Question

A solid cylinder of mass $3\text{kg}$ is rolling without slipping on a smooth horizontal surface with velocity $4\text{m/s}$. It collides with a horizontal spring of spring constant $200{\text{Nm}}^{-1}$. The maximum compression produced in the spring will be

• A. $0.7\text{m}$
• B. $0.2\text{m}$
• C. $0.5\text{m}$
• D. $0.6\text{m}$
  

Answer 1

The correct option is D $0.6\text{m}$


For rolling without slipping,
$v=r\omega ...\left(i\right)$
Moment of inertia of solid cylinder about axis passing through its $COM$ is
$I_{\text{CM}}=\frac{M{r}^2}{2}$
$\Rightarrow$Hence rotational $K.E$ associated with cylinder is,
${K.E}_{\text{Rot}}=\frac{1}{2}{I}_{\text{CM}}{\omega }^2=\frac{1}{2}\times \frac{M{r}^2}{2}\times \left(\frac{v^2}{r^2}\right)$
$\therefore {K.E}_{\text{Rot}}=\frac{M{v}^2}{4}$
Translational $K.E$ associated with cylinder is,
${K.E}_{\text{Trans}}=\frac{1}{2}M{v}^2$
$\Rightarrow$Applying mechanical energy conservation on system of (solid cylinder+spring)
$\text{Loss in}{K.E}_{\text{Trans}}+\text{Loss in}{K.E}_{\text{Rot}}=\text{Gain in Elastic P.E of spring}$
$\frac{1}{2}M{v}^2+\frac{M{v}^2}{4}=\frac{1}{2}k{x}^2...\left(ii\right)$
Where $x=$ maximum compression in spring, such that cylinder stops there(i.e $v=0,\omega =0$)

From Eq. $\left(ii\right)$,
$\frac{1}{2}\times 3\times (4{)}^2+\frac{3\times 4^2}{4}=\frac{1}{2}\times 200\times x^2$
$x^2=\frac{36}{100}$
$\therefore x=\sqrt{\frac{36}{100}}=0.6\text{m}$