CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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.


flag
Suggest Corrections
thumbs-up
1
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Integrating Solids into the Picture
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon