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 -
89.​

Answer 1

Answer:

23886 correct answer uijo*--5**5/5*5*56665

Answer 2

Given that,

Moment of inertia = I

frequency = n

The angular frequency is

$\omega=2\pi n$

We need to calculate the initial kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}I\omega^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}I\times(2\pi n)^2$

$K,E_{i}=2I\pi^2n^2$

If the frequency is doubled then,

$f=2n$

So, The angular frequency is

$\omega=2\pi n$

We need to calculate the final kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}I\omega^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}I\times(4\pi n)^2$

$K,E_{i}=8I\pi^2n^2$

We need to calculate the work done

Using formula of work done

$W=\Delta K.E$

$W=K.E_{f}-K.E_{i}$

Put the value into the formula

$W=8I\pi^2n^2-2I\pi^2n^2$

$W=6I\pi^2n^2$

Hence, The work done  is $6I\pi^2n^2$