Question 609: Centre of Mass and Rotational Motion

Question

When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from 10 Hz to 20 Hz . The moment of inertia of flywheel about its axis of rotation is $\left(\pi^2=10\right)$

Accepted Answer

Given, work done, $W=12000 \mathrm{~J}$,
Initial frequency, $f_1=10 \mathrm{~Hz}$
and final frequency, $f_2=20 \mathrm{~Hz}$
Angular velocity for rotational motion is given by
$$\begin{aligned}
& \omega=2 \pi f \
& 	herefore \quad \omega_1=2 \pi t_1=2 \pi 	imes 10=20 \pi \mathrm{rad} / \mathrm{s} \
& 	ext { and } \omega_2=2 \pi f_2=2 \pi 	imes 20=40 \pi \mathrm{rad} / \mathrm{s}
\end{aligned}$$
According to work-energy theorem,
work done in rotation = change in rotational kinetic energy
$$\begin{aligned}
\Rightarrow \quad W & =\frac{1}{2} l \omega_2^2-\frac{1}{2} l \omega_1^2 \quad\left[\because K E_{	ext {rotational }}=\frac{1}{2} / \omega^2ight] \
& =\frac{1}{2} l\left(\omega_2^2-\omega_1^2ight)\quad	ext{.... (i)}
\end{aligned}$$
where, $I=$ moment of inertia of the flywheel.
Substituting given values in Eq. (i), we get
$$\begin{aligned}
12000 & =\frac{1}{2} l\left(1600 \pi^2-400 \pi^2ight) \
\Rightarrow \quad & =\frac{1}{2} l(1200 	imes 10) \quad\left[\because \pi^2=10ight] \
\Rightarrow \quad I & =\frac{12000 	imes 2}{12000}=2 \mathrm{~kg} \mathrm{~m}^2
\end{aligned}$$

Answer

$1 \mathrm{~kgm}^2$

Answer

$2 \mathrm{~kgm}^2$

Answer

$1.688 \mathrm{~kgm}^2$

Answer

$1.5 \mathrm{~kgm}^2$

Centre of Mass and Rotational Motion

When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from 10 Hz to 20 Hz . The moment of inertia of flywheel about its axis of rotation is \left(\pi^2=10\right)

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