Question

In a playground, there is a small merry-go-round of radius 1.20 mand mass 180 kg. Its radius of gyration (see Problem 79 of Chapter 10) is 91.0 cm.A child of mass 44.0 kgruns at a speed of 3.00 m/salong a path that is tangent to the rim of the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round. Calculate (a) the rotational inertia of the merry-go-round about its axis of rotation, (b) the magnitude of the angular momentum of the running child about the axis of rotation of the merry-go-round, and (c) the angular speed of the merry-go-round and child after the child has jumped onto the merry-go-round.

Short answer

1. Rotational Inertia of merry go round about its axis of rotation is 149 kg.m².
2. The magnitude of the angular momentum of the running child about the axis of rotation of the merry-go-round is 158 kg.m²/s.
3. The angular speed of the merry-go-round and child after the child has jumped onto the merry-go-round is 0.744 rad/s².

Step-by-step solution

Step 1: Given

$$\begin{gathered} \mathrm{m}=180\mathrm{kg} \\ \mathrm{k}=91\mathrm{cm} \\ \mathrm{M}=44\mathrm{kg} \\ \mathrm{r}=1.20\mathrm{m} \\ \mathrm{v}=3\mathrm{m}/\mathrm{s} \end{gathered}$$

Determining the concept

Use the formula for rotational inertia in terms of mass and radius of gyration to find the rotational inertia. Using the formula for angular momentum in terms of mass, velocity and radius, find the angular momentum. Finally, use conservation of angular momentum to find angular velocity.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formulaare as follow:

$$\begin{gathered} \mathbf{I}\mathbf{=}\mathbf{m}\mathbf{\times }{\mathbf{k}}^{\mathbf{2}} \\ \mathbf{L}\mathbf{=}\mathbf{m}\mathbf{\times }\mathbf{v}\mathbf{\times }\mathbf{r} \end{gathered}$$

where,r is radius, v is velocity, m is mass, Lis angular momentum, l is moment of inertia and risradius of gyration.

Determining the rotational Inertia of merry go round about its axis of rotation

(a)

To find rotational inertia, use the following formula:

$$\begin{gathered} \mathrm{I}=\mathrm{m}\times {\mathrm{k}}^2 \\ \mathrm{I}=180\times 0.{91}^2 \\ \mathrm{I}=149\mathrm{kg}.{\mathrm{m}}^2 \end{gathered}$$

Hence, rotational Inertia of merry go round about its axis of rotation is 149 kg.m².

 Determining the magnitude of the angular momentum of the running child about the axis of rotation of the merry-go-round

(b)

Now, to find angular momentum,

$$\begin{gathered} \mathrm{L}=\mathrm{M}\times \mathrm{r}\times \mathrm{v} \\ \mathrm{L}=44\times 3\times 1.2 \end{gathered}$$

So,

data-custom-editor="chemistry" $\mathrm{L}=158\mathrm{kg}.{\mathrm{m}}^2/\mathrm{s}$

Hence, the magnitude of the angular momentum of the running child about the axis of rotation of the merry-go-round is 158 kg.m²/s.

 Determining the angular speed of the merry-go-round and child after the child has jumped onto the merry-go-round

(c)

Now, to find angular velocity, use conservation of momentum,

$$\begin{gathered} {\mathrm{L}}_{\mathrm{f}}={\mathrm{L}}_{\mathrm{child}} \\ {\mathrm{L}}_{\mathrm{f}}=\left(\mathrm{I}+{\mathrm{mr}}^2\right)\mathrm{\omega } \\ {\mathrm{L}}_{\mathrm{child}}=\mathrm{mvr} \\ \mathrm{mvr}=\left(\mathrm{I}+{\mathrm{mr}}^2\right)\mathrm{\omega } \\ \mathrm{\omega }=\frac{\mathrm{mvr}}{\mathrm{I}+{\mathrm{mr}}^2} \end{gathered}$$

So,

$$\begin{gathered} \mathrm{\omega }=\frac{158}{149+44+1.2^2}=0.744\mathrm{rad}/\mathrm{s} \\ \mathrm{\omega }=0.744\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence,the angular speed of the merry-go-round and child after the child has jumped onto the merry-go-round is 0.744 rad/s².

Therefore, use the formula for rotational inertia in terms of mass and radius of gyration to find the rotational inertia. Using the formula for angular momentum in terms of mass, velocity and radius, the angular momentum can be found. Finally, conservation of angular momentum can be used to find angular velocity.