Question

A circular ring of wire of mass M and radius
R is making n revolutions/sec about an axis
passing through a point on its rim and
perpendicular to its plane. The kinetic energy
of rotation of the ring is given by-
(A) 472MR2n2
(B) 272MR2n2
(C) 1,2MR2n2
(D) 8r2MR2n2

Answer 1

kinetic energy of rotation = 8π²m²R²n²

rotational kinetic energy is given as K.E = 1/2 Iω²

where I is moment of inertia about axis of rotation and ω is angular frequency.

A circular ring rotating about an axis passing through its rim and perpendicular to plane.

moment of inertia, I = mR² + mR² = 2mR² [ from parallel axis theorem]

as number of revolution per second is n

so, angular frequency, ω = 2πn rad/s

now kinetic energy of rotation, K.E = 1/2 × (2mR)² × (2πn)²

= 1/2 × 4mR² × 4π²n²

= 8mR²π²n²

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Answer 2

$\huge\bold\purple{Answer:-}$

8π²m²R²n²