The correct option is A 3π5 sec
Let us rotate solid cylinder from its stable equilibrium position by the small angle θ. Let vc be the instantaneous speed of the COM (C) of the solid cylinder.
Solid cylinder is in pure rolling, hence its angular velocity about its own centre C is
ω=vcr ......(1)
Since C moves along a circle of radius (R−r), the speed of C along the circular path can be written as
vc=ω′(R−r) ....(2)
where ω′ is the angular velocity of cylinder about the centre of trough.
Thus from eqns. (1) and (2), we get
ω=ω′(R−r)r ......(3)
Mechanical energy of oscillation of the solid cylinder is conserved, i.e. E=K+U= constant
So, 12mv2c+12Icω2+mg(R−r)(1−cosθ)=constant
(where m is the mass of solid cylinder and Ic is the moment of inertia of the solid cylinder about an axis passing through its COM and perpendicular to the plane of figure of the solid cylinder)
⇒12mω2r2+12mr22ω2+mg(R−r)(1−cosθ)=constant
Using (3) in the above equation, we get
34mω2(R−r)2+mg(R−r)(1−cosθ)=constant
Differentiating w.r.t. time on both sides,
32mω(R−r)2α+mgω(R−r)sinθ=0
Since R≠r and ω≠0, we get
α=−2g3(R−r)sinθ
For small values of θ, sinθ≈θ
∴α=−2g3(R−r)θ
Comparing this with α=−ω20θ, we get
ω0=√2g3(R−r) is the angular frequency of oscillations.
Hence, time period T=2πω0=2π√3(R−r)2g
From the data given in the question,
R=0.8 m and r=0.2 m
T=2π√3×(0.8−0.2)2×10=3π5 sec
Thus, option (a) is the correct answer.