Torque in a parallel combination of springs

[Aurelius120]

Homework Statement

In a given parallel spring rod system find the value of $K_{net}$

Relevant Equations

$k_{net}=k_1+k_2$

In deriving the $k_{net}$ of the given system, it is taken that the extension produced by both springs is equal but their force is different. Therefore $(k_1+k_2)x=k_{net}x \implies k_1+k_2=k_{net}$.
In absence of pivot, an object rotates around an axis through COM and perpendicular to plane.

Heres the problem:
Force exerted by each spring is different and therefore the object will experience a net torque about its own axis.(torque of weight is zero).
Therefore the object should rotate to such a position that both rotational and translational equilibrium are achieved and then the premise of derivation is false as extension is unequal.
How then can we replace the set of springs by a single spring if its torque isn't zero?

In the second case b, the derivation holds as torque of all forces about its own axis are zero.

Is is taken for granted that the net torque is zero? If my derivation is incorrect, then what is the correct derivation??

 

[Aurelius120]

In absence of pivot, an object rotates around an axis through COM and perpendicular to plane.

While we are at it . If it is within the limits of the question, please try to include a brief explanation for this fact stated by my teacher with no proof

 

[Lnewqban]

Something (a sliding guide, for example) must be keeping the mass from rotating in order to satisfy the “it is taken that the extension produced by both springs is equal” initial condition.

If allowed to rotate, then both forces should be equal for a balanced position.

 

[Aurelius120]

Something (a sliding guide, for example) must be keeping the mass from rotating in order to satisfy the “it is taken that the extension produced by both springs is equal” initial condition.

If allowed to rotate, then both forces should be equal for a balanced position.

So an external force is necessary for the condition to hold.
Nevertheless,
What happens to the system in absence of an external force(e.g. sliding guide) in case the weight of the rod is larger than the force provided when both springs have equal force?

Does it oscillate between the position of rotational equilibrium $(\tau_{net}=0, F_{springs}<Mg)$ and position of translational equilibrium $(\tau_{net}\neq 0 , F_{net}=0)$ ?

In other words if the system is released from position $(k_1+k_2)x=Mg$, does it tilt up due to torque and then down due to $Mg-F_{net}$ and so on

 

[kuruman]

How then can we replace the set of springs by a single spring if its torque isn't zero?

Are you sure the net torque is not zero? If that's the case there would be angular acceleration of some sort, no? That is not the case. The rod will not be horizontal, but tilted at some angle and at equilibrium. You can find how much each spring extends by demanding that the sum of all the forces and all the torques be zero.

If you displace this tilted rod from the equilibrium and then let it go, it will oscillate with frequency $\omega$. The equivalent single spring will have constant $k_{eq}=m\omega^2$.

If you don't believe me, hang a meter stick from equal lengths of twine (stiff spring) and rubber band (loose spring). What do you see?

 

[Aurelius120]

@kuruman but a single spring system won't have angular SHM the way the two spring system had right. In other words it can't keep the rod in exactly same position and orientation as tilted parallel spring system, right?

 

[kuruman]

Right. However, the problem asks you to find $k_{net}.$ How do you think that is defined? If the definition is, as you seem to imply in "relevant equations", $k_{net}=k_1+k_2$, then add the two and be done with it. Assume that the two springs are close to each other so that the tilt is negligible.

I think you are overthinking this problem. It looks like the goal of this question is for you to compare cases (a) and (b) in which the mass just bobs up and down with some frequency $\omega$ and ask yourself the question: If I were to replace the two springs in each case with a single spring $k_{eq}$, what would this equivalent constant be so that the frequency doesn't change?

 

[haruspex]

How then can we replace the set of springs by a single spring if its torque isn't zero?

Even if the two springs had the same constant, replacing them by a single spring is not going to be equivalent with regard to all motions of the system. You need to start by defining the sense in which they are to be equivalent.
An obvious choice is to say that a given vertical force applied at the mass centre produces the same vertical displacement in both models. Even to make that work, it may be necessary to assume that in the two spring case they are attached asymmetrically in such a way that the block does not rotate.

 

[Aurelius120]

I think you are overthinking this problem. It looks like the goal of this question is for you to compare cases (a) and (b) in which the mass just bobs up and down with

Yep. In this problem we have to find the time period.
It's just that they teach parallel combination of springs as if $x$ is always constant in parallel (like $V$ in parallel resistors ). So I thought that formula was ever true then this torque perplexed me.