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The moment of inertia of a body about a certain axis is $1.2 \mathrm{~kg} \mathrm{~m}^2$. Initially, the body is at rest. In order to produce rotational kinetic energy of $1800 \mathrm{~J}$, an angular acceleration of $30 \mathrm{rads}^{-2}$ must be applied about the given axis for a duration of.........s.
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The correct answer is:
$\frac{\sqrt{30}}{3}$
Given that, moment of inertia, $I=1.2 \mathrm{~kg}-\mathrm{m}^2$
Initial angular velocity, $\omega_0=0$
Angular acceleration, $\alpha=30 \mathrm{rad} \mathrm{s}^{-2}$
Final rotational kinetic energy, $K_f=1800 \mathrm{~J}$
Let final angular velocity $=\omega$
Using, $\quad K_f=\frac{1}{2} I \omega^2$
By substituting the values, we get
$1800=\frac{1}{2} \times 1.2 \times \omega^2$
$\Rightarrow \quad \omega=10 \sqrt{30} \mathrm{rad} / \mathrm{s}$
We know that, $\omega=\omega_0+\alpha t$
$\Rightarrow \quad 10 \sqrt{30}=0+30 t$
$t=\frac{\sqrt{30}}{3} \mathrm{~s}$
Initial angular velocity, $\omega_0=0$
Angular acceleration, $\alpha=30 \mathrm{rad} \mathrm{s}^{-2}$
Final rotational kinetic energy, $K_f=1800 \mathrm{~J}$
Let final angular velocity $=\omega$
Using, $\quad K_f=\frac{1}{2} I \omega^2$
By substituting the values, we get
$1800=\frac{1}{2} \times 1.2 \times \omega^2$
$\Rightarrow \quad \omega=10 \sqrt{30} \mathrm{rad} / \mathrm{s}$
We know that, $\omega=\omega_0+\alpha t$
$\Rightarrow \quad 10 \sqrt{30}=0+30 t$
$t=\frac{\sqrt{30}}{3} \mathrm{~s}$
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