Question

A uniform thin ring of mass $0.4$ kg rolls without slipping on a horizontal surface with a linear velocity of $10$ cm/s. The kinetic energy of the ring is?

• A. $4\times {10}^{-3}$ joules
• B. $4\times {10}^{-2}$ joules
• C. $2\times {10}^{-3}$ joules
• D. $2\times {10}^{-2}$ joules

Answer 1

The correct option is A $4\times {10}^{-3}$ joules

${\text{K.E}}_{Total}={\text{K.E}}_{translatory}+{\text{K.E}}_{rotation}{\text{K.E}}_{translatory}=\frac{1}{2}m{v}^2{\text{K.E}}_{rotation}=\frac{1}{2}I{\omega }^2$

Where moment of inertia of ring$=m{r}^2$

Reducing the above values we get

${K.E}_{Total}=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2=m{v}^2$

${K.E}_{Total}=0.4\times (0.1{)}^2=4\times {10}^{-3}joules.$