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6.System of Particles and Rotational Motion
hard
A fly wheel of moment of inertia $I$ is rotating at $n$ revolutions per $sec$. The work needed to double the frequency would be
A
$2{\pi ^2}I{n^2}$
B
$4{\pi ^2}I{n^2}$
C
$6{\pi ^2}I{n^2}$
D
$8{\pi ^2}I{n^2}$
Solution
Lets consider,
$n =$ initial frequency
$\omega=2 \pi n$
Initial kinetic energy is
$K _{ i }=\frac{1}{2} I \omega^2=\frac{1}{2} I \times 2 \pi n \times 2 \pi n$
$K _{ i }=2 I \pi^2 n ^2$
When frequency is double to the initial frequency, the the kinetic energy will be
$K _{ f }=\frac{1}{2} I \omega^2=\frac{1}{2} I \times 4 \pi n \times 4 \pi n$
$K _{ f }=8 I \pi^2 n ^2$
By work-energy theorem,
$W = K _{ f }- K _{ i }$
$W =8 I \pi^2 n ^2-2 I \pi^2 n ^2=6 I \pi^2 n ^2$
Standard 11
Physics
