Spring Motion (1st Semester Physics)

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SUMMARY

The discussion centers on a physics problem involving a spring with a force of (4.2 N/m)x, where x is the extension beyond its normal length, and a mass of 400 g rotating with an angular velocity of 8 rad/s. The correct approach to find the extension of the spring is to use the centripetal acceleration formula a = w²r, leading to the equation 4.2x = (.4(8²)(.6+x))/r. The final correct answer for the spring's extension is 0.87 meters, confirming the importance of proper application of physics equations in circular motion.

PREREQUISITES
  • Understanding of centripetal acceleration and its formulas
  • Knowledge of Hooke's Law and spring force calculations
  • Familiarity with angular velocity and its relationship to linear motion
  • Basic algebra skills for solving equations
NEXT STEPS
  • Study the derivation and application of centripetal acceleration formulas
  • Learn about Hooke's Law and its implications in oscillatory motion
  • Explore the relationship between angular velocity and linear velocity in circular motion
  • Practice solving physics problems involving forces and motion on a horizontal plane
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for examples of problem-solving in dynamics.

glascupio
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Hey guys,

Got a problem here that is getting quite annoying...

A spring is 0.6 m long. One end is permanently attached to a pivot on a horizontal table top. The force that it exerts on a body attached to the other end is (4.2 N/m)x, where x is the distance that the spring is stretched beyond its normal length. Suppose a mass of 400 g is attached to the other end of the spring and the whol system is set in circular motion. (The entire motion takes place on the horizontal table top.) How far will the spring be stretched if the mass rotates with an angular velocity of 8 rad/s?

All right, first off a = m(w^2)r in circular motion, where m is the mass, w (omega) is the angular velocity and r is the radius. However, in this equation the radius is .6 + x, where x is the change in distance. Since this is the only force involved in the problem, it must be going toward the center of the circle (because centripetal acceleration is only the result of other accelerations). This means that 4.2x = m(w^2)r. However, this leaves me with,

4.2x = .4(8^2)(.6+x)
(4.2x)/(.6+x) = 25.6
(4.2x)/(.6+x) - 25.6 = 0

Problem is, this does not yield a correct answer...

Then I tried again using mv^2/r = a and wR=v... this yielded 1.2 meters, the answer should be .87 meters. I know there's an error somewhere, I just can't find it. Can anybody lend me a hand?

Thanks in advance,
Glascupio
 
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Whoops,

Still having the same problem, but I realized why my acceleration from v and my acceleration from w weren't the same, simply because I forgot to leave r as .6 + x. When I put this in it gives me the same answer as the w formula, though the answer is still wrong. Thoughts?

Thanks in advance,
Glascupio
 


Hi Glascupio,

I can see where you may have made a mistake in your calculations. The equation for centripetal acceleration is a = v^2/r, not mv^2/r. So, instead of using 4.2x = .4(8^2)(.6+x), you should use 4.2x = (.4(8^2)(.6+x))/r. This will give you the correct answer of 0.87 meters.

Also, just a small correction, the equation for centripetal acceleration is a = w^2r, not m(w^2)r. The mass is already accounted for in the equation for force, which is F = ma.

Hope this helps! Keep up the good work in your physics class.
 

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