Question

A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Figure). A toy train of mass mis placed on the track and, with the system initially at rest, the train’s electrical power is turned on. The train reaches speed$\mathbf{0}\mathbf{.15}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$with respect to the track. What is the angular speed of the wheel if its mass is$\mathbf{\text{1.1m}}$and its radius is$\mathbf{0}\mathbf{.43}\mathbf{}\mathbf{m}\mathbf{?}$(Treat the wheel as a hoop, and neglect the mass of the spokes and hub.)

Short answer

The angular speed of the wheel is$0.17\mathrm{rad}/\mathrm{s}$

Step-by-step solution

Given

1. The mass of the train is$\mathrm{m}$
2. The mass of the wheel is$,\mathrm{M}=1.1\mathrm{m}.$
3. The radius of the wheel is$,r=0.43m.$
4. The velocity of train with respect to wheel (track)${\text{v}}_{\mathrm{tw}}=0.15\text{m}/\text{s}$.
5. Initial velocity of the system is,${\mathrm{v}}_{\mathrm{i}}=0\mathrm{m}/\mathrm{s}$.

To understand the concept

Find angular momentum of the wheel by using the given values. Then by using the concept of relativity, find the velocity of the train with respect to the observer. This can be used to calculate the angular momentum of the train. From angular momentum of the wheels and train, get the final angular momentum of the system. After equating it to the initial angular momentum of the system, the angular speed of the wheel can be found.

Formula:

According to the law of conservation of angular momentum,${\mathbf{L}}_{\mathbf{i}}\mathbf{}\mathbf{=}{\mathbf{L}}_{\mathbf{f}}$

Calculate the angular speed of the wheel

The initial angular momentum of the system is zero as no torque is acting on the system of train and wheel.

The final momentum of the wheel is,

$\begin{array}{c}{\mathrm{L}}_{\mathrm{wf}}=\mathrm{I\omega }\\ \Rightarrow {\mathrm{L}}_{\mathrm{wf}}=-{\mathrm{Mr}}^2\mathrm{\omega }\end{array}$

The velocity of the train with respect to the observer = velocity of train relative to the wheel + velocity of the wheel with respect to observer

$\begin{array}{l}{\mathrm{v}}_{\mathrm{to}}={\mathrm{v}}_{\mathrm{tw}}+{\mathrm{v}}_{\mathrm{wo}}\\ {\mathrm{v}}_{\mathrm{to}}={\mathrm{v}}_{\mathrm{tw}}-\mathrm{\omega r}\end{array}$

So, the final angular momentum of the train is

localid="1660981312139" ${\mathrm{L}}_{\mathrm{tf}}={\mathrm{mv}}_{\mathrm{to}}=\mathrm{m}({\mathrm{v}}_{\mathrm{tw}}-\mathrm{\omega r})\mathrm{r}$

So, the final angular momentum of the system is

${\mathrm{L}}_{\mathrm{wf}}+{\mathrm{L}}_{\mathrm{tf}}$

The law of conservation of angular momentum gives

$\begin{array}{c}{\text{L}}_{\text{i}}={\text{L}}_{\text{f}}\\ \Rightarrow 0=-{\mathrm{Mr}}^2\mathrm{\omega }+\mathrm{m}({\mathrm{v}}_{\mathrm{tw}}-\mathrm{\omega r})\mathrm{r}\\ \Rightarrow {\mathrm{Mr}}^2\mathrm{\omega }={\mathrm{mv}}_{\mathrm{tw}}\mathrm{r}-\mathrm{m\omega r}^2\\ \Rightarrow {\mathrm{Mr}}^2\mathrm{\omega }+{\mathrm{m\omega r}}^2={\mathrm{mv}}_{\mathrm{tw}}\mathrm{r}\end{array}$

$\begin{array}{c}\Rightarrow ({\mathrm{Mr}}^2+{\mathrm{mr}}^2)\mathrm{\omega }={\mathrm{mv}}_{\mathrm{tw}}\mathrm{r}\\ \Rightarrow \mathrm{\omega }=\frac{{\mathrm{mv}}_{\mathrm{tw}}\mathrm{r}}{{\mathrm{Mr}}^2+{\mathrm{mr}}^2}\\ \Rightarrow \mathrm{\omega }=\frac{{\mathrm{mv}}_{\mathrm{tw}}\mathrm{r}}{(\mathrm{M}+\mathrm{m}){\mathrm{r}}^2}\\ \Rightarrow \mathrm{\omega }=\frac{{\mathrm{mv}}_{\mathrm{tw}}}{(\mathrm{M}+\mathrm{m})\mathrm{r}}\end{array}$

$\begin{array}{l}\Rightarrow \mathrm{\omega }=\frac{\mathrm{m}\left(0.15\right)}{(1.1\mathrm{m}+\mathrm{m})0.43}\\ \Rightarrow \mathrm{\omega }=0.166~0.17\frac{\mathrm{rad}}{\mathrm{s}}\end{array}$

Therefore, the angular speed of the wheel is $0.17\mathrm{rad}/\mathrm{s}$.