AP EAPCET 25-Aug-2021 Shift 1 Paper

Given, mass of disc =m
Radius of disc =a
Angular speed =ω
Moment of inertia of the disc, I=

1

2

ma2
Initial angular momentum, Li=I×ω=

1

2

ma2ω
Let ω′ be the angular momentum of the disc when the particle of mass m is attached at the rim.
Final angular momentum, Lf=Iω′+mav′
=

1

2

ma2ω′+ma2ω′ [∴v′=ω′a]
Applying conservation of angular momentum,
Li=Lf

1

2

ma2ω=

1

2

ma2ω′+ma2ω′
⇒ω′=

ω

3