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

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)

the skeletons of most mammals reach a certain size and then stop growing. however, many animals, including some mammals, keep growing throughout their lives. kangaroos, for example, just keep growing and growing until they die. also, the longer an animal lives the greater the chance it might catch some deadly disease.