Question

A ring of mass ${}^{\prime }{m}^{\prime }$ is rolling without slipping with velocity of centre of mass $\left(v\right)$ as shown in figure. Four particles each of mass ${}^{\prime }{m}^{\prime }$ are also attached at points $A,B,C$ and $D$. Total kinetic energy of system is:

• A. $5m{v}^2$
• B. $3m{v}^2$
• C. $2m{v}^2$
• D. $8m{v}^2$

Answer 1

The correct option is A $5m{v}^2$

For the combination of translation and rotation of system (ring $\&4$ masses),
$K{E}_{\text{Total}}=K{E}_{\text{Trans}}+K{E}_{\text{Rot}}...\left(i\right)$
Since $v_{\text{CM}}=v,m_{\text{system}}=(4m+m)=5m$
$K{E}_{\text{Trans}}=\frac{1}{2}\times \left(5m\right)\times v_{\text{CM}}^2=\frac{5}{2}m{v}^2...\left(ii\right)$

Similarly,
$K{E}_{\text{Rot}}=\frac{1}{2}{I}_{\text{CM}}{\omega }^2$
$\text{M.O.I}$ of whole system about it's $\text{COM}$ is,
$I_{\text{CM}}=m{r}^2+4\times \left(m{r}^2\right)=5m{r}^2$
$\&$ for pure rolling,
$v=r\omega \text{or},\omega =\frac{v}{r}$
$\Rightarrow K{E}_{\text{Rot}}=\frac{1}{2}\times 5m{r}^2\times {\left(\frac{v}{r}\right)}^2$
$\therefore K{E}_{\text{Rot}}=\frac{5}{2}m{v}^2...\left(iii\right)$

From Eq $\left(i\right)\left(ii\right)\left(iii\right)$:
$K{E}_{\text{Total}}=\frac{5}{2}m{v}^2+\frac{5}{2}m{v}^2$
$\therefore K{E}_{\text{Total}}=5m{v}^2$