A uniform, solid disk with mass ?m? and radius ?R? is pivoted about a horizontal axis through its center. A small object of the same mass ?m? is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

Solution 78P This question can be solved by using the principle of conservation of energy. Total initial mechanical energy of the system is equal to the total final mechanical energy. The disk is released from rest, hence initial kinetic energy is zero. Initial potential energy = mgR, where R is given to be the radius of the disk. Initial total mechanical energy = mgR…..(1) Let the angular speed be . 2 So, kinetic energy = I2 , I is the moment of inertia of the whole system. Now, I = moment of inertia of the disk + moment of inertia of the small object 1 2 2 I = 2R + mR I = mR 2 2 1 3 2 2 3 2 2 So, kinetic energy = 2 × (2mR ) = mR 4 When the small object is directly below the axis, its potential energy will be zero. 3 2 2 Therefore, total final mechanical energy = mR4 …..(2) Equating equations (1) and (2), we get 3 2 2 mgR = mR4 2 4g = 3R 4g = 3R This is the required angular speed.