Mass suspended from spring in ship - Gravity problem

But when the ship is sailing, the weight is not only affected by the Earth's gravity, but also the centripetal force of the rotating Earth. This can be calculated as ω²r (where r is the radius of the Earth). Therefore, the weight when the ship is sailing will be close to mg ± ω²r. Since we are dealing with small values of v compared to ωr, we can use binomial expansion to simplify the equation and get 1 ± 2ωv/g. This explains the ± sign and where the mass and radius went.
  • #1
mtong
33
0
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
 
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  • #2
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 [itex]v << \omega R[/itex] you can use the binomial expansion and eliminate the higher order term. The radius is implicit in [itex]\omega = v_{earth}/R[/tex]. It might be useful to think in terms of [itex]\omega R \pm v = R(\omega \pm \Delta \omega)[/itex].

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

AM
 
Last edited:
  • #3


Hello Lucas,

Thank you for your question. Let me try to explain the given equation and the ± sign.

First, let's consider the weight of the mass when the ship is at rest. As you mentioned, it will be equal to mg - the centripetal force of the rotating Earth. This centripetal force is given by F = mv^2/r, where m is the mass, v is the velocity of the ship, and r is the radius of the Earth.

Now, let's consider the weight of the mass when the ship is sailing. In this case, the weight will be close to wo, but it will differ depending on the direction the ship is sailing. This is because the ship's velocity will either add to or subtract from the Earth's rotation velocity, ω. This is why we have the ± sign in the equation.

To understand where the 1 ± 2ωv/g comes from, we need to look at the forces acting on the mass when the ship is sailing. We have the weight of the mass, mg, acting downwards, and the centripetal force, mv^2/r, acting outwards. In order for the mass to remain suspended from the spring balance, these two forces must balance each other out. This can be represented by the equation: mg = mv^2/r.

Now, let's substitute the value of the centripetal force, mv^2/r, in the equation with the given value for v/r, which is ω±v/r. This gives us mg = m(ω±v/r)^2/r. Simplifying this equation, we get mg = m(ω^2 ± 2ωv/r + v^2/r^2). Now, we can see that the only term that depends on the ship's velocity, v, is 2ωv/r. This is where the 1 ± 2ωv/g comes from. And since the weight of the mass at rest, wo, is equal to mg, we can write the equation as wo = m(1 ± 2ωv/g)g, which is the same as the given equation.

I hope this explanation helps you understand the given equation and the ± sign. Let me know if you have any further questions. Good luck!
 

Related to Mass suspended from spring in ship - Gravity problem

1. How does the mass suspended from a spring in a ship affect the gravity problem?

The mass suspended from a spring in a ship does not affect the gravity problem. Gravity is a fundamental force that acts on all objects regardless of their mass or location.

2. What is the purpose of suspending a mass from a spring in a ship?

The purpose of suspending a mass from a spring in a ship is to create a simple harmonic motion that can be used to measure the acceleration due to gravity. This method is often used in experiments to determine the value of g.

3. How does the spring affect the mass and its motion in the ship?

The spring provides a restoring force that is proportional to the displacement of the mass from its equilibrium position. This means that as the ship moves, the spring will stretch or compress, causing the mass to oscillate back and forth.

4. Does the mass suspended from a spring in a ship experience the same gravitational force as a mass at rest?

Yes, the mass suspended from a spring in a ship experiences the same gravitational force as a mass at rest. This is because gravity acts on all objects equally, regardless of their motion or location.

5. Can the mass suspended from a spring in a ship be used to determine the ship's velocity?

No, the mass suspended from a spring in a ship cannot be used to determine the ship's velocity. The spring simply measures the acceleration due to gravity, and the ship's velocity is influenced by many other factors such as wind, currents, and engine power.

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