What is the spring constant in terms of mass and unstretched length?

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SUMMARY

The spring constant \( k \) can be derived from the forces acting on a mass \( m \) connected to a spring with an unstretched length \( L \). When the mass is swung in a circular motion, it does not follow a circular path due to the oscillation of the spring, resulting in an egg-shaped trajectory. The correct expression for the spring constant is \( k = \frac{m(g + a)}{\Delta s} \), where \( \Delta s \) is the stretch of the spring, which is half of its unstretched length at the bottom of the path. Understanding the forces at play and applying conservation of energy principles are crucial for deriving \( k \) accurately.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with Hooke's Law and spring constants
  • Knowledge of circular motion dynamics
  • Basic principles of energy conservation
NEXT STEPS
  • Study the derivation of spring constants using Hooke's Law
  • Learn about the dynamics of oscillatory motion
  • Explore the relationship between centripetal force and circular motion
  • Investigate energy conservation in mechanical systems
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking to explain the concepts of spring dynamics and circular motion.

cosmo1993
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Homework Statement


A mass m is connected to a spring with unstretched length L. You hold on to one end of the spring and swing the mass around. You practice getting the mass going until the spring just goes slack at the top of the path. At the bottom of the path. the spring stretches by an amount equal to half of its unstretched length.
Why doesn't the mass make a circular path?
Find the spring constant k of the spring

Homework Equations


F = -kΔs
at the bottom of the circle
Normal force = mg + ma

The Attempt at a Solution


I figured that the mass does not make a circular path because the spring is oscillating as it travels around the circle, therefore, the path of the mass is more egg shaped.

I said that K = (m(g+a))/Δs would be the spring constant but I am not sure if I am correct. Any help?
 
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No, that's not correct. What's ##a##? What's ##\Delta s##? What about ##L##? You haven't expressed ##k## in terms of known quantities.

By the way, there is no normal force. A normal force is a force exerted by a surface in contact with an object. The mass isn't in contact with any surface, so there can be no normal force.

Start by drawing a free-body diagram for the mass at the top of the path and at the bottom of the path.
 
cosmo1993 said:

Homework Statement


A mass m is connected to a spring with unstretched length L. You hold on to one end of the spring and swing the mass around. You practice getting the mass going until the spring just goes slack at the top of the path. At the bottom of the path. the spring stretches by an amount equal to half of its unstretched length.
Why doesn't the mass make a circular path?
Find the spring constant k of the spring

Homework Equations


F = -kΔs
at the bottom of the circle
Normal force = mg + ma

The Attempt at a Solution


I figured that the mass does not make a circular path because the spring is oscillating as it travels around the circle, therefore, the path of the mass is more egg shaped.

I said that K = (m(g+a))/Δs would be the spring constant but I am not sure if I am correct. Any help?

You need to give the spring constant in terms of L and m. What is a and what is its direction?

You are right, the mass does not follow a circular path. What forces act on it at the highest and at the lowest points?

At these points, you can consider the path as piece of some circle. There must be some force pointing towards the centre to make the path. How those forces are related to the speed and the distance from the centre?

You can also assume conservation of energy.

ehild
 

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