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6.System of Particles and Rotational Motion
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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
$1\, J$
(AIIMS-2007)
Solution
When a body rolls over a smoooth surface, it has linear$K.E$. and rotational$K.E$
$\therefore E = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}$
$where\,\omega = \frac{V}{r}\,and\,I = \frac{2}{5}m{r^2}\,for\,solid\,sphere.$
$\therefore K.E. = \frac{1}{2}m{v^2} + \frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right).\frac{{{V^2}}}{{{r^2}}}$
$ = \frac{1}{2}m{v^2} + \frac{1}{5}m{v^2} = \frac{7}{{10}}m{v^2} = \frac{7}{{10}} \times 1 \times {1^2}$
$ = \frac{7}{{10}}J$
Standard 11
Physics
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