Question

A uniform disk of radius R and mass M is pivoted about a horizontal axis parallel to its symmetry axis and passing through its edge so that it can swing freely in a vertical plane. It is released from rest with its center of mass at the same height as the pivot. R MPivot g What is the angular velocity of the disk when its center of mass is directly below the pivot

Answer 1

Step-by-step explanation:

Moment of inertia of the disc = 1/2 m r² .

moment of inertia of the disc about an axis parallel to symmetrical axis through C.M. and passing through a point on its circumference

= Ig + mr² ( theorem of parallel axis )

= 1/2 mr² + mr²

I = 3/2 mr²

During motion given , center of mass is lowered by distance equal to radius

loss of potential energy

= mgr

gain of rotation kinetic energy = 1/2 I ω² , ω is angular velocity , I is moment of inertia and ω is angular velocity attained at lowest point

1/2 I ω² = mgr

(1/2 x 3/2) mr² ω² = mgr

(3 / 4 ) r ω² = g

ω² = 4g / 3 r

ω = √(4g / 3 r)