Question

A uniform disk of radius 0.529 m and unknown mass is constrained to rotate about a perpendicular axis through its center. A ring with same mass as the disk\'s is attached around the disk\'s rim. A tangential force of 0.223 N applied at the rim causes an angular acceleration of 0.103 rad/s2. Find the mass of the disk.

Answer 1

Here, a ring has been attached around the rim of the disk, and has a mass same as that of the disk. Now, we know that a force of 0.223 N produces and acceleration of 0.103 rad/s^2

That is a total torque of FxR = 0.223 x 0.529 = 0.117967 N-m produces the given acceleration

We know that, angular acceleration = Torque / Moment of inertia

That is, moment of inertia of the system = Torque / acc = 0.117967 / 0.103 = 1.1453 Kg-m²

Also, the inertia for a disk is given as MR²/2 where as that for a ring is MR². For the given two masses, mass and radii are same.

hence, MR² + MR²/2 = 3 x M x R² / 2 = 1.1453

or, M = 1.145 x 2 / 3 x 0.529 x 0.529 = 2.728 Kg

Therefore the required mass is 2.728 Kgs