A uniform circular disc has radius `R` and mass `m`. A particle, also of mass `m`, if fixed at a point A on the edge of the disc as shown in the figure. The disc can rotate freely about a horizontal chord PQ that is at a distance `R//4` from the centre `C` of the disc. The line AC is perpendicular to `PQ`. Initially the disc is held vertical with the point A at its highest position. it is then allowed to fall, so that it starts rotation about PQ. Find the linear speed of the particle as it reaches its lowest position.

Explanation

initial and final position are shown below.
Decrease in potential energy of mass
`=mg{2xx(5R)/(4)}=(5mgR)/(2)`
Decrease in potential energy of disc
`=mg{2xx(R)/(4)}=(mgR)/(2)`
Therefore, total decrease in potential energy of system
`=(5mgR)/(2)+(mrR)/(2)=3mgR`
Therefore, total decrease in potential energy of system
`=(5mgR)/(2)+(mgR)/(2)=3mgR`
Gain in kinetic eergy of system `=(1)/(2)Iomega^(2)`
where `I=` moment of inertial of system (disc `+` mass) about axis `PQ`
`=` moment of inertia of disc `+` moment of inertia of mass
`={(mR^(2))/(4)+m((R)/(4))^(2)}+m((5R)/(4))^(2)`
`I=(15mR^(2))/(8)`
from conservation of mechanical energy,
Decrease in potential energy `=` gain in kinetic energy
`3mgR=(1)/(2)((15mR^(2))/(8))omega^(2)`
`impliesomega=sqrt((16g)/(5R))`
Therefore, linear speed of particle at its lowest point
`v=romega=((5R)/(4))`
`omega=(5R)/(4)sqrt((16g))/(5R)`
`impliesv=sqrt(5gR)`