Equations of Motion for a mass on a circular track with a linear spring

In summary, the conversation discusses a problem involving a collar sliding along a circular track with a linear spring attached to it. The goal is to determine the differential equation of motion for the collar in terms of the angle θ and the reaction force exerted by the track on the collar as a function of θ. The solution involves setting up a reference frame and using the transport theorem to calculate velocity and acceleration. The forces acting on the collar are gravity, normal, and the linear spring. After setting up the equations of motion and normal force, the solution is found to be correct with the only suggestion being to double check the units.
  • #1
tlonster
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0

Homework Statement



A collar of mass m slides without friction along a circular track of radius R as
shown in THE ATTACHMENT. Attached to the collar is a linear spring with spring constant K and
unstretched length zero. The spring is attached at the fixed point A located a distance
2R from the center of the circle. Assume gravity acts down and determine (a) the differential equation of motion for the collar in terms the angle θ and (b) the reaction force exerted by the track on the collaras a function of the angle θ.


Homework Equations



The transport theorem

F = ma


The Attempt at a Solution



I have worked through this problem and am not confident in my answer. A second look would be much appreciated. Here's my answer:

I created reference frame E: {er, eθ, ez} at mass m such that it's position is

r = Rer

By using the transport theorem for velocity and acceleration as seen by the intertial frame I came up with:

v(m/I) = 0 + θ(dot)R[eθ]
a(m/I) = θ(double dot)R[eθ] - (θ(dot))^2R[er]

Next the forces acting on my mass are gravity, normal, and linear spring

∴Ʃ F = mg[JI] + N[er] + K(3R-0)[-er + JI]

where JI = sinθ[er] + cosθ[eθ]

when I substitute JI into my ƩF equation, set equal to ma(m/I), and drop the unit vectors, I get these two equations:

1. mgsinθ + N -3KR + 3KRsinθ = -m(θ(dot))^2R
2. mgcosθ + 3KRcosθ = mθ(double dot)R

For my equation of motion in terms of θ, I only need to use the second equation, correct? Where θ(double dot) = (mgcosθ + 3KRcosθ)/(mR)

and for the normal force (N) I just use the first equation and move everything to one side for N = -mgsinθ +3KR - 3KRsinθ -m(θ(dot))^2R

Can someone please take a look at this? I feel as though my spring force
Fs = K(3R-0)[-er + JI] is not correct, but I don't know what else it should be. Any help would be appreciated.

Thank you.

 

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  • #2





Your solution looks correct to me. The only thing I would suggest is to double check your units and make sure they are consistent throughout the equations. Other than that, your approach and equations seem to be correct. Keep up the good work!
 

Related to Equations of Motion for a mass on a circular track with a linear spring

1. What is the equation for the displacement of a mass on a circular track with a linear spring?

The equation for displacement (x) of a mass on a circular track with a linear spring is given by x = A*cos(ωt), where A is the amplitude of the motion and ω is the angular frequency.

2. What is the relationship between the angular frequency and the spring constant in this system?

The angular frequency (ω) is directly proportional to the square root of the spring constant (k) in this system. This means that as the spring constant increases, the angular frequency also increases, resulting in faster oscillations of the mass.

3. How does the mass affect the period of motion in this system?

The period of motion (T) is inversely proportional to the square root of the mass (m) in this system. This means that as the mass increases, the period of motion decreases, resulting in faster oscillations of the mass.

4. Can the amplitude of motion be changed in this system?

Yes, the amplitude of motion can be changed by adjusting the initial conditions of the system. This can be done by changing the initial displacement or velocity of the mass, or by changing the initial position of the spring.

5. What happens to the energy of the system as the mass oscillates on the circular track?

The energy of the system is conserved as the mass oscillates on the circular track. The potential energy of the spring is converted into kinetic energy as the mass moves, and then back into potential energy as the mass returns to its original position. This continuous exchange of energy results in a constant total energy for the system.

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