Relationship Between Angular Velocity and the Spring Constant

In summary: Look at it from a reverse perspective: if v were to double, and r were to remain the same, what would happen to the centripetal force, mv^2 / r? By what factor would it increase? How would this increase in centripetal force be achieved by changing the spring constant?Alternatively, solve your equation for v. Since k is the only variable, and all the others are constants, you can see more easily what needs to be done to increase v by a factor of two. In this equation, v is the only variable that needs to be increased. To do so, you would need to change k.
  • #1
dbakg00
21
1

Homework Statement



A mass attached to a spring which is mounted on a turntable. If you wanted to double the angular velocity, but keep the radius of the circle the same, by how much would you change the spring constant?

Homework Equations



[itex]\omega = \frac{v}{r}[/itex]




The Attempt at a Solution



I'm a bit confused on how these two are related.

If the radius is staying the same & the goal is to double [itex]\omega[/itex], then the part of the equation which would change would be, v, the tangential velocity. I just can't figure out how k is related to tangential velocity. Can someone hel point me in the right direction? Thanks
 
Physics news on Phys.org
  • #2
dbakg00 said:

Homework Statement



A mass attached to a spring which is mounted on a turntable. If you wanted to double the angular velocity, but keep the radius of the circle the same, by how much would you change the spring constant?

Homework Equations



[itex]\omega = \frac{v}{r}[/itex]

The Attempt at a Solution



I'm a bit confused on how these two are related.

If the radius is staying the same & the goal is to double [itex]\omega[/itex], then the part of the equation which would change would be, v, the tangential velocity. I just can't figure out how k is related to tangential velocity. Can someone hel point me in the right direction? Thanks

For the mass to move in a circle, it must have a centripetal force acting on it. What's the equation for centripetal force? What must happen to the centripetal force if the velocity doubles, but the radius remains the same?

Now, the centripetal force is just a requirement for circular motion. Something has to be physically present to provide that centripetal force, and if it can't, then the circular motion will cease. For example, with a planet orbiting a star, the force of gravity acts as the centripetal force that keeps the planet in its orbit, and if it is not strong enough given the speed of the object, that object will escape, instead of being in a bound orbit. Similarly, if you twirl a mass on a string in a circle around your head, the tension in the string is what acts as the centripetal force in this problem. If you twirl too fast, the motion will require more centripetal force than the string can provide, and it will snap.

What is providing the centripetal force in THIS problem? The answer to that question will tell you how k comes into play.
 
  • #3
Well, I think the equation for the centripetal force would be [itex]\Delta r * k[/itex] , but I'm not sure if that is the formula you're looking for.

If that is what you are referring to then the k would have to double because the [itex]\Delta r[/itex] is not changing.

Am I on the right track?
 
  • #4
By what factor will the centripetal force increase if you double the angular velocity at constant radius?
 
  • #5
by two??
 
  • #6
dbakg00 said:
Well, I think the equation for the centripetal force would be [itex]\Delta r * k[/itex] , but I'm not sure if that is the formula you're looking for.

If that is what you are referring to then the k would have to double because the [itex]\Delta r[/itex] is not changing.

Am I on the right track?

Yes, but you didn't directly answer the question I asked you:

cepheid said:
What is providing the centripetal force in THIS problem? The answer to that question will tell you how k comes into play.

The answer is: the spring. The restoring force of the spring is what provides the centripetal force that keeps the mass moving in a circle in this situation.

dbakg00 said:
by two??

No. Doubling the angular speed at constant radius will not increase the centripetal force by a factor of two. To see why not, answer THIS question: what is the equation for centripetal force?
 
  • #7
[itex] k*x=m*r*\omega^{2}[/itex]

and [itex]\omega^{2}= \frac{v^{2}}{r^{2}}[/itex]

so i think the force, kx, is equal to [itex]\frac{m*v^{2}}{r}[/itex]

If the above is true, then I can see how k is related to the velocity, but I am not quite sure how to get v to double. Maybe multiply k by (1/2)??
 
  • Like
Likes Simanto Rahman
  • #8
dbakg00 said:
[itex] k*x=m*r*\omega^{2}[/itex]

and [itex]\omega^{2}= \frac{v^{2}}{r^{2}}[/itex]

so i think the force, kx, is equal to [itex]\frac{m*v^{2}}{r}[/itex]

If the above is true, then I can see how k is related to the velocity, but I am not quite sure how to get v to double. Maybe multiply k by (1/2)??

Look at it from a reverse perspective: if v were to double, and r were to remain the same, what would happen to the centripetal force, mv^2 / r? By what factor would it increase? How would this increase in centripetal force be achieved by changing the spring constant?

Alternatively, solve your equation for v. Since k is the only variable, and all the others are constants, you can see more easily what needs to be done to k in order to get v to double, when the equation is written in this form.
 
  • #9
cepheid said:
Look at it from a reverse perspective: if v were to double, and r were to remain the same, what would happen to the centripetal force, mv^2 / r? By what factor would it increase? How would this increase in centripetal force be achieved by changing the spring constant?

Alternatively, solve your equation for v. Since k is the only variable, and all the others are constants, you can see more easily what needs to be done to k in order to get v to double, when the equation is written in this form.

Ok, now I see, it would take an increase of 4k to double the velocity with a constant radius. I was over thinking it...thanks for you help everyone
 

1. What is angular velocity?

Angular velocity is a measure of how quickly an object is rotating around an axis. It is typically measured in radians per second.

2. How is angular velocity related to the spring constant?

The relationship between angular velocity and the spring constant is that the angular velocity of an object attached to a spring is directly proportional to the spring constant. This means that as the spring constant increases, the angular velocity also increases.

3. How does the spring constant affect the motion of an object attached to a spring?

The spring constant determines how stiff or soft a spring is. A higher spring constant means that the spring will exert a stronger force on the object attached to it, resulting in a faster angular velocity and greater amplitude of motion.

4. Can the angular velocity of an object attached to a spring change?

Yes, the angular velocity of an object attached to a spring can change depending on the spring constant and the amplitude of the motion. As the amplitude increases, the angular velocity also increases, and vice versa.

5. How can the relationship between angular velocity and the spring constant be applied in real-world situations?

The relationship between angular velocity and spring constant can be seen in many real-world examples, such as the motion of a pendulum, the vibration of a guitar string, or the rotation of a fan. By understanding this relationship, engineers can design and optimize various systems and machines that involve rotational motion and springs.

Similar threads

  • Introductory Physics Homework Help
Replies
29
Views
925
  • Introductory Physics Homework Help
Replies
8
Views
337
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
789
  • Introductory Physics Homework Help
Replies
32
Views
1K
  • Introductory Physics Homework Help
Replies
31
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
909
  • Introductory Physics Homework Help
Replies
9
Views
837
  • Introductory Physics Homework Help
Replies
2
Views
472
  • Introductory Physics Homework Help
Replies
8
Views
1K
Back
Top