A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2 . Calculate the elongation of the wire when the mass is at the lowest point of its path.

Asked by Abhisek | 1 year ago |  65

#### 1 Answer

##### Solution :-

Mass, m = 14.5 kg

Length of the steel wire, l = 1 m

Angular velocity, v = 2 rev/s

Cross-sectional area of the wire, A = 0.065 x 10-4 m2

Total pulling force on the steel wire when the mass is at the lowest point of the vertical circle,

F = mg + mr ω2

= 14.5×9.8+14.5×1×(12.56)2

=2429.53 N

Young’s modulus = Stress / Strain

Y=$$\dfrac{F}{A}\dfrac{l}{\Delta l}$$

$$\Delta l=\dfrac{F}{A}\dfrac{l}{Y}$$

Δl=$$\dfrac{(2429.53×1)}{(0.065×10^{−4}) × (2×10^{11})}$$

= 1.87×10−3m

Hence, the elongation of the wire when the mass is at the lowest is 1.87×10−3m.

Answered by Pragya Singh | 1 year ago

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