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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. 10 years ago
solution 4=> 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