Question

If a solid sphere of mass $1 \,kg$ and radius $0.1\, m$ rolls without slipping at a uniform velocity of $1\, m/s$ along a straight line on a horizontal floor, the kinetic energy is

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

Answer 1

Answer: (C)

$\frac{7}{10} J$

Answer 2

K.E of rolling $=\left(\frac{1}{2}mv^{2}+\frac{1}{2}I. \frac{v^{2}}{R^{2}}\right)$ $K.E.=\frac{1}{2}mv^{2}+\frac{1}{2}. \frac{2}{5}. \frac{mR^{2}.v^{2}}{R^{2}}$ $K.E.=\frac{1}{2}mv^{2}\times\frac{7}{5}$ $m=1\,kg , v=1\,m/s$ $K.E.=\frac{1}{2}\times1\times 1\times \frac{7}{5}=\frac{7}{10}\,J$