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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 $1.2 \;kg m^{2}$. 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
A
$4$
B
$2$
C
$8$
D
$10$
(AIPMT-1990)
Solution
Rotational kinetic energy $\, = \,\,\frac{1}{2}\,I{\omega ^2}\,\, = \,\,1500\,\,\, \Rightarrow \,\,\frac{1}{2}\, \times \,1.2\, \times \,{\omega ^2}\,\,$
$ = \,\,1500\,\,\, \Rightarrow \,\,{\omega ^2}\,\, = \,\,\frac{{3000}}{{1.2}}\,\, \Rightarrow \,\,\omega \,\, = \,\,50\,rad/s$
$\omega_{0}=0$
$\omega = \omega_0 + \alpha t ⇒ 50 = 0 + 25 \times t ⇒ t = 2\;s$
Standard 11
Physics
