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
B
mr2ω2
C
12mr2ω2
D
12mrω2

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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