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Law of Conservation of Mechanical Energy

A 5m long c...

Question

A $5$ m long cylindrical steel wire with radius $2 \times 10^{- 3}$ m is suspended vertically from a rigid support and carries a bob of mass $100$ kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring radiation losses.
(For the steel wire: Youngs Modulus $= 2.1 \times 10^{11} P a$ ; Density $= 7860 k g / m^{3}$ ; Specific heat $= 420$ J/kg-K).

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Solution

$m s Δ θ = Δ U = (1 / 2) \times$ stress $\times$ strain $\times$ volume
$(A . L . ρ) s Δ θ = (M g)^{2} L / (2 Y A)$ neglecting the strain due to the weight of wire
$Δ θ = (M g)^{2} / (2 ρ A Y s) = 0.00457$ K.

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Similar questions

5 m

2 \times 10^{- 3}

(T a k e g = 10 m / s^{2})

m o d u l u s = 2.1 \times 10^{11} N / m^{2}

= 7860 k g / m^{3}

420 J / k g C)

5 m

2 \times 10^{- 3} m

100 kg

{(x \times 10^{- 3})}^{\circ} C

x

= 2.1 \times 10^{11} {N/m}^{2}

= 7860 {kg/m}^{3}

= 420 {J/kg-}^{\circ} C

L

μ

A

Y

M

10

^{- 6}

2 m

210

16

8000

^{- 3}

420

^{- 1}

^{- 1}

g = 3.1 π m s^{- 2} .

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