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6.System of Particles and Rotational Motion
medium
A flywheel is in the form of solid circular disc of mass $72\,\, kg$ and radius of $0.5\,m$ and it takes $70\, r.p.m.$ , then the energy of revolution approximately is ....... $J.$
A
$24$
B
$240$
C
$2.4$
D
$2400$
Solution
Rotational $K.E. = \frac {1}{2} I\omega ^2$
$ = \frac{1}{2}\left[ {\frac{1}{2} \times 72 \times {{(0.5)}^2}} \right] \times {\left( {\frac{{70 \times 2\pi }}{{60}}} \right)^2} = \,240\,J$
Standard 11
Physics