• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Problem on Hooke's force law

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!!!!?? =)
 

HallsofIvy

Science Advisor
Homework Helper
41,709
876
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.)
 
85
0
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
 
85
0
a correction...

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

Sridhar
 
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:
 
1,037
1
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.
 

Related Threads for: Problem on Hooke's force law

  • Posted
Replies
4
Views
9K
  • Posted
Replies
4
Views
407
  • Posted
Replies
10
Views
2K
  • Posted
Replies
3
Views
4K
  • Posted
Replies
3
Views
1K
  • Posted
Replies
2
Views
2K
  • Posted
Replies
7
Views
5K
  • Posted
Replies
2
Views
22K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top