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6.System of Particles and Rotational Motion
hard
A cord is wound round the circumference of wheel of radius $r$. The axis of the wheel is horizontal and the moment of inertia about it is $I. \,A$ weight $mg$ is attached to the cord at the end. The weight falls from rest. After falling through a distance $ 'h '$, the square of angular velocity of wheel will be ..... .
A
$\frac{2 mgh }{ I +2 mr ^{2}}$
B
$\frac{2 mgh }{ I + mr ^{2}}$
C
$2 gh$
D
$\frac{2 gh }{ I + mr ^{2}}$
(JEE MAIN-2021)
Solution
$mgh =\frac{1}{2} I \omega^{2}+\frac{1}{2} mv ^{2}$
$v =\omega r$
$mgh =\frac{1}{2} I \omega^{2}+\frac{1}{2} m \omega^{2} r ^{2}$
$\frac{2 mgh }{\left( I + mr ^{2}\right)}=\omega^{2}$
Standard 11
Physics
