Question

A thin disk of mass 8.6 kg and radius 34.1 cm is suspended by a horizontal axis perpendicular to the disk through its rim. The disk is displaced slightly from equilibrium and released. The acceleration of gravity is 9.81 m/s 2 . Find the period of the subsequent simple harmonic motion. Answer in units of s.

Answer 1

m = mass of the disk = 8.6 kg

r = radius of the disk = 34.1 cm = 0.341 m

Moment of inertia of disk about the center of mass is given as

I_cm = (0.5) m r² = (0.5) (8.6) (0.341)² = 0.5 kgm²

d = distance between center of mass and axis of rotation = r = 0.341 m

I = moment of inertia about the axis of rotation

I = I_cm + m d²

I = 0.5 + (8.6) (0.341)²

I = 1.5 kgm²

Time period is given as

T = 2 sqrt(I/(mgd))

T = 2 (3.14) sqrt(1.5/((8.6) (9.8) (0.341)))

T = 1.435 sec