9.11 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.

3 Views | Posted 5 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

Mass, m = 14.5 kg

Length of the steel wire, l = 1.0 m

Angular velocity, $ω$ = 2 rev/s

Cross sectional area of the wire, A = 0.065 ${c m}^{2}$ = 0.065 $\times 10^{- 4} m^{2}$

Let Δl be the elongation of the wire

When the mass is placed at the position of the vertical circle, the total force on the mass is

F = mg + ml $ω^{2}$ = 14.5 $\times (9.81 + 1 \times 2^{2})$ = 200.25 N

Young’s modulus for steel, $Y_{s}$ = Stress / Strain = 2 $\times 10^{11}$ Pa

Stress = F/A = 200.25/0.065 $\times 10^{- 4}$

Strain = (Δl/l) = (Δl/1) = Δl

Δl = (200.25/0.065 $\times 10^{- 4}) /$ 2 $\times 10^{11}$ = 1.54 $\times 10^{- 4}$ m

Mass, m = 14.5 kgLength of the steel wire, l = 1.0 mAngular velocity,  <math><mi>ω</mi></math> = 2 rev/sCross sectional area of the wire, A = 0.065 <math><msup><mrow><mrow><mi>c</mi><mi>m</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = 0.065 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>4</mn></mrow></mrow></msup><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math>Let Δl be the elongation of the wireWhen the mass is placed at the position of the vertical circle, the total force on the mass isF = mg + ml <math><msup><mrow><mrow><mi>ω</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = 14.5 <math><mo>×</mo><mfenced separators="|"><mrow><mrow><mn>9.81</mn><mo>+</mo><mi></mi><mn>1</mn><mo>×</mo><msup><mrow><mrow><mn>2</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfenced></math> = 200.25 NYoung’s modulus for steel,  <math><msub><mrow><mrow><mi>Y</mi></mrow></mrow><mrow><mrow><mi>s</mi></mrow></mrow></msub></math> = Stress / Strain = 2 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>11</mn></mrow></mrow></msup></math> PaStress = F/A = 200.25/0.065 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>4</mn></mrow></mrow></msup></math>Strain = (Δl/l) = (Δl/1) = ΔlΔl = (200.25/0.065 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>4</mn></mrow></mrow></msup><mo>)</mo><mo>/</mo></math> 2 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>11</mn></mrow></mrow></msup></math> = 1.54 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>4</mn></mrow></mrow></msup></math> m

0 Upvotes 0 Downvotes

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question