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```				   USING ENERGY CONSERVATION ON THE PARTICLE ONLY WE CAN THE SPEED WHEN IT IS AT THE BOTTOM POSITION.. BUT THE SOLUTION WHICH I SAW WAS A LENGTHY ONE  CHANGE IN PE=2*5/4R=CHANGE IN KINETIC ENERGY=1/2*MV^2 HENCE V=(5GR)^1/2 PLZ CONFIRM.. 4. A uniform circular disc has radius R and mass m. A particle, also of mass m, is fixed t a point A on the edge of the disc as shown in the figure. The disc can rotate freely about a fixed 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 the then allowed to fall so that it starts rotating about PQ. Find the linear speed of the particle as it reaches its lowest position.
```

5 years ago

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```										solution4=>
initially total energy (TE)  = PE + KE
PE = PE of disk + PE of particle
taking PQ line as x axis then
PE of disk initially = mgR/4
PE of particle = mg(R+R/4) = 5mgR/4
KE initial = 0
TE =  0 + mgR/4 = 5mgR/4 = 3mgR/2       .................1

finally system is rotating with some angular velocity (w)
PE of disk = -mgR/4         (position of center of disk is below the x axis)
PE of particle = -5mgR/4   (position of particle is below x axis)
KE = IW2/2                  (I is moment of inertia of system about PQ)
TE = -3mgR/2 + IW2/2               ....................2

since total energy is conserved so equatn 1 = 2
W = [ 3mgR/I ]1/2                        ......................3

now I = Id (disk) + Ip(particle)
=[mR2/4 + mR2 /16 ] (I of disk) + [ 25mR2/16 ] (I of particle)
= 30/16mR2
on putting this value in eq 3
W = [8g/5R]1/2
V = Wr                    (for particle , r = 5R/4)
V = [8g/5R]1/2 (5R/4) = [5gR/2]1/2
this is the required ans
approve if u like my ans
```
5 years ago

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