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Question

A solid uniform cylinder of mass M performs small oscillations in the horizontal plane if slightly displaced from its mean position as shown in the figure. Initially, the springs are at their natural lengths and the cylinder does not slip on the ground during oscillations due to friction between the ground and cylinder. The force constant of each spring is k. What is the time period of this oscillation?


A
π3Mk
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B
π3M4k
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C
π23Mk
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D
3π2Mk
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Solution

The correct option is C π23Mk
In the situation given in the problem, the cylinder is in its equilibrium position when springs are unstrained. When it is slightly rolled and released, it starts executing SHM and due to friction, the cylinder is in pure rolling motion. Now during oscillation, let us consider the centre of the cylinder is at a distance x from the mean position and moving with a speed v as shown in figure.
As the cylinder is in pure rolling, its angular speed of rotation can be given as ω=vR.


The total energy of the system at this instant,
ET=12Mv2+12Icylω2+12k(2x)2×2
[springs are compressed and elongated by x each]
=12Mv2+12(12MR2)(vR)2+12k(2x)2×2
[Icyl=12MR2]
Differentiating with respect to time, we get
dETdt=12M(2vdvdt)+12(12MR2)1R2(2vdvdt)+4k(2xdxdt)=0
Mva+12Mva+8kxv=0
a=163kxM
Comparing the equation with basic differential equation of SHM,
We get, the angular frequency of SHM as,
ω=16k3M
Thus, the time period of this oscillation is
T=2πω=2π3M16k=π23Mk

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