Mass and spring in circular motion.

In summary, the conversation discussed solving a problem involving a mass moving in a circular orbit under the influence of a three dimensional harmonic potential. The frequency of the orbit and subsequent small oscillations were calculated using equations and assumptions based on known information. The question of the equilibrium length r0 was clarified and ultimately determined to have a value of zero in this particular problem.
  • #1
AbigailM
46
0
Just preparing for a physics prelim and working through previous exam questions.

Homework Statement


A mass m moving in a circular orbit about the origin is attracted by a three dimensional harmonic potential,

[itex]U(r)=\frac{1}{2}kr^{2}[/itex]

What is the frequency of the orbit? If a small kick is supplied in the radial direction, what will be the frequency of the ensuing small oscillations in r?

The Attempt at a Solution



[itex]k(r-r_{0})=m\omega^{2}_{0}r[/itex]

[itex]\frac{k}{m}\frac{(r-r_{0})}{r}=\omega^{2}_{0}[/itex]

[itex]\mathbf{Frequency\hspace{1 mm} of\hspace{1 mm} orbit\hspace{1 mm}}\omega_{0}=\sqrt{\frac{k}{m}\frac{(r-r_{0})}{r}}[/itex]

[itex]\ddot{r}=-\omega^{2}r \hspace{5 mm} \omega^{2}=\frac{k}{m}[/itex]

[itex]\omega^{2}=\frac{k}{m}=\frac{r\omega^{2}_{0}}{r-r_0}[/itex]

[itex]\mathbf{Frequency\hspace{1 mm} of\hspace{1 mm} oscillation\hspace{1 mm}}\omega=\omega_0\sqrt{\frac{r}{r-r_{0}}}[/itex]

Just wondering if my solution looks correct. Thanks for the help.
 
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  • #2
Hi AbigailM :smile:

Your method looks correct to me, but you aren't given r0 in the question... :wink:
 
  • #3
Ok, so where I'm confused is that as the spring-mass is rotated, it extends from its equilibrium length due to centripetal force. So why is [itex](r-r_{0})[/itex] wrong?

Thanks for the help.
 
  • #4
AbigailM said:
Ok, so where I'm confused is that as the spring-mass is rotated, it extends from its equilibrium length due to centripetal force. So why is [itex](r-r_{0})[/itex] wrong?

Thanks for the help.

It isn't wrong. The 'equilibrium length' r0 isn't a given information in your question, you have assumed it(unless you wrote an incomplete question here :uhh:). And normally, we give answers in terms of stuff that's known to us. What do you think would be the value of r0 in known terms? :smile:
 
  • #5
Ahhh ok I see now. The restoring force is balanced by the centrifugal force from the rotation. We can set [itex]r_{0}=0[/itex]. The reason I included [itex]r_{0}[/itex] is because for a harmonic oscillator, the mass oscillilates around the origin [itex]r_{0}[/itex]. Any many cases it's defined to be zero. And I think your right, if we were to consider it in this problem it would have been given.

Thanks again for the help Infinitum.
 
  • #6
AbigailM said:
Ahhh ok I see now. The restoring force is balanced by the centrifugal force from the rotation. We can set [itex]r_{0}=0[/itex]. The reason I included [itex]r_{0}[/itex] is because for a harmonic oscillator, the mass oscillilates around the origin [itex]r_{0}[/itex]. Any many cases it's defined to be zero. And I think your right, if we were to consider it in this problem it would have been given.

Thanks again for the help Infinitum.

Looks correct now :smile:
 

1. What is the relationship between mass and spring in circular motion?

In circular motion, the centripetal force acting on an object is directly proportional to its mass and the square of its velocity. This means that as the mass of an object increases, the force required to maintain circular motion also increases. The spring in circular motion acts as the centripetal force, so its force will also increase with mass.

2. How does the spring constant affect circular motion?

The spring constant, also known as the stiffness of the spring, determines how much force is required to stretch or compress the spring. In circular motion, a higher spring constant means that a larger force is needed to maintain the same radius of circular motion. Therefore, a higher spring constant will result in a higher centripetal force and a higher velocity for the object in circular motion.

3. How does the radius of circular motion affect the behavior of the mass and spring system?

The radius of circular motion is directly related to the centripetal force acting on the object. As the radius decreases, the centripetal force required to maintain circular motion increases. This means that the spring will be stretched or compressed more, depending on the direction of the circular motion. A larger radius will result in a smaller centripetal force and less stretch or compression of the spring.

4. Can the mass and spring system be in equilibrium during circular motion?

Yes, it is possible for the mass and spring system to be in equilibrium during circular motion. This occurs when the centripetal force provided by the spring is equal to the force required to maintain circular motion. In this case, the spring is neither stretched nor compressed, and the mass moves at a constant velocity in a circular path.

5. How does changing the mass or spring constant affect the period of circular motion?

The period of circular motion, which is the time taken for one complete revolution, is affected by both the mass and spring constant. Increasing the mass will result in a longer period, as more force is required to maintain circular motion. Similarly, increasing the spring constant will also result in a longer period, as a larger force is needed to maintain the same radius of circular motion.

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