Question

A sphere of mass 20 kg is suspended by a metal wire of unstretched length 4 m and diameter 1 mm. When in equilibrium, there is a clear gap of 2 mm between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle θ with the vertical and is released. Find the maximum value of θ so that the sphere does not rub the floor. Young's modulus of the metal of the wire is 2.0 x 10¹¹ N/m². Make appropriate approximations.

Answer 1

Let us assume that when oscillating the sphere traverses a circular path. If the elongation at equilibrium position be l, then 

 l/L = mg/AY

 l = mgL/AY 

The length in the equilibrium position L' = L+l

When oscillating the sphere will have a horizontal speed v at vertical position of the wire. This v will depend on the angle θ. P.E. of the sphere at this displacement angle =mg(L'-L'.cosθ) 

{Taking the lowest position as zero potential energy level} 

K.E. of the sphere at the vertical position of the wire =mv²/2  

Equating both we get  

mv²/2 = mgL'(1-cosθ) 

mv² = 2mgL'(1-cosθ) 

The total tension in the wire at vertical position 

= mg + mv²/r = mg + 2mgL'(1-cosθ)/r  

= mg{r + 2L'(1-cosθ)}/r 

The stress in the wire at this time 

= Tension in the wire/cross-sectional area  

= (mg/rA){r+2L'(1-cosθ)}

 Allowed elongation= l + 0.002 m (so that the sphere does not touch the floor) 

Strain at this time = (l + 0.002)/L  

Hence (l + 0.002)/L = (mg/rAY){r+2L'(1-cosθ)} 

→mgL/AY + 0.002 = (mgL/rAY){r+2L'(1-cosθ)} 

[substituting l with its expression] 

→{r+2L'(1-cosθ)} = r + 0.002*rAY/mgL  

→1-cosθ   = 0.001*rAY/mgLL' 

→cosθ = 1 -0.001*rAY/mgLL' 

→cosθ=1-0.001*(L+l+0.002){π(0.001)²/4}*2x10¹¹/{20*10*4*(L+l)} 

{Let us consider the 0.002 m length negligible in comparision to L', so L+l+0.002≈L+l, i.e  r ≈ L'} 

→cosθ = 1 - 0.196 

→cosθ = 0.804 

→θ = 36.4°

Answer 2

At equilibrium => T = mg
When it moves to an angleθ, and released, the tension the T' at lowest point is

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