Question

A cord is wound round the circumference of wheel of radius r. the axis of the wheel is horizontal and moment of inertia about it is I A weight mg is attached to the end of the cord and falls from the rest. After falling through the distance h. the angular velocity of the wheel will be….

(A) √[2gh / (I + mr)]
(B) [2mgh / (I + mr^2)]^1/2
(C) [2mgh / (I + mr^2)]^1/2
(D) √(2gh)

Answer 1

The angular velocity of the wheel after a weight mg falls through a distance h is given by ω = √(2mgh / (I + mr^2)). This follows from the law of conservation of energy, relating gravitational potential energy to kinetic energy of rotation and translation.

Step-by-step explanation:

The question involves the concept of energy conservation and the relationship between the linear motion of a falling weight and the rotational motion of a wheel. When a weight mg falls through a distance h, it loses potential energy (mgh) which is converted into rotational kinetic energy (1/2 I ω^2) of the wheel and the linear kinetic energy (1/2 mv^2) of the weight itself. The angular velocity (ω) of the wheel, considering the wheel's moment of inertia (I) and the radius (r), can be determined by equating the lost potential energy to the gained kinetic energy.

Using the work-energy theorem, the energy conservation equation is written as: mgh = 1/2 I ω^2 + 1/2 mv^2. Since the cord is wrapped around the wheel, the linear velocity of the weight (v) is related to the angular velocity of the wheel (ω) through the relationship v = rω. Substituting this into the equation and solving for ω gives us ω = √(2mgh / (I + mr^2)). Among the answer options provided, the correct answer is (C) [2mgh / (I + mr^2)]^1/2.