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 ${\omega }_o$.

• A. $R{\omega }_o\sqrt{\left(2m/3k\right)}$
• B. $R{\omega }_o\sqrt{\left(3m/2k\right)}$
• C. $R{\omega }_o\sqrt{\left(m/k\right)}$
• D. None of these

Answer 1

The correct option is B $R{\omega }_o\sqrt{\left(3m/2k\right)}$

The initial energy of the disc,

$E_i={\left(K.E.\right)}_{\text{translational}}+{\left(K.E.\right)}_{\text{rotational}}$

$E_i=\frac{1}{2}I{{\omega }_0}^2+\frac{1}{2}m{v}^2$

$E_i=\frac{1}{2}\left(\frac{m{R}^2}{2}\right){{\omega }_0}^2+\frac{1}{2}m{\left(R{\omega }_0\right)}^2$

$E_i=\frac{3}{4}m{R}^2{\omega }_0^2$

Now, the final energy, $E_f=\frac{1}{2}k{x}^2$

Therefore,

$\frac{1}{2}k{x}^2=\frac{3}{4}m{R}^2{\omega }_0^2$

$\Rightarrow x^2={\left(R{\omega }_0\right)}^2\frac{3m}{2k}$

$\Rightarrow x=\left(R{\omega }_0\right)\sqrt{\frac{3m}{2k}}$