# Mass suspended from spring in ship - Gravity problem

Hello all just a quick question,

A mass is suspended from a spring balance in a ship sailing along the equator with a speed v.
(a) Show that the scale reading will be very close to wo = (1 ± 2ωv/g), where ω is the angular
speed of Earth, and wo is the weight of the mass with the ship at rest.
(b) Explain the ± sign
.

I have wrapped my head around the question but am having trouble coming up with the given equation.
-The weight of the mass at rest will be equal to mg- the centripetal force of the rotating earth.
-The weight of the mass when the ship is sailing will be close to wo, but differ depending of the direction the ship is sailing, hence the ±.
-As this is the case I recognize that the velocity when calculating the centripetal force on the mass will be ω±v/r.

What I do not see is how 1 came into 1 ± 2ωv/g, nor where the mass or radius went

Thank you for any help,
Lucas

Andrew Mason
Homework Helper
mtong said:
Hello all just a quick question,

A mass is suspended from a spring balance in a ship sailing along the equator with a speed v.
(a) Show that the scale reading will be very close to wo = (1 ± 2ωv/g), where ω is the angular
speed of Earth, and wo is the weight of the mass with the ship at rest.
(b) Explain the ± sign
.

I have wrapped my head around the question but am having trouble coming up with the given equation.
-The weight of the mass at rest will be equal to mg- the centripetal force of the rotating earth.
-The weight of the mass when the ship is sailing will be close to wo, but differ depending of the direction the ship is sailing, hence the ±.
-As this is the case I recognize that the velocity when calculating the centripetal force on the mass will be ω±v/r.

What I do not see is how 1 came into 1 ± 2ωv/g, nor where the mass or radius went
Your approach is right. The vector sum of the spring force and gravity has to equal the centripetal force. Where $v << \omega R$ you can use the binomial expansion and eliminate the higher order term. The radius is implicit in $\omega = v_{earth}/R[/tex]. It might be useful to think in terms of [itex]\omega R \pm v = R(\omega \pm \Delta \omega)$.

The answer: 1 ± 2ωv/g is the acceleration. Multiply by m to give the normal force or weight read by the scale.

AM

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