A soil cylinder of mass `M` and radius `R` is connected to a spring as shown in fig. The cylinder is placed on a rough horizontal surface. All the parts except the cylinder shown in the figure are light. If the cylinder is displaced slightly from its mean position and released, so that it performs pure rolling back and forth about its equilibrium position, determine the time period of oscillation?

Options

A. `2pisqrt((M)/(k))`
B. `2pisqrt((3M)/(2k))`
C. `2pisqrt((3M)/(k))`
D. none of these

Answer

B

Explanation

Let us say in displaced position, the axis of cylinder is at a distance x from its mean position and its velocity of center of mass if x and angular velocity is `omega`. Then, as cylinder is not slipping `v=Romega`. In this position the spring elongates by x. Using energy method we can find frequency of oscillation very easily.
Total energy of oscillation is
`E=(Iomega^2)/(2)+(Mv^2)/(2)+(kx^2)/(2)`
We have `I=(MR^2)/(2)`, so `E=(3)/(4)Mv^2+(kx^2)/(2)`
`E=v^2+((2)/(3)(k)/(M))x^2=`constant
Comparing with `v^2+omega^2x^2=`constant
So, `omega=sqrt((2k)/(3M))impliesT=2pisqrt((3M)/(2k))`