Physics, asked by PhysicsHelper, 1 year ago

Two small balls each of mass m are connected by a light rigid rod of length L. The system is suspended from its center by a thin wire of torsional constant k. The rod is rotated about the wire through an angle θ₀ and released. Find the tension in the rod as the system passes through the mean position.

Answers

Answered by tiwaavi

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).

Hope it helps.

Attachments:

devanayan2005: Class answer buoy

Answered by Anonymous

Answer ⇒ √(k²θ₀⁴/L² + m²g²)