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.

Open in App

Solution

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 cm²

Let δl be the elongation of the wire when the mass is at the lowest point of its path.

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

F = mg + mlω²

= 14.5 × 9.8 + 14.5 × 1 × (2)²

= 200.1 N

Young’s modulus for steel = 2 × 10¹¹ Pa

Hence, the elongation of the wire is 1.539 × 10^–4 m.

Suggest Corrections

Similar questions

Q. 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

Q. A

15 k g

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 r e v

s^{- 1}

at the bottom of the circle. The cross-section of the wire is

0.05 {c m}^{2}

. The elongation of the wire when the mass is at the lowest point of its path is
(Take

g = 10 m s^{- 2}, Y_{s t e e l} = 2 \times 10^{11} N m^{- 2}

)

Q. 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

c m^{2}

. Calculate the elongation of the wire when the mass is at the lowest point of its path.

Q. A 6 kg weight is fastened to the end of a steel wire of unstretched length 60 cm. It is whirled in a vertical~~ circle and has an angular velocity of 2 rev/s at the bottom of the circle. The area of cross-section of the. i wire is

0.05 c m^{2}

. Calculate the elongation of the wire when the weight is at the lowest point of the path . Young's modulus of steel

= 2 \times 10^{11}

N/m^{2} .

A 6-kg weight is fastened to the end of a steel wire of unstreched length 60 cm. It is whirled in a vertical circle and has an angular velocity of 2 revolution per second at the bottom of the circle. The area of cross-section of the wire is $0.05 c m^{2}$ . The elongation of the wire when the weight is at the lowest point of the path is $(Y_{s t e e l} = 2 \times 10^{11} P a)$

Join BYJU'S Learning Program