Question
State work energy theorem for rotational motion.
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Answer
According to work energy theorem, work done will be reflected as change in rotational kinetic energy. Work done for a small angular displacement $\text{d}\theta\text{ is },\text{dW}=\tau\text{d}\theta$
$\text{W}=\int\tau\text{ d }\theta=\int\text{I }\alpha\text{ d }\theta$
$=\int\text{I }\text{d }\omega\frac{\text{d }\theta}{\text{dt}}=\int\text{I }\omega\text{ d }\omega$
$\text{W}=\int^\limits{\omega\text{f}}_\limits{\omega\text{f}}\text{I }\omega\text{ d }\omega=\text{I}\Bigg|\frac{\omega^2}{2}\Bigg|^{\omega\text{f}}_{\omega\text{i}}$
$\text{W}=\frac{1}{2}\text{I}(\omega^2_{\text{f}}-\omega^2_{\text{i}})$
It is the change in rotational kinetic energy.
$\text{W}=\int\tau\text{ d }\theta=\int\text{I }\alpha\text{ d }\theta$
$=\int\text{I }\text{d }\omega\frac{\text{d }\theta}{\text{dt}}=\int\text{I }\omega\text{ d }\omega$
$\text{W}=\int^\limits{\omega\text{f}}_\limits{\omega\text{f}}\text{I }\omega\text{ d }\omega=\text{I}\Bigg|\frac{\omega^2}{2}\Bigg|^{\omega\text{f}}_{\omega\text{i}}$
$\text{W}=\frac{1}{2}\text{I}(\omega^2_{\text{f}}-\omega^2_{\text{i}})$
It is the change in rotational kinetic energy.
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