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

Answer 1

Answer: (B)

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