{R^2}$

Substitute 15 kg for M, 12 kg for m, 2.4 m for R, 1.2 m for r, 230 kg for ${M_0}$ .

$\begin{array}{c}\\I = \left( {15{\rm{ kg}}} \right){\left( {2.4{\rm{ m}}} \right)^2} + \left( {12{\rm{ kg}}} \right){\left( {1.2{\rm{ m}}} \right)^2} + \frac{1}{2}\left( {230{\rm{ kg}}} \right){\left( {2.4{\rm{ m}}} \right)^2}\\\\ = 766{\rm{ kg}} \cdot {{\rm{m}}^2}\\\end{array}$

The rotational kinetic energy ${E_r}$ expression is,

${E_r} = \frac{1}{2}I{\omega ^2}$

Replace I with the equation $M{R^2} + m{r^2} + \frac{1}{2}{M_0}{R^2}$ in the above equation.

${E_r} = \frac{1}{2}\left( {M{R^2} + m{r^2} + \frac{1}{2}{M_0}{R^2}} \right){\omega ^2}$

Substitute 15 kg for M, 12 kg for m, 2.4 m for R, 8700 J for ${E_r}$ , 0.75 rev/s for $\omega$ , 230 kg for ${M_0}$ and solve for r.

$\begin{array}{c}\\\left( {8700{\rm{ J}}} \right) = \frac{1}{2}\left( {\left( {15{\rm{ kg}}} \right){{\left( {2.4{\rm{ m}}} \right)}^2} + \left( {12{\rm{ kg}}} \right){r^2} + \frac{1}{2}\left( {230{\rm{ kg}}} \right){{\left( {2.4{\rm{ m}}} \right)}^2}} \right){\left( {0.75{\rm{ rev/s}}} \right)^2}{\left( {\frac{{2\pi {\rm{ rad}}}}{{1{\rm{ rev}}}}} \right)^2}\\\\r = 1.7{\rm{ m}}\\\end{array}$

Ans: Part A

The moment of inertia of Max and Maya is $104{\rm{ kg}} \cdot {{\rm{m}}^2}$ .