A solid cylinder of mass `m` is attached to a horizontal spring with force constant `k`. The cylinder can roll without slipping along the horizontal plane. (See the accompanying figure.) Show that the center of mass of the cylinder executes simple harmonic motion with a period `T = 2pisqrt((3m)/(2k))`, if displaced from mean position.

Explanation

Let at any position, x be the extension in the spring and v be the velocity of centre of mass of the cylinder. Then,
P.E. of spring `U=(1)/(2)kx^(2)`
K.E. of translation, `K_(T)=(1)/(2)Mv^(2)`
K.E. of rotation, `K_(R)=(1)/(2)Iomega^(2)=(1)/(4)Mv^(2)`
[ `:'I=(1)/(2)Mr^(2)` and `v=omegar`]
Total mechanical energy of the sysem is
`E=U+K_(T)+K_(R)=(1)/(2)kx^(2)+(1)/(2)Mv^(2)+(1)/(4)Mv^(2)`
`=(1)/(2)kx^(2)+(3)/(4)Mv^(2)`...(i)
As per question , `v=0, ` if `x=0.25m` , then
`E+(1)/(2)xx3xx(0.25)+0=(3)/(32)J`
At equilibrium position, `U=0` [as `x=0`],
so from (i),
`(3)/(32)=0+(3)/(4)Mv^(2)` or `Mv^(2)=(1)/(8)J`
At equilibrium position
(a) Translational K.E.
`=(1)/(2)Mv^(2)=(1)/(2)((1)/(8))=(1)/(16)J`
(b) Rotational K.E.,
`=(1)/(4)Mv^(2)=(1)/(4)((1)/(8))=(1)/(32)J`
(c) In SHM, the total energy is conserved,
so `(dE)/(dt)=0`
From (i), we have
`0=(1)/(2)k.2x (dx)/(dt)+(3)/(4)M2v(dv)/(dt)`
or `(3)/(4)M(d^(2)x)/(dt^(2))=-(1)/(2)kx`
`[:' (dx)/(dt)=v` and `(dv)/(dt)=(d^(2)x)/(dt^(2))]`
or `(d^(2)x)/(dt^(2))=(-2k)/(3M)x` ...(ii)
From (ii), we note that acceleration `(d^(2)x//dt^(2)) prop x` and is directed towards equilibrium position. Hence, the cylinder will execute SHM. Comparing (ii) with the equation
`(d^(2)x)/(dt^(2))=-omega^(2)x,` we have
`omega^(2)=(2k)/(3M)`
or `omega=sqrt((2k)/(3M))`
Time period of SHM is,
`T=(2pi)/(omega)=2pisqrt((3M)/(2k))`