Question

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. $8{\pi }^{2}I{n}^2$

Answer 1

The correct option is C $6{\pi }^{2}I{N}^2$

Lets consider,

$n=$ initial frequency

$\omega =2\pi n$

Initial kinetic energy is

$K_i=\frac{1}{2}I{\omega }^2=\frac{1}{2}I\times 2\pi n\times 2\pi n$

$K_i=2I{\pi }^2{n}^2$

When frequency is double to the initial frequency , the the kinetic energy will be

$K_f=\frac{1}{2}I{\omega }^{\prime 2}=\frac{1}{2}I\times 4\pi n\times 4\pi n$

$K_f=8I{\pi }^2{n}^2$

By work-energy theorem,

$W=K_f-K_i$

$W=8I{\pi }^2{n}^2-2I{\pi }^2{n}^2=6I{\pi }^2{n}^2$

The correct option is C.