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Question

The kinetic energy of a ring of mass $$m$$ and radius $$r$$ which rotates about an axis passing through its centre and perpendicular to the plane with angular velocity $$\omega$$, is :


A
mrω2
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B
mr2ω2
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C
12mr2ω2
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D
12mrω2
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Solution

The correct option is D $$\dfrac{1}{2}mr^2\omega^2$$
kinetic energy= $$\dfrac{1}{2} mv^2$$

Now for rotational motion we know, 
$$v=r\omega$$
Substituting the value we get 
kinetic energy= $$\dfrac{1}{2} m \omega^2 r^2$$

Physics

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