Question

Two small balls, each of mass m are connected by a light
rigid rod of length L. The system is suspended from its
centre by a thin wire of torsional constant k. The rod is
rotated about the wire through an angle 0, and released.
Find the force exerted by the rod on one of the balls as
the system passes through the mean position.​

Answer 1

Answer:

Answer ⇒ √(k²θ₀⁴/L² + m²g²)

Explanation ⇒ The Rod is rotated through an angle θ₀

, thus the torsion in the wire = kθ₀

Now, Total energy = P.E. stored at this position

Also, Potential energy stored at that position = 1/2 × kθ₀².

At the mean position, this total energy will be converted to K.E.

∴ K.E. at the mean position = 1/2 × I⍵²

where ⍵ is the angular velocity of the ball system at the mean position.

Equating both the equations, we get

I⍵² = kθ₀²

⍵² = kθ₀²/I

where I is the moment of Inertia about the centre of the rod.

∴ I = m(L/2)² + m(L/2)²

∴ I = mL²/2

Therefore,

⍵² = kθ₀²/(mL²/2)

∴ ⍵² = 2kθ₀²/mL²

∴ (v/distance)² = 2kθ₀²/mL²

∴ v²/(L²/4) = 2kθ₀²/mL²

∴ 4v²/L² =2kθ₀²/mL²

∴ 2v² = kθ₀²/m

∴ v² = kθ₀²/2m

Therefore, Centrifugal force = mv²/r = 2mv²/L

= kθ₀²/L

Now Refer to the attachment.

TSinθ = kθ₀²/L

TCosθ = mg

∴ T²Sin²θ + T²Cos²θ = k²θ₀⁴/L² + m²g²

∴ T² = k²θ₀⁴/L² + m²g²

∴ T = √(k²θ₀⁴/L² + m²g²)

Hence, the tension in string is √(k²θ₀⁴/L² + m²g²).