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(B) According to the law of conservation of energy,the potential energy lost by the falling weight is converted into the kinetic energy of the system

Vedclass · Accepted Answer

$\sqrt{\frac{2mgh}{I + mr^2}}$

Vedclass · Answer

(B) According to the law of conservation of energy,the potential energy lost by the falling weight is converted into the kinetic energy of the system (rotational kinetic energy of the wheel and translational kinetic energy of the weight).$mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2$Since the cord is wound around the wheel,the linear velocity $v$ of the weight is related to the angular velocity $\omega$ of the wheel by $v = r\omega$.Substituting $v = r\omega$ into the energy equation:$mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}m(r\omega)^2$$mgh = \frac{1}{2}(I + mr^2)\omega^2$Solving for $\omega$:$\omega^2 = \frac{2mgh}{I + mr^2}$$\omega = \sqrt{\frac{2mgh}{I + mr^2}}$

$A$ cord is wound round the circumference of a 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 end of the cord and falls from rest. After falling through a distance $h$,the angular velocity of the wheel will be

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