Question

A uniform disc of mass m and radius r is suspended through a wire attached to its center. If the time period of the torsional oscillations be T, what is the torsional constant of the wire?

Answer 1

Answer ⇒ 2π²mr²/T²

Explanation ⇒ Let the torsional constant be k.

If the torque in the wire be for an angular displacement of θ, then

= kθ

Also, By the definition of Torque, = Iα, where I is the moment of inertia and α is the angular acceleration.

Now, moment of inertia of the disc about the axis passing through its centre = mr²/2

∴ α = /Iα

∴ α = kθ/mr²/2

∴  α = 2kx/mr³      [∵ θ =x/r]

∴ a/r = (2k/mr³)x           [∵ α = a/r]

∴ a = (2k/mr²)x

Since the acceleration is proportional to the displacement, this means the motion for this tortion is an SHM.

Comparing it with a = ω²x

∴ ⍵ = √(2k/mr²)

Now, time period  =2π/⍵

∴ T = 2π√(mr²/2k)

∴  k = 2π²mr²/T²

Hence, the torsional constant of the wire is 2π²mr²/T².

Hope it helps.

Answer 2

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