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

 

[negation]

Part of the challenge is to come up with your own illustration.You are mixing up two versions of the formula for centripetal acceleration:
a_c = v²/r = ω²r

That was costly. Can't trust memory.
S = Θ/r = (dΘ.r - Θdr)/r² = vr/r² = v/r
ω=v/r
∴v = ω²r²
a_c = v₂/r = ω²r²/r = ω²r
F_C = m[ω²r] = 2kg[(2πf)²r] = 2kg(39.4784176 rad s^-1)(0.7m)
Good to go?

Edit: I tried working backwards from the answers I was provided.

In my final equation:
F_C = m[ω²r] = 2kg[(2πf)²r] = 2kg(39.4784176 rad s^-1)(0.7m), this gives
55.26978465 N
55.26978465 N = |-k|(0.4m) = 138.1744616

But should I divide 55.26978465 N by 2, and equating this to |-k|(0.4m) I get ~69 which is correct.
I can only rationalize that dividing the value 55.26978465 N by 2 is such that only the force acting on one mass can be determined.

 

[Doc Al]

That was costly. Can't trust memory.
S = Θ/r = (dΘ.r - Θdr)/r² = vr/r² = v/r
ω=v/r
∴v = ω²r²
a_c = v₂/r = ω²r²/r = ω²r
F_C = m[ω²r] = 2kg[(2πf)²r] = 2kg(39.4784176 rad s^-1)(0.7m)
Good to go?

Except that you added the masses together for some reason. Think in terms of Newton's 2nd law, which you are applying to each mass separately.

 

[ehild]

Why do you use 2kg for the mass? Consider one of the masses. It moves along a circular orbit, with given radius and time period. The spring exerts force on the mass and that force provides the centripetal force.
The same force acts on the other end of the spring on the other mass, and it moves along the same orbit.

ehild

 

[negation]

Except that you added the masses together for some reason. Think in terms of Newton's 2nd law, which you are applying to each mass separately.

I did an edit in the previous post

 

[negation]

Why do you use 2kg for the mass? Consider one of the masses. It moves along a circular orbit, with given radius and time period. The spring exerts force on the mass and that force provides the centripetal force.
The same force acts on the other end of the spring on the other mass, and it moves along the same orbit.

ehild

It hit me that since we considered the total change in length of the spring due to the 2 masses, we give consideration to the total mass.

 

[ehild]

The force acts in a point. If you speak about an extended body, with forces acting at different points, you can sum up the individual equations m_ia_i=F_i, and arrive at the equation for the acceleration of the centre of mass: a_CM∑m_i=∑Fi . If there are only forces of interaction (no external force) the sum of forces is zero. The system, which has 2 kg mass altogether, has zero acceleration. But the individual masses feel the force of the spring and have centripetal acceleration.

ehild

 

[ehild]

It hit me that since we considered the total change in length of the spring due to the 2 masses, we give consideration to the total mass.

A stretched spring has tension. To keep it stretched there must be forces equal to that tension acting at both ends. You can not exert force on a spring if the other end is free. Hook's law refers to the tension which is equal to kΔL, and it is equal to the force acting on one end of the spring (but force of the same magnitude and opposite direction has to act on the other end).
Again, the net force on the stretched spring is zero, as the forces acting at the ends are opposite and of equal magnitude.

ehild