Question

A small ball is attached to one end of a spring that has an unstrained length of 0.248 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.05 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.0104 m. By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless?

Answer 1

Unstrained length of the string, L = 0.248 m

Stretch in the string when whirled around a horizontal circle, = 0.0104 m

By spinning horizontally, we are told that gravity plays little in the deformation (stretching) of the spring.

Therefore, all of the forces in the spring are centripetal.

Centripetal acceleration, a = v²/r = v² / (L + ) = (3.05 m/s)²/(0.248 m + 0.0104 m) = 36.00 m/s²

Since the spring constant is, well, constant, the stretch due to gravity would be -

∆x = (9.8m/s² / 36.00 m/s²) * 0.0104 m = 0.0028 m (Answer)

Hope, you understand the solution !