Spring Motion (1st Semester Physics)

In summary, the conversation is about a problem involving a spring with a length of 0.6 m and a force of (4.2 N/m)x acting on a mass of 400 g attached to one end. The system is set in circular motion on a horizontal table, with an angular velocity of 8 rad/s. The goal is to find the distance the spring will be stretched. Two equations are used to solve the problem, but the answer is incorrect. The conversation ends with the realization that the radius in the equations should be .6 + x, but the correct answer is still not obtained.
  • #1
glascupio
2
0
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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  • #2
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
 
  • #3


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.
 

1. What is spring motion?

Spring motion is the movement of an object attached to a spring as it oscillates back and forth due to the force of the spring.

2. What factors affect the motion of a spring?

The motion of a spring is affected by the mass of the object attached to it, the stiffness of the spring (known as the spring constant), and the amplitude of the motion.

3. How is the period of spring motion calculated?

The period of spring motion can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass of the object, and k is the spring constant.

4. What is the relationship between the period and frequency of spring motion?

The period and frequency of spring motion are inversely proportional. This means that as the frequency increases, the period decreases, and vice versa.

5. How does the amplitude of spring motion affect its energy?

The amplitude of spring motion affects the energy of the system. As the amplitude increases, the energy also increases, and vice versa. This is because the potential energy stored in the spring is directly proportional to the amplitude of the motion.

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