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Q. A wheel of mass $ 10\,kg $ has a moment of inertia of $ 160\,kg-m^2 $ about its own axis, the radius of gyration will be

Punjab PMETPunjab PMET 2001System of Particles and Rotational Motion

Solution:

Here : Mass of wheel $M=10\, kg$
Moment of inertia $=160\, kg\, m ^{2}$
(where $K$ is radius of gyration)
Using relation,
$I=M K^{2}$
$160=10 \times K^{2}$ or $K^{2}=16$
or $K=4\, m$