Hooke's law and circular motion.

[negation]

Homework Statement

In a zero gravity experiment a spring has a rest length of 1.0 metre. It is attached at
each end to 1.0 kg masses. The combination is then set rotating about its centre of
mass at 1.0 revolution per second. Each mass undergoes uniform circular motion with
a radius of 70 cm. Neglecting the mass of the spring and assuming Hooke’s Law, the
spring constant must be approximately:

The Attempt at a Solution

F_c = F_H

(1kg)(2pi(f))/(1m) = k(1m)
I tried equating them and solving for k but it's wrong.

 

[Doc Al]

Not sure what formulas you are using.

How much is the spring stretched? What force does it exert? What's the centripetal acceleration?

 

[negation]

Not sure what formulas you are using.

How much is the spring stretched? What force does it exert? What's the centripetal acceleration?

I was equating the centripetal force to hook's law.

I do no know how much is the spring being stretched but I know that there is a centripetal force pulling the spring to the center.
The centripetal acceleration is v²/r.

 

[Doc Al]

I was equating the centripetal force to hook's law.

That's what you should have been doing, but your posted equation does not seem to match that.

I do no know how much is the spring being stretched

You are told the radius of each mass's revolution. (Draw a picture for yourself.) By comparing that to the unstretched length, you should be able to figure out how much the spring is stretched.

but I know that there is a centripetal force pulling the spring to the center.

The spring exerts a force on the masses. That force is the centripetal force.

The centripetal acceleration is v²/r.

That works. But you might find the alternate expression, in terms of ω, more suited for this problem.

 

[negation]

That's what you should have been doing, but your posted equation does not seem to match that.


You are told the radius of each mass's revolution. (Draw a picture for yourself.) By comparing that to the unstretched length, you should be able to figure out how much the spring is stretched.


The spring exerts a force on the masses. That force is the centripetal force.


That works. But you might find the alternate expression, in terms of ω, more suited for this problem.

That's the problem! I cannot even begin to figure out the physical interpretation.
Well, ω = 2πf

 

[Doc Al]

That's the problem! I cannot even begin to figure out the physical interpretation.

Imagine the two masses swinging around in a circle, the center of which is at the middle of the spring.

Well, ω = 2πf

OK, you'll need that.

 

[dauto]

You can just plug 1m for things you don't know yet. You have to find theirs values and then plug it in.

 

[HallsofIvy]

You are told that the spring has a rest length of 1 m. You are told that the two masses are rotating in a circle with a radius of 0.7 m. What is the distance between the masses? How much has the spring been stretched?

(I think dauto meant to say "you cannot just plug in 1m for thing you don't know yet.")

 

[negation]

You are told that the spring has a rest length of 1 m. You are told that the two masses are rotating in a circle with a radius of 0.7 m. What is the distance between the masses? How much has the spring been stretched?

(I think dauto meant to say "you cannot just plug in 1m for thing you don't know yet.")

If the spring has a rest length of 1m and during the rotation the radius is 0.7m then could it be that there is a change in length of 0.3m?
By this, F_H = k(r-x) = k(r-0.3m) where r is the original radius (rest length of the spring and x is the change in radius)

 

[Doc Al]

If the spring has a rest length of 1m and during the rotation the radius is 0.7m then could it be that there is a change in length of 0.3m?

If the radius is 0.7 m, what is the full length of the stretched spring?

 

[dauto]

If the spring has a rest length of 1m and during the rotation the radius is 0.7m then could it be that there is a change in length of 0.3m?
By this, F_H = k(r-x) = k(r-0.3m) where r is the original radius (rest length of the spring and x is the change in radius)

Not quite. Draw a picture following the description in the problem.

 

[negation]

If the radius is 0.7 m, what is the full length of the stretched spring?

I really have no idea. The next best is to guess but I never like blundering my way through.

 

[CAF123]

Consider a spring of relaxed length 1m with two masses at either end. The CoM is in the middle of the spring, a distance 0.5m from each mass. We are told that each mass undergoes uniform circular motion with a radius of 0.7m (i.e each mass is now at a distance of 0.7m from the middle of the spring). How much did the spring stretch?

I think this is the only possible interpretation, since in your analysis the force from the spring on the masses would not be pointing radially inward, and hence there would be no possibility for circular motion.

 

[Doc Al]

I really have no idea. The next best is to guess but I never like blundering my way through.

Did you draw yourself a picture?

 

[negation]

Consider a spring of relaxed length 1m with two masses at either end. The CoM is in the middle of the spring, a distance 0.5m from each mass. We are told that each mass undergoes uniform circular motion with a radius of 0.7m (i.e each mass is now at a distance of 0.7m from the middle of the spring). How much did the spring stretch?

I think this is the only possible interpretation, since in your analysis the force from the spring on the masses would not be pointing radially inward, and hence there would be no possibility for circular motion.

I'm starting to develop a physical intepretation. Tje spring stretch 0.2m.

 

[negation]

Did you draw yourself a picture?

I did but it doesn't capture fully the demand of the question. I'm refining the physical intepretation as I go.

 

[Doc Al]

I'm starting to develop a physical intepretation. Tje spring stretch 0.2m.

How did you determine this?

 

[Doc Al]

I did but it doesn't capture fully the demand of the question. I'm refining the physical intepretation as I go.

What does your diagram show?

 

[negation]

What does your diagram show?

Initially it the mass is positioned at 0.5m from the origin. The hint was the COM. During the spin, it is displaced to a position 0.7m from the origin which is also the COM.
Tell me if I am wrong

 

[Doc Al]

Initially it the mass is positioned at 0.5m from the origin. The hint was the COM. During the spin, it is displaced to a position 0.7m from the origin which is also the COM.

Excellent. So what's the total length of the stretched spring? Then, how much has it stretched from its original length?

 

[ehild]

It is difficult to discuss about non-existing pictures... I attached one. Both masses orbit around the CM along the red circle, the stretched spring connecting them.

ehild

 

[negation]

Excellent. So what's the total length of the stretched spring? Then, how much has it stretched from its original length?

It has stretched 0.7m from the origin. But by symmetry, the total length of the spring, in the strictest sense, is 1.4m during the circular rotation.

Edit: I'll try to go further.
Taking into account only the case of one mass, the spring has stretched a distance of 0.2m and so Δr = 0.2
By Hooke's law, F_H = -kΔr = (-0.2m)k.
This stretch in distance is due to the centripetal force
F_c = ma_c = m(2πf)^2/r = m[(ω)^2]/2

 

[negation]

It is difficult to discuss about non-existing pictures... I attached one. Both masses orbit around the CM along the red circle, the stretched spring connecting them.

ehild

I appreciate the effort:thumbs:

 

[Doc Al]

It has stretched 0.7m from the origin. But by symmetry, the total length of the spring, in the strictest sense, is 1.4m during the circular rotation.

Good! Keep going: How much has it stretched from its original length?

 

[negation]

Good! Keep going: How much has it stretched from its original length?

I'll try to go further.
Taking into account only the case of one mass, the spring has stretched a distance of 0.2m and so Δr = 0.2
By Hooke's law, F_H = -kΔr = (-0.2m)k.
This stretch in distance is due to the centripetal force
F_C = mac = m(2πf)^2/r = m[(ω)^2]/2

 

[Doc Al]

I'll try to go further.
Taking into account only the case of one mass, the spring has stretched a distance of 0.2m and so Δr = 0.2

There is only one spring. The amount of force it exerts depends on the total stretch of the spring. (Don't just consider half the spring.)

 

[negation]

There is only one spring. The amount of force it exerts depends on the total stretch of the spring. (Don't just consider half the spring.)

Alright. That means during the circular motion, the spring stretched a distance of 0.4m in totality. 0.2m on each side.
F_H = -k(0.4m) = (-0.4m)k
F_H = F_C

 

[Doc Al]

Alright. That means during the circular motion, the spring stretched a distance of 0.4m in totality. 0.2m on each side.
F_H = -k(0.4m) = (-0.4m)k
F_H = F_C

Now you're cooking.

 

[negation]

Now you're cooking.

I wish physical illustration were provided. It's such a huge handicap for my mental stipulation.

F_H =(-0.4m)k
F_C = M[(2πf)^(2)/r] = (2kg)[(39.4784176 rads^-1)/0.7m] = 17.95195802 N
Edited some confusing notations.

 

[Doc Al]

I wish physical illustration were provided. It's such a huge handicap for my mental stipulation.

Part of the challenge is to come up with your own illustration.

F_H =(-0.4m)k
F_C = M[(2πf)^(2)/r] = (2kg)[(39.4784176 rads^-1)/0.7m] = 17.95195802 N
Edited some confusing notations.

You are mixing up two versions of the formula for centripetal acceleration:
a_c = v²/r = ω²r