A uniform disk with a 25 cm radius swings without friction about a nail through the rim. If it is released from rest from a position with the center level with the nail, then what is its angular velocity as it swings through the point where the center is below the na

Explanation:

During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy

mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .

mgr  = 1/2 x ( 1/2 m r²+ mr²) x ω²

gr  = 1/2 x 1/2  r² x ω² + 1/2r² x ω²

g = 1 / 4 x ω² r + 1 / 2 x ω² r

g = 3  x ω² r/ 4

ω² = 4g /3 r

= 4 x 9.8 /  3 x  .25

= 52.26

ω = 7.23  rad / s .

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