Question

A disc of mass $M$ and radius $R$ has a spring of constant k attached to its centre, the other end of the spring being fixed to a vertical wall. If the disk rolls without slipping on a level floor, how far to the right does the centre of mass move, if initially the spring was unstretched and the angular speed of the disc was ${\omega }_0$

• A. $\text{R}\omega \sqrt{\left(2\text{M}/\text{3}\text{k}\right)}$
• B. $\text{R}\omega \sqrt{\left(\text{3}\text{M}/\text{2}\text{k}\right)}$
• C. $\text{R}\omega \sqrt{\left(\text{M}/\text{k}\right)}$
• D. $\text{R}\omega \sqrt{\left(\text{M}/2\text{k}\right)}$

Answer 1

Answer: (B)

Due to pure rolling ${\text{E}}_{t=0}={\text{E}}_{t=1}$
$\text{I}=\frac{{\text{MR}}^2}{2}$
${\text{V}}_{\text{cm}}=\omega \text{R}$
${\text{E}}_{\text{t}=0}=\frac{1}{2}{\text{MV}}_{\text{cm}}^2+\frac{1}{2}{\text{I}}_{\text{cm}}{\omega }^2$
$\frac{1}{2}\text{M}{\omega }^2{\text{R}}^2+\frac{1}{2}\frac{{\text{MR}}^2}{\text{2}}\times {\omega }^2$
$\frac{1}{2}\times \frac{3}{2}\text{M}{\omega }^2{\text{R}}^2$
${\text{E}}_{\text{t}=\text{T}}=\frac{1}{2}{\text{kx}}^2$
$\frac{1}{2}\times \frac{3}{2}\text{M}{\omega }^2{\text{R}}^2=\frac{1}{2}{\text{kx}}^2$
$\text{x}=\omega \text{R}\sqrt{\frac{3\text{M}}{2k}}$