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Question

A uniform disc of mass m and radius R is pivoted smoothly at its centre of mass. A light spring of stiffness k is attached with the disc tangentially as shown in the figure. If the disc is rotated through a small angle and released, the angular frequency of oscillation of the disc is (in rad/s)
[(Take m=5 kg and k=10 N/m]


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Solution

Let us twist (rotate) the disc through a small clockwise angle θ. Then, the spring will be deformed (compressed) by a distance x=Rθ. Hence, the spring force Fs=kx=k(Rθ) will produce a restoring torque.


Restoring torque, τ=FsR where Fs=kRθ
This gives, τ=kR2θ .....(1)
It means that after removing the external, (applied) torque, the restoring torque rotates the disc with an angular acceleration α which will bring the spring disc system back to its original state.
Applying Newton's second law for rotational motion, we have
τ=Iα
Using (1) in the above formula, we get
kR2θ=Iα
This gives α=kR2θI
Moment of inertia of a disc about an axis passing through its COM is given by
I=mR22
Thus, α=2kmθ
Comparing the above equation with α=ω2θ, we get
ω=2km
From the data given in the question,
ω=2×105ω=2 rad/sec

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