Equations of Motion of a Mass Attached to Rotating Spring

In summary, the student attempted to solve the equations of motion for a particle attached to a light spring, but was not able to get the correct results.
  • #1
SaraM
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0
1. Homework Statement

A particle of mass m is attached to the end of a light spring of equilibrium length a, whose other end is fixed, so that the spring is free to rotate in a horizontal plane. The tension in the spring is k times its extension. Initially the system is at rest and the particle is given an impulse that starts its movement at right angles to the spring with velocity v. Write down the equations of motion in polar co-ordinates.

2. Homework Equations

Radial Force Fr= -k⋅r
Radial acceleration ar = mv2/r
Tangential Force Fθ = Torque = r⋅F

3. The Attempt at a Solution

I tried applying Newton's second law for both the radial and tangential components.
In r : mar= -k⋅r ⇒ r = a⋅cos(ωt) (which is not the correct answer)
In θ: 1- mr⋅aθ= r⋅F (got me nowhere)
2- vθ= v/r=ω ; ω=√k/m (how do I recover θ(t) from it?)

Thank you in advance
 
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  • #2
Welcome to PF!
SaraM said:
2. Homework Equations

Radial Force Fr= -k⋅r
This does not account for the equilibrium length of the spring.
Radial acceleration ar = mv2/r
Does the expression on the right have the correct dimensions for an acceleration?
In polar coordinates, the radial acceleration ar has two terms. For a review of velocity and acceleration in polar coordinates, see http://faculty.etsu.edu/gardnerr/2110/notes-12e/c13s6.pdf
Tangential Force Fθ = Torque = r⋅F
Force and torque do not have the same dimensions. So, a force cannot equal a torque.
 
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  • #3
Thank you for your reply!

Following your hints, here's what I ended up with:
In ##r##: ##~m( \ddot r -r \dot \theta^2)=- k(r-a)##
In ##\theta##: ##~\dot\theta=\frac{v}{r}=-\frac{k}{mv}(r-a) ## (which I got from the centripetal force ##m\frac{v^2}{r}=-k(r-a)##)

What's left to do is plug in ##\dot\theta## in the equation in ##r## and find ##r(t)##
Is there still something incorrect in my reasoning?
 
Last edited:
  • #4
SaraM said:
##m( \ddot r -r \dot \theta^2)=- k(r-a)##
OK, this looks good. It expresses ##\sum F_r = ma_r##.
##\dot\theta=\frac{v}{r} ##
This is not quite correct. Note that ##r \dot{\theta}## represents the ##\theta## component of ##\vec{v}##.
##m\frac{v^2}{r}=-k(r-a)##
No, the correct relation is the relation that you got above: ##m( \ddot r -r \dot \theta^2)=- k(r-a)##.

What's left to do is plug in ##\dot \theta## in the equation in ##r## and find ##r(t)##.
OK. To get the correct expression for ##\dot \theta## in terms of ##r## you can do either of the following:
(1) Set up ##\sum F_\theta = ma_\theta##. Consult the link for the expression for ##a_\theta##.

(2) Decide if there is any torque acting on the particle relative to the origin. If not, what quantity is conserved?
 
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  • #5
OK, got it. Thank you so much for your help! Cleared things up.
 

Related to Equations of Motion of a Mass Attached to Rotating Spring

1. What is the equation of motion for a mass attached to a rotating spring?

The equation of motion for a mass attached to a rotating spring is given by:
m(d^2x/dt^2) = -kx + mrω^2 cos(ωt + δ)
where m is the mass of the object, x is its displacement from equilibrium, k is the spring constant, r is the distance from the spring's axis of rotation to the mass, ω is the angular velocity of the rotating spring, and δ is the phase angle.

2. What is the significance of the phase angle in the equation of motion?

The phase angle, δ, in the equation of motion represents the initial position of the mass when the rotating spring starts its motion. It is important because it determines the starting point of the oscillation and can affect the amplitude and frequency of the motion.

3. How does the angular velocity of the rotating spring affect the motion of the mass?

The angular velocity, ω, of the rotating spring directly affects the frequency of the oscillation of the mass. A higher angular velocity results in a higher frequency and a shorter period of oscillation. It also affects the amplitude of the motion, with a higher ω resulting in a larger amplitude.

4. Can the equation of motion be used to analyze the motion of any mass attached to a rotating spring?

Yes, the equation of motion for a mass attached to a rotating spring can be used to analyze the motion of any object that is attached to a spring and is rotating about a fixed axis. It is a generalized equation that can be applied to various scenarios, such as a mass attached to a rotating pendulum or a mass attached to a rotating wheel.

5. How can the equation of motion be solved to find the position of the mass at any given time?

The equation of motion is a second-order differential equation, which can be solved using various methods such as the method of undetermined coefficients or the Laplace transform. Once solved, the resulting equation can be used to find the position of the mass at any given time by plugging in the appropriate values for the variables.

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