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

Answer: (A)

Kinetic energy of ring,
$K=\frac{1}{2} m v^{2}\left(1+\frac{K^{2}}{R^{2}}\right)$
$=\frac{1}{2} 0.4 \times\left(\frac{10}{100}\right)^{2} \times(1+1)$
$\left[\therefore \frac{K^{2}}{R^{2}}=1\right]$
$\Rightarrow K =4 \times 10^{-3}$ Joule