The moment of inertia of a body about a given | vedclass

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Question

Rotational kinetic energy $\, = \,\,\frac{1}{2}\,I{\omega ^2}\,\, = \,\,1500\,\,\, \Rightarrow \,\,\frac{1}{2}\, 	imes \,1.2\, 	imes \,{\omega ^2}\,

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Rotational kinetic energy $\, = \,\,\frac{1}{2}\,I{\omega ^2}\,\, = \,\,1500\,\,\, \Rightarrow \,\,\frac{1}{2}\, 	imes \,1.2\, 	imes \,{\omega ^2}\,\,$

$ = \,\,1500\,\,\, \Rightarrow \,\,{\omega ^2}\,\, = \,\,\frac{{3000}}{{1.2}}\,\, \Rightarrow \,\,\omega \,\, = \,\,50\,rad/s$

$\omega_{0}=0$

$\omega = \omega_0 + \alpha t &rArr; 50 = 0 + 25 	imes t &rArr; t = 2\;s$

The moment of inertia of a body about a given axis is $1.2 \;kg m^{2}$. Initially, the body is at rest. In order to produce a rotational kinetic energy of $1500\; joule$, an angular acceleration of $25 \;rad s^{-2}$ must be applied about that axis for a duration of

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