A solid sphere of mass M and Radius R is attached to a massless spring of force constant k as shown in the figure. The cylinder is pulled towards right slightly and released so that it executes simple harmonic motion of time period. If the cylinder rolls on the horizontal surface without slipping, calculate its time period of oscillation.

Explanation

Total energy = translation K.E. + rotational k.E. + P.E. stored in spring
`E = (1)/(2)mv^(2) + (1)/(2)Iomega^(2) + (1)/(2)kx^(2)`
`E = (1)/(2)mv^(2) + (1)/(2)((2)/(5)mR^(2))((v)/(R ))^(2) + (1)/(2)kx^(2)`
`rArr E = (7)/(10) mv^(2) + (1)/(2)kx^(2)`
Now, the total energy of the sytem must remain constant, i.e., `(dE)/(dt) = 0`
We have `(dE)/(dt) = (7)/(10)m (2v (dv)/(dt)) + (1)/(2) k(2x (dx)/(dt)) = 0`
Now acceleration `a = (dv)/(dt)` and velocity `v = (dx)/(dt)`
Therefore, `(7)/(5) mva + kvx = 0`
or ` v((7)/(5)ma + kx) = 0`
Since `v ne 0` we have
`(7)/(5) ma + kx = 0`
or ` a = -((5)/(7) (k)/(m))x = -omega^(2)x`
where `omega = sqrt((5k)/(7m))`
Hence, time period `T = 2pi sqrt((7m))/(5k)`