# Pt. Mass and Spring rotating on table

• d843
In summary, a 3 kg mass attached to a massless spring with a spring constant of 14 N/m and an unstretched length of zero is pulled a distance of 0.5 m from the center and given a speed of 2 m/s perpendicular to the spring on a frictionless table. The angular momentum of the system about the fixed end of the spring is 3 kg m^2/s and the total energy is 7.75 J. Using equations for conservation of angular momentum and energy, the minimum distance of the mass from the center of the ellipse is 0.93 m, and the maximum distance is unknown. The minimum speed of the object can be found by using the equation L=Iw^2
d843
A mass of m = 3 kg is tied to a point by a massless spring with spring constant k = 14 N/m. Assume the spring has an unstretched length of zero. The mass is pulled a distance of d0 = 0.5 m from the center and given a speed of v0 = 2 m/s perpendicular to the spring. (The mass slides on a frictionless table.)

My Diagram (best I can do with ASCII)
Axis-(SPRING)-(MASS)
X-/\/\/\/\/\/\/-[3kg]

a) What is the angular momentum of the system about the fixed end of the spring?
b) What is the total energy of the system? (Assume the potential energy in the spring is zero when the spring is unstretched)
c) Find the minimum distance of the mass from the center of the ellipse.
d) Find the maximum distance of the mass from the center.
e) Find the minimum speed of the object.

I've solved for a-c (3 kg m2/s, 7.75 J, 0.5 m)
I'm really stuck on d. I know that once I find the answer to d I can solve e pretty quickly (I think the minimum speed will be at the maximum distance-so I can use L=3=Iw^2 and then w=v/r to solve for the velocity)

The hint it gave was "Set up equations for the conservation of angular momentum and energy."

It's entirely possible that I just need to do something else and look at this again later and I'll figure it out, but any help would be greatly appreciated.

Ok,
Well somehow I figured it out. d) 0.93m

Though it was by complete chance that I got the right answer. I was taking my conservation of momentum equation and was playing w/ the potential energy in the spring in the KE conservation eqn. So, it's not so big of a deal now, but if someone could explain the proper way of doing this problem it would be great. Thanks!

To solve for d and e, we can use the conservation of energy and angular momentum equations. The angular momentum of the system is conserved, so we can set the initial angular momentum (L0) equal to the final angular momentum (Lf):

L0 = Lf

We can also use the conservation of energy equation:

E0 = Ef

Since the potential energy in the spring is zero when it is unstretched, the initial energy (E0) is equal to the kinetic energy (K0) of the mass:

E0 = K0 = 1/2 * m * v0^2

The final energy (Ef) is equal to the potential energy (Uf) in the spring plus the kinetic energy (Kf) of the mass:

Ef = Uf + Kf

Since the mass is at its maximum distance from the center of the ellipse, the potential energy in the spring is at its maximum:

Uf = 1/2 * k * dmax^2

And since the mass is at its minimum distance from the center of the ellipse, the kinetic energy of the mass is at its minimum:

Kf = 1/2 * m * vmin^2

Now we can set these equations equal to each other and solve for dmax and vmin:

E0 = Ef

1/2 * m * v0^2 = 1/2 * k * dmax^2 + 1/2 * m * vmin^2

Solving for dmax:

dmax = √(v0^2 / k)

And solving for vmin:

vmin = √(v0^2 - (k/m) * dmax^2)

Plugging in the given values, we get:

dmax = √(2^2 / 14) = √(4/14) = 0.37 m

vmin = √(2^2 - (14/3) * (0.37)^2) = √(4 - 2.24) = √1.76 = 1.33 m/s

Therefore, the maximum distance of the mass from the center is 0.37 m and the minimum speed of the mass is 1.33 m/s.

## 1. What is the concept of "Pt. Mass and Spring rotating on table"?

The concept refers to a physical system consisting of a point mass attached to a spring and placed on a rotating table. The rotation of the table causes the point mass to undergo circular motion, while the spring provides a restoring force that leads to oscillatory motion.

## 2. How does the rotation of the table affect the motion of the point mass and spring?

The rotation of the table causes the point mass to experience a centrifugal force, which is directed away from the center of rotation. This force affects the equilibrium position of the spring and leads to a change in the amplitude and frequency of the oscillations.

## 3. What factors influence the behavior of the point mass and spring on a rotating table?

The behavior of the system is influenced by several factors such as the angular velocity of the table, the mass of the point mass, the stiffness of the spring, and the initial conditions of the system (i.e. amplitude and phase of oscillations).

## 4. Can the point mass and spring rotating on a table be used to model real-world systems?

Yes, this system can be used to model various real-world systems such as the motion of a pendulum on a rotating ship, the behavior of a swing ride at an amusement park, or the oscillations of a satellite in orbit.

## 5. What are some applications of studying the behavior of "Pt. Mass and Spring rotating on table"?

Studying this system can help in understanding the dynamics of rotating systems and how they affect the behavior of objects attached to them. It also has applications in fields such as mechanical engineering, physics, and astronomy.

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