Gujarati
Hindi
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

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