How can a spring attached to a mass cause it to do circular motion in real life? I have the following question :

I am having trouble imagining this that how this configuration will be achieved?
How can a spring with a mass on a plane can do circular motion?
Like if it is elongated, won't the spring just pull it towards center and then only do motion along that line?
 A: Physically, and theoretically, this is no different from a satellite in orbit where gravity constantly pulls inwards. No motor would be needed, ideally, just an initial sideways speed. Practically, there might be a lot more factors to take into account with a spring, not least friction forces.
When an object from rest is pulled inwards then it will fall inwards. But if that object is moving sideways initially, then the inwards pull adds an inwards velocity component so the velocity as a whole is angled and not directly inwards. This will trace out an ellipsis.
With higher sideways speeds, it is angled even more and the ellipsis widens. The spring will then not be contracting as much. At some limiting sideways speed, the direction is angled just perfectly so that the ellipsis become a circle and the spring ideally does not contract at all.
This would be comparable to when a satellite is in circular orbit - it is constantly falling towards earth, but it constantly "misses" due to its sideways speed.
A: Particle will want to "escape" the spring by "wanting to go" in a tangential direction, but as far as spring is attached to some pivot point- it will not let particle to do that, and thus will prolong it's length, with this action altogether forcing a particle to rotate about spring pivot point.
A spring elongation can be found from the fact that in this scenario centripetal force will be Hooke's force, so :
$$ k\Delta l = \frac {mv^{2}}{l_0 + \Delta l} \tag 1$$
Now solve eq. (1) for elongation $\Delta l$, and you'll get quadratic equation, like in the form :
$$ \Delta l = a~ \sqrt{b~ + l_0^{~2}} - c~l_0 \tag 2$$
What are an exact constants $a,b,c$,- I'll leave you to figuring out yourself.
A: Seismographs: an earthquake recording instrument.
Mass is lifted by spring, pen is attached to the mass.
When earth shakes side to side, mass resists motion with the aid of spring making the pen draws earthquake wave graph.
In case of plane, the pen or any prepheral device can be attached by the side of the mass.
A: You just need:

*

*a spring providing the centripetal force


*an initial condition with tangential velocity $\mathbf{v} =v_{\theta}\boldsymbol{\hat{\theta}}$ and initial position $\mathbf{x}$ s.t. the spring elongation provides the centripetal force with the intensity required for the uniform circular motion with constant tangential velocity $v_{\theta}$,
$m \dfrac{v_{\theta}^2}{r} = F_c = F_{spring}$,
being $r = |\mathbf{x}|$, the distance of the mass from the origin at the center of the circle.
If the constitutive law of the spring reads $F_{spring} = k (r-r_0)$, the (initial) distance of the mass from the center of the circular trajectory and the tangential velocity are related by
$m \dfrac{v_{\theta}^2}{r} = k ( r - r_0 )$.
For linear springs with $r_0$, the relation becomes $v_{\theta} = r \sqrt{\dfrac{k}{m}}$.
Once you match this relation with the initial conditions, if friction is negligible, you get a uniform circular motion.
A: *

*Embed the spring in a tube with rods sticking out on both ends.

*Place the tube on a low friction table (material, lubricants, etc)

*Attach a motor below the table, drill a hole, run the axis of the motor through the table and attach one end of the tube/rod.

*Attach a mass to the other end of the tube/rod. Minimize friction between the mass, spring and everything else

*Turn on motor

A: A particle of mass it attached to one end of a spring and moving with constant speed $v$ in circular path, while other end of spring is attached to a fixed end consider as a centre of circular path, letting spring to freely rotate about it.
Now the force exerted by a particle on spring is centrifugal force that causes elongation $l$ in a spring. Where radius of circular path is natural length of spring, as elongation is small so we can neglect increase in radius by elongation. Thus,
$\dfrac{mv^2}{l_0}=kl\Rightarrow l=\dfrac{mv^2}{kl_0}\tag*{}$
Now question of how a particle attached with spring moves in circular path. Attach spring's one end with vertical shaft connected to motor as in a cieling fan. Then motion of a particle is in horizontal plane.
