The radius of the smaller disk is r, the mass of the smaller disk is m, the radius of larger disk is R, the mass of the larger disk is M and the angle at which assembly is rotated is θ.

Consider the figure for the given situation.

Figure (1)

The loss in the potential energy at S is converted into kinetic energy at Q.

Write the expression for the height of the smaller disk from the centre point O.

    h=OB−OA

Here, h is the height of the smaller disk from the centre point O.

Substitute R for OB and Rcosθ for OA in above equation to find h.

    h=R−Rcosθ=R(1−cosθ)

Here, R is the radius of larger disk and θ is the angle at which assembly is rotated.

Write the expression for the loss in potential energy.

    Epotential=mgh

Here, Epotential is the loss in potential energy, m is mass of the smaller disk and g is the acceleration due to gravity.

Substitute R(1−cosθ) for h in above equation to find Epotential.

    Epotential=mgR(1−cosθ)

Write the expression for the moment of inertia of the larger disk about the cylinder axis.

    Ilarger disk=MR22

Here, Ilarger disk is the moment of inertia of the larger disk about the cylinder axis, M is mass of the larger disk and R is the radius of larger disk.

Write the expression for the moment of inertia of the smaller disk about the cylinder axis.

    Ismall disk=mr22

Here, Ismall disk is the moment of inertia of the smaller disk about the cylinder axis and r is the radius of smaller disk.

Write the expression for the moment of inertia of the smaller disk about the diameter.

    I′=mR2

Here, I′ is the moment of inertia of the smaller disk about the diameter.

Write the expression for the net moment of inertia of the two disk system.

    I=Ilarger disk+Ismall disk+I′

Here, I is the net moment of inertia of the two disk system.

Substitute MR22 for Ilarger disk, mr22 for Ismall disk and mR2 for I′ in the above equation to find I.

    I=MR22+mr22+mR2

Write the expression for the angular velocity of the disk.

    ω=vR

Here, ω is the angular velocity of the disk and v is the linear velocity of the disk.

The gain in kinetic energy of the system is equal to the sum of the center of mass of the small disk, the rotational energy of the larger disk and the rotational energy of the smaller disk about O.

 Write the expression for the gain in kinetic energy of the system.

    Ekinetic=12Iω2

Here, Ekinetic is the gain in kinetic energy of the system.

Substitute (MR22+mr22+mR2) for I and vR for ω in the above equation to find Ekinetic.

    Ekinetic=12(MR22+mr22+mR2)(vR)2

Apply conservation law of energy.

    Epotential=Ekinetic

Substitute 12(MR22+mr22+mR2)(vR)2 for Ekinetic and mgR(1−cosθ) for Epotential in the above equation.

    mgR(1−cosθ)=12(MR22+mr22+mR2)(vR)2(vR)2=2mgR(1−cosθ)(MR22+mr22+mR2)v2=2mgR(1−cosθ)(MR22+mr22+mR2)R2v=[2mgR(1−cosθ)(MR22+mr22+mR2)]1/2R

Further solve the above equation.

    v=[2gR(1−cosθ)R22(Mm+mr2mR2+mm)]1/2R=[4gR(1−cosθ)(M/m)+(r/R)2+2]1/2=2[Rg(1−cosθ)(M/m)+(r/R)2+2]1/2

Conclusion:

Therefore, the speed of the center of the small disk as it passes through the equilibrium position is 2[Rg(1−cosθ)(M/m)+(r/R)2+2]1/2.

As the value of angle at which assembly is rotated is very small.

    sinθ≈θ

From the figure, write the expression for the equation of motion.

    Id2θdt2=−mgRsinθ

Substitute θ for sinθ in above equation.

    Id2θdt2=−mgRθd2θdt2+mgRIθ=0        (1)

Write the expression for the equation of motion.

    d2θdt2+ω2θ=0        (2)

Compare equations (1) and (2).

    ω2=mgRIω=mgRI

Formula to calculate the period of the motion is,

    ω=2πTT=2πω

Here, T is the period of the motion.

Substitute mgRI for ω in the above equation.

    T=2πmgRI

Substitute (MR22+mr22+mR2) for I in the above equation.

    T=2πmgR(MR22+mr22+mR2)=2π[MR2+mr2+2mR22mgR]1/2=2π[(M+2m)R2+mr22mgR]1/2

Conclusion:

Therefore, the period of the motion is 2π[(M+2m)R2+mr22mgR]1/2.