Question

A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis. Consider a toy train of mass M placed on the track with the system initially at rest, the train's electrical power is turned on. The train reaches speed v with respect to the track. What is the wheel's angular speed if its mass is m and its radius is R? (Treat it as a hoop and neglect the mass of the spokes and hub).

• A. $\frac{\mathrm{v}}{\left({\displaystyle \frac{\mathrm{m}}{\mathrm{M}}+1}\right)\mathrm{R}}$
• B. $\frac{\mathrm{v}}{\left({\displaystyle \frac{\mathrm{m}}{\mathrm{M}}+2}\right)\mathrm{R}}$
• C. $\frac{\mathrm{v}}{\left({\displaystyle \frac{\mathrm{M}}{\mathrm{m}}-1}\right)\mathrm{R}}$
• D. $\frac{\mathrm{v}}{\left({\displaystyle \frac{\mathrm{M}}{\mathrm{m}}-2}\right)\mathrm{R}}$

Answer 1

The correct option is A

$\frac{\mathrm{v}}{\left({\displaystyle \frac{\mathrm{m}}{\mathrm{M}}+1}\right)\mathrm{R}}$

Step 1: Given

Wheel mass, ${\mathrm{m}}_1=\mathrm{m}$

Wheel Radius, $\text{Radius}=\mathrm{R}$

Train mass, ${\mathrm{m}}_2=\mathrm{M}$

Step 2: To find

We have to find the heel's angular speed.

Step 3: Calculate the wheel's angular speed by using conservation of angular momentum

$$\begin{gathered} {\mathrm{L}}_{\mathrm{wheel}}={\mathrm{L}}_{\mathrm{train}} \\ {\mathrm{mR}}^2\mathrm{\omega }=\mathrm{M}\left(\mathrm{v}-\mathrm{R\omega }\right)\mathrm{R} \\ \mathrm{\omega }=\frac{\mathrm{v}}{\left({\displaystyle \frac{\mathrm{m}}{\mathrm{M}}+1}\right)\mathrm{R}} \end{gathered}$$

Hence, Option(A) is correct.