Question

A ring and a disc roll on the horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is $4J$ then total kinetic energy of the disc is

• A. $3J$
• B. $4J$
• C. $5J$
• D. $6J$

Answer 1

The correct option is A $3J$

Total kinetic energy of the body $K.E_{total}=(\frac{1}{2}m{v}^2{)}_{translational}+(\frac{1}{2}I{w}^2{)}_{rotational}$

Total kinetic energy of the ring $K.E_{ring}=\frac{1}{2}m{v}^2+\frac{1}{2}\left(M{r}^2\right)w^2$

$K.E_{ring}=\frac{1}{2}m{v}^2+\frac{1}{2}M{v}^2=M{v}^2$ (for pure rolling $v=rw$)

$⟹$ $m{v}^2=4$

Total kinetic energy of the ring $K.E_{ring}=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{1}{2}M{r}^2\right)w^2$

$K.E_{disc}=\frac{1}{2}m{v}^2+\frac{1}{4}M{v}^2=\frac{3}{4}M{v}^2$ (for pure rolling $v=rw$)

$⟹$ $K.E_{disc}=\frac{3}{4}\times 4=3J$