Moment of inertia of a body about a given axis is 2.4 kg-m^2 . To produce a rotational kinetic

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6.System of Particles and Rotational Motion

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The moment of inertia of a body about a given axis is $2.4\ kg-m^2$. To produce a rotational kinetic energy of $750\ J$, an angular acceleration of $5\ rad/s^2$ must be applied about that axis for.......... $\sec$

A

$6$

B

$5$

C

$4$

D

$3$

Solution

$\frac{1}{2}I{\omega ^2} = 750\,J$

==>${\omega ^2} = \frac{{750 \times 2}}{{2.4}} = 625 $

$\Rightarrow \omega = 25\,rad/s$

$\alpha = \frac{{{\omega _2} – {\omega _1}}}{t} $

$\Rightarrow 5 = \frac{{25 – 0}}{t} $

$\Rightarrow t = 5\,s$

Standard 11

Physics

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