Why does a ball tied to a spring move outwards when rotated (ground frame)?

Question

Suppose a bob or a ball is tied to a spring which in turn is pivoted to a certain point on a table as shown, if the ball is given velocity $v_o$ perpendicular to spring it moves in circular motion and also the spring elongates(even if the spring is in natural length initially), what force causes it to move out? I believe spring force and centripetal force acts inward and not outward. I am looking for the explanation in ground frame (or any other frame but not the rotating frame, because there you can explain it using centrifugal force ig). Assume no friction. Sorry for the messy diagram.

Answer 1

Your ground frame is an inertial frame. In an inertial frame, all objects move in a straight line unless affected by a force. In your diagram, the object is moving with a velocity of $v_0$, in the +y direction. Your question asks about moving "outward." That means the radius is trying to increase. So long as your object is not in the -y half of the coordinate plane (which, as drawn, it is not), a positive y velocity has an "outward" component to it. This is easily seen by converting to polar coordinates ($r=\sqrt{x^2+y^2}$ and $\theta=\tan^{-1} \frac y x$). Note this is not thinking in a rotating frame, but merely changing how our vectors are being measured -- the coordinates have changed. It's pretty clear to see that if $y\ge0$, $\sqrt{x^2+y^2}$ must increase as $y$ gets larger.

In your example image, this motion will always be "outward." However, it is easy to construct a related situation where that motion is inward. Allow me to replace the spring with a string, in favor of clarity. And instead of starting at the point where the ball is crossing the x axis ($y=0$), I'd like to start the thought experiment 45 degrees earlier ($y=-x$). We're spinning the ball on the string, just like normal. However, at this 45 degree point, we give the string a sharp pull (practically: we bend our wrist in, pulling the ball slightly closer, then as we approach that 45 degree point, we quickly let our wrist out and then "snap" the string applying much more force than is usually being applied in the balanced steady-state configuration). I'm not going to worry about the physics of the sharp pull itself (which could be complicated), but rather what happens after that. If you think about such a situation, you could apply the correct impulse to cancel out all of the positive +x velocity, leaving the ball on a +y path. Intuitively, I snapped the string to cause the ball to go straight up.

This situation is similar to your situation, except now $y<0$. When $y<0$, increasing $y$ decreases r. The result is the same in cartesian coordinates: the ball moves directly upwards, but in polar coordinates we see that the radius is decreasing first, until it reaches a minimum at $y=0$, and then starts increasing.

Intuitively you've probably felt this. If you give such a snapping motion, the string goes slack for a short while, as the ball takes a straight path through your circle rather than a curved path along it. Once it reaches the circle on the other side, you get an impulse as you "catch" it and return it to circular motion.

Answer 2

you have 2 generalized coordinates $~ r(t)~,\varphi(t)$

the EOM's obtained in inertial system are:

$$ m\,\ddot r=m\,r\,\dot{\varphi}^2-k\,(r-l_0)+m\dot{\varphi}^2\,v_0(t)$$

and

$$m\, r^2\ddot\varphi=-m\,(2\,r\,\dot r\dot\varphi+v_0(t)\,\dot r+\dot{v}_0\,r)$$

where $~l_0~$ is the spring initial length

obviously is that the velocity $~v_0(t)~$ caused force $~F_r=m\dot{\varphi}^2\,v_0(t)~$ towards the radial direction and $~F_\varphi=m\,(v_0(t)\,\dot r+\dot{v}_0\,r)~$ towards $~\varphi~$ direction

the path of the mass is then

Answer 3

"I believe spring force and centripetal force act[s] inward and not outward." You believe correctly... At first the mass moves in a straight line at right angles to the initial position of the spring, as the (almost) unstretched spring exerts (almost) no force. The straight line takes it further from the point of attachment of the spring to the table. Therefore the spring becomes stretched, and so exerts an inward pull on the mass. This pull is the required centripetal force for the mass to move in a curved path.

Working in polar co-ordinates, the radial equation of motion is $$k(r-a)=-m(\ddot r-r \dot \theta^2)$$ in which $a$ is the initial value of $r$, which is also the unstretched length of the spring. The angular equation integrates to give us conservation of angular momentum, so $$r^2\dot\theta=av_0$$ in which $v_0$ is the magnitude of the (tangential) launch velocity.

Eliminating $\dot\theta$ from the radial equation and integrating it: $$\dot r^2=v_0^2(1-a^2/r^2)-\omega_1^2(r-a)^2$$ Here we have written $\omega_1^2$ for $k/m$ and have evaluated the constant of integration by imposing the initial condition that $\dot r=0$ when $r=a$.

I thought it might be interesting to investigate what the equation yields when $\dot r=0$. Re-assuringly we find that one root is $(r-a)=0$. If $r\neq a$ we may divide through by $(r-a)$, giving $$0=v_0^2\ \frac {r+a}{r^2}-\omega_1^2(r-a)$$ I haven't yet tried to solve this equation for $r$, but one significant feature is easy to show... None of its roots is the 'steady state' value of $r$ that we'd get if the mass eventually settled into a circular orbit maintaining its angular momentum at launch. For such an orbit, $$k(r-a)=-mr\dot\theta^2\ \ \ \ \ \text{so}\ \ \ \ \omega_1^2(r-a)=\frac{a^2v_0^2}{r^3}$$ What happens, I believe, is that the mass, when launched as described, overshoots the steady state path, so $\dot r\neq 0$ when it crosses the steady state circular orbit. Its radial distance then oscillates (perhaps in the manner of Eli's diagram?). An oscillatory motion is indeed to be expected, since no damping force has been included in my treatment.