Question

In FIGURE CP12.88, a 200 g toy car is placed on a narrow 60-cm-diameter track with wheel grooves that keep the car going in a circle. The 1.0 kg track is free to turn on a frictionless, vertical axis. The spokes have negligible mass. After the car’s switch is turned on, it soon reaches a steady speed of 0.75 m/s relative to the track. What then is the track’s angular velocity, in rpm?

Short answer

Angular velocity of the track is 3.974 rev /min

Step-by-step solution

Given information

Mass of the toy =200 g =0.2 kg
Diameter of the track =60 cm =0.6 m
So radius of track =0.3 m
Mass of the frictionless track= 1 kg
Speed relative to track (v)=0.75m/s

Explanation

First find the angular speed

$$\begin{gathered} \omega =\frac{v}{r} \\ \Rightarrow \omega =\frac{0.75(\mathrm{m}/\mathrm{s})}{(0.3\mathrm{m})}=2.5\mathrm{rad}/\sec \dots \dots \dots \dots \dots \dots \dots \left(1\right) \end{gathered}$$

From The law of conservation of angular momentum is constant

L = L^'

Momentum = I x angular velocity

$I\omega =I^{\text{'}}{\omega }^{\text{'}}$

$$\begin{gathered} \Rightarrow m{r}^2\omega =\left(M{r}^2+m{r}^2\right){\omega }^{\text{'}} \\ \Rightarrow {\omega }^{\text{'}}=\frac{m{r}^2\omega }{(M+m)r^2}..............................\left(2\right) \end{gathered}$$

Now substitute the given values

$$\begin{gathered} \Rightarrow {\omega }^{\text{'}}=\frac{(0.2kg)\left(2.5\frac{\mathrm{rad}}{\mathrm{s}}\right)}{(1\mathrm{kg}+0.2\mathrm{kg})} \\ \Rightarrow {\omega }^{\text{'}}=0.416\frac{\mathrm{rad}}{\mathrm{s}} \\ \Rightarrow {\omega }^{\text{'}}=\frac{0.416}{2\pi }\times 60 \\ \Rightarrow {\omega }^{\text{'}}=3.974\text{revolution}/\min \end{gathered}$$