A uniform plank of mass m, free to move in the horizontal direction only , is placed at the top of a solid cylinder of mass `2 m` and radius R. The plank is attached to a fixed wall by mean of a light spring constant k. There is no slipping between the cylinder and the planck , and between the cylinder and the ground. Find the time period of small oscillation of the system.

Explanation

Method 1: When the plank is displaced slightly (say x) from equilibrium , as there is no sliding at any surface , the friction will be of static nature . As cylinder is rolling we can take the motion of the cylinder as pure rotation about point of contact.
Free body diagrams of plank and cylinder

Equation of motion for plank
`kx - f = m(a)`
Writing torque equation to cylinder about plank of contact
`f - (2 R) = i_(p). alpha [(1)/(2) (2m) R^(2) + 2m R^(2)] alpha`
`= 3 mR^(2) alpha`
or `f = (3)/(2) m Ra`
As cylinder is rolling we can write, the acceleration of topmost point of cylinder (or plank)
`a = alpha (2 R) implies alpha R = (a)/(2)`
Substituting the value of (iii) and (ii), we get `f = (3)/(4) ma`
Now, from Eqs (i) and (iv),
`kx - ((3)/(4) ma) = ma`
`implies a = (4 k)/(7 m) x`
Wtiting above equation with proper vector relation
`bar a = - ((4 k)/(7 m)) bar x`
Comparing with stander equation `bar a = - omega^(2) bar x`
which gives `omega^(2) = (4 k)/(7 m) or T = pi sqrt ((7m)/(k))`
Method 2: Suppose that the plank is displaced from its equilibrium position by x at time t , the center of the cylinder is therefore , displaced by `x//2`.
Therefore the mechanical energy of the system is given by
`E = KE ("plank") + PE ("spring") + KE ("cylinder")`
`= (1)/(2) mv^(2) + (1)/(2)kx^(2) + (1)/(2) I_(p) omega^(2)`
`I_(P) = I_(G) + (2 m) R^(2) = (1)/(2) (2 m) R^(2) + (2 m) R^(2)`
`= 3 mR^(2)`
As cylinder is rolling velocity of center of cylinder will be half of the velocity of the plank (top most point of the cylinder)
`nu_(cylinder) = (nu)/(2) = omega R implies omega = (nu)/(2 R)`
Substituting the value of `omega` in Eq. (i) we get
`E = (1)/(2) mv^(2) + (1)/(2)kx^(2) + (1)/(2) (3 m R^(2)) ((nu)/(2 R))`
`= (1)/(2) mv^(2) + (1)/(2)kx^(2) + (3)/(8)mv^(2)`
`= (7)/(8) mv^(2) + (1)/(2)kx^(2)`
Differentiating Eq. (ii) w.r.t. time
`(dE)/(dt) = 0 = (7)/(8) m.2v ((d nu)/(dt)) + k.2x, ((dx)/(dt)`
`= (7)/(8) mnu. a + kx. nu`
Here `(7)/(4) ma + kx = 0 or a = - (4 k)/(7 m) x`
Comparing with `a = - omega^(2) x implies T = 2 pi sqrt((7m)/(4k)) = pi sqrt((7m)/(k))`
Method 3: we can write Eq (ii)as `E = nu^(2) + (4k)/(7m) s^(2) = `constant
as total energy of sysytem performing SHM is constant. Comparing with standard equation `nu^(2) + omega^(2) x^(2) = constant`
which gives `omega^(2) = (4k)/(7m) or T = pi sqrt ((7m)/(k))`