So... What is the complication with the force in this setup?

  • Thread starter FrostScYthe
  • Start date
  • Tags
    Force Law
In summary, the spring in a cage has two ends attached to it. The spring is stretched when it is rotated around the center of the cage. The centripetal force is exerted towards the center of the circle when the cage and the spring are rotating.
  • #1
FrostScYthe
80
0
A cage holds a 150g mass, which is attached to one end of a spring the other end of the spring is attached to the opposite end of th ecage. The spring is streched 4.0 cm as the mass is whirled at 25.0 rev/s in a circle of radius 6.0 cm about the center of the cage on a horizontal frictionless surface as indicated in Figure P.6 (Hopefully you can understand it without the image ;\). What is the spring constant?

anyway, I'm trying to use the formula

Fspring = -Kx k is the constant.

I think xf - xi = .04m - 0m = .04 = x

I don't know how to get the force though... I split it as F=ma
then I know the mass.. .150kg, but I don't know how I can get the acceleration from the 25.0 rev/s

I'm confused [b(] Can someone help me!?? =)
 
Physics news on Phys.org
  • #2
You should have learned a formula that says if something moves in a circle, with radius R cm, at a constant ω revolutions per second then then the acceleration is toward the center at 4π2ω2R cm/s2

(ω is "omega", π is "pi". The "4π2" is because you have to convert from revolutions per second to radians per second.)
 
  • #3
The solution

Hi,

This is the soln from what I understood of the problem...
Just equate kx = mrw2
w = 25 rev/s
m = .15 kg
r = 0.06m
x = 0.04m

U will get the value of k now...


How this helps u...

Sridhar
 
  • #4
a correction...

in my previous solution, just convert w from rev/s to rad/s and proceed.

Sridhar
 
  • #5
Aha! Thanx guys =) I think I got it right now that is...

.150kg(4(pi)^2(25rev/s)^2(.06m)) = K(.04m)

K = 5.55 x 10^3 N/m , That sounds right ;\, all the problems seem to be in the thousands or 10 thousands..

Anyway, THANNKKK YOUUUUUUUUUUU, I hope I just don't get it wrong when I get to class :wink:
 
  • #6
There is one complication that nobody has mentioned here. Maybe you're supposed to ignore it for this problem -- it's hard to be sure without having the illustration and the exact text.

You said that the spring is attached to both ends of the cage. So if I'm understanding that correctly, you have triangular configuration where the cage forms the base of the triangle, and the middle of the spring is being held by someone or something, so that middle point is the axis of rotation and also the apex of the triangle. You didn't mention the length of the cage.

The problem is this:
You need a centripetal force directed along a line from the center of mass of the cage to the center of the circle. The magnitude of that force is what you computed above. But the way your setup is configured, half of your force is exerted radially inward from the front end of the cage, and half from the back end of the cage. Neither one is exerted at the center of mass.

Therefore, each end of the spring exerts a force that has one component that acts along the length of the cage, equal and opposite to the corresponding force exerted by the other end of the spring. And, each end of the spring has a component of force acting perpendicular to the length of the cage, directed into the circle, but not exactly toward the center. (How far off-center depends on the length of the cage.) It is probably true that the sum of those two "inward" components are equivalent to one force equal to their sum, acting at the center of mass, so it is the sumof those components, rather than the overall total spring force, that is providing the centripetal acceleration. And that would be equal to the total spring force multiplied by the cosine of an angle which is one-half of the angle at the apex of the triangle.

Now, if the length of the cage is negligible compared to the length of the spring, the difference between this result and your result would be negligible, so maybe you are supposed to ignore it in this problem. But if this cage is moving in a circle with a radius of only 6 cm., it is hard to imagine a cage so small that its length would be negligible compared to the length of the spring.

Just thought I'd mention it.
 

FAQ: So... What is the complication with the force in this setup?

What is Hooke's force law?

Hooke's force law, also known as Hooke's law, is a principle in physics that describes the relationship between the force applied to an elastic object and the resulting displacement or deformation of the object. It states that the force applied is directly proportional to the displacement, as long as the object remains within its elastic limit.

Who discovered Hooke's force law?

Hooke's force law was discovered by English scientist Robert Hooke in the 17th century. He first observed this relationship while studying the behavior of springs and elastic materials.

What is the significance of Hooke's force law?

Hooke's force law is significant because it helps us understand and predict the behavior of elastic materials, such as springs, rubber bands, and even human tissues. It is also an essential principle in engineering and is used in the design of many structures and machines.

What is the formula for Hooke's force law?

The formula for Hooke's force law is F = -kx, where F is the force applied, k is the spring constant, and x is the displacement or deformation of the object. This formula can be rearranged to find any of the three variables, as long as the other two are known.

What are some real-life applications of Hooke's force law?

Hooke's force law has many real-life applications, such as in the design of shock absorbers, car suspension systems, and bungee jumping equipment. It is also used in medical devices, such as prosthetics and braces. In addition, Hooke's law is used in materials testing to determine the strength and elasticity of different materials.

Similar threads

Replies
9
Views
2K
Replies
8
Views
3K
Replies
16
Views
1K
Replies
5
Views
2K
Replies
2
Views
2K
Replies
3
Views
867
Replies
3
Views
10K
Back
Top