Question

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is -

• A. $\frac{2}{5}$
• B. $\frac{2}{7}$
• C. $\frac{1}{5}$
• D. $\frac{7}{10}$

Answer 1

Answer: (B)

$\frac{2}{7}$

Answer 2

The correct answer is (B): $\frac{2}{7}$

KER = $\frac{1}{2}$ lω²

= $\frac{1}{2}$ × $\frac{2}{5}$ x ω² × (mR²)

KEtotal = $\frac{1}{2}$ x $\frac{7}{5}$ x mR² x ω²

∴$\frac{KE_R}{KE_{total}}$ = $\frac{2}{7}$