Question

A ring and a disc of same mass and radius roll without slipping along a horizontal surface with the same velocity. If the $K.E.$ of the ring is $8\text{J}$, then $K.E.$ of the disc is

• A. $2\text{J}$
• B. $4\text{J}$
• C. $6\text{J}$
• D. $16\text{J}$

Answer 1

The correct option is C $6\text{J}$

Let $m$ be the mass and $r$ be the radius for both objects
$\Rightarrow$Condition for rolling without slipping,
$v=r\omega$
$K{E}_{Total}=K{E}_{Rot}+K{E}_{Trans}$
$\Rightarrow K{E}_{Total}=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2...\left(i\right)$
Where $v=$ translational velocity of COM of body
$I=$ Moment of inertia about axis passing through COM
$I=m{r}^2$ For ring
$I=\frac{m{r}^2}{2}$ For disc
Total kinetic energy for disc:
$K{E}_{Disc}=\frac{1}{2}\times \left(\frac{m{r}^2}{2}\right)\times \frac{v^2}{r^2}+\frac{1}{2}m{v}^2$
$\Rightarrow K{E}_{Disc}=\frac{3}{4}m{v}^2...\left(ii\right)$
Total kinetic energy for Ring:
$K{E}_{Ring}=\frac{1}{2}\times \left(m{r}^2\right)\times \frac{v^2}{r^2}+\frac{1}{2}m{v}^2$
$\therefore K{E}_{Ring}=m{v}^2$
$\because K{E}_{Ring}=8\text{J}$
Hence from Eq $\left(ii\right)$:
$K{E}_{Disc}=\frac{3}{4}m{v}^2=\frac{3}{4}\times 8=6\text{J}$