The initial trajectory of a ball released from uniform circular motion

This short paper presents a simple analytical model for the abrupt termination of circular motion, as discussed in the ‘The Most Mind-Blowing Aspect of Circular Motion’. The model confirms that when a string is released, a ball at the far end of the string continues to move in a near-circular motion for a short time.


Introduction
In the web channel program The Most Mind-Blowing Aspect of Circular Motion on All Things Physics, the presenter examines the circular motion of a ball attached to a string.The central question posed at the beginning of the program is: 'What path does the ball take immediately after releasing the string?' Three possible answers are offered: (a) the ball continues in circular motion, (b) it moves tangentially to the circle, and (c) it moves normally to the circle.Through various experiments, the presenter demonstrates that the correct answer is (a)-the ball continues to move in a circular motion.The program was inspired by experimental and numerical investigations of a slinky's motion after release [1].Additional information about the slinky problem can be found in [2,3], and the related citations.
As some commentators of the program have pointed out, it is worth noting that after release, we observe the motion of both the ball and the string.If the ball instead detached from the string's end, it would move tangent to the orbit upon release.However, since release occurs from a point at the rotation axis, the motion involves both the ball and the string.
To explore this problem analytically, we have developed a simple analytical model of the ball-spring system.This model aims to demonstrate that answer (a) can be derived not only by experimentation and numerical simulations but also by analytical methods.Therefore, we do not use more sophisticated spring models, such as the one in [4], as they are limited to numerical simulations.Instead, we describe the spring using a lumped model in which half of the mass is attached each end, treating the sphere as a point mass.Our focus is on understanding the initial motion of the sphere after the spring is released, and to achieve this we use a series solution method as described in [5].A more detailed treatment of the solutions to the model equations is given in the appendix.

The model
We assume the ball is connected to a Kelvin-Voight spring with stiffness k, damping coefficient c, and relaxed length ℓ .
0 The ball's mass is m , b while the mass of the spring is m .s The motion occurs in the xy plane, with one end of the spring initially at the coordinate origin (figure 1).The ball rotates at a constant angular velocity w .
0 The plane on which the ball and spring rotate is frictionless.In addition, we neglect air resistance in our analysis.
After release, the coordinates of the ball are ( ) x y , , 2 2 and those of the other end of the spring are ( ) x y , .The force F acting between the endpoints is: Geometry and kinematics of the problem. where represents the stretched length of the spring and represents the spring's stretching rate.Using Newton's second law of motion, the equations of motion are: Define the y-axis such that the spring becomes detached from the centre at an instant in time when it is located along the y-axis.Therefore, the initial conditions are: Here, r 0 represents both the initial length of the spring and the initial radius of the circular orbit of the ball.This length can be obtained from the dynamic equilibrium equation which gives the value Stationary solution is only possible when w < k m .0 2 However, even if this condition is met, damping of the spring is necessary to achieve circular motion-unless, the circular motion happened to be in a stationary orbit.Without damping, the ball would oscillate in a stationary orbit.

Solution
In general, the equation (4) have no analytical solution and must be, therefore, integrated numerically.However, given our focus solely on the initial motion, we derive the solution in the form of a power series.The coordinates of interest are those of the ball and are is the reduced mass of the system, w = k m n is the natural frequency of the ball-spring system, and z w = c m 2 n is the damping ratio.By eliminating time from the last two equations, we arrive at Using formulae for curvature, we determine that immediately after the release, the curvature is, using (9) So, the curvature of the ball's trajectory does not change immediately after release.
Let us now address whether the curve given by (10) coincides with the initial circle.The equation for circle is: It agrees with equation (10) up to second order which confirms that the ball will, for a short time, continue to move on the initial circle's path, as claimed in the program.This agreement happens because the first two coefficients in (9) and (10) do not depend on model parameters, ensuring that the initial motion is circular.However, that is not the case for massless and rigid strings, as shown in the appendix.

Discussion
From the equation (10), it agreement happens that the initial motion of the ball is not precisely circular but only approximately so.Achieving perfect circular motion is not feasible within this model due to its governing equations' absence of time-delay terms.Consequently, as the centre of mass immediately begins to move uniformly in a straight line, the ball promptly responds to the spring disconnection, causing a disturbance in its initial circular motion.To estimate the time at which the ball's trajectory noticeably deviates from its initial circular motion, we can introduce a tolerance level e and monitor when the trajectory lies outside this threshold.Alternatively, we can use the polar radius of the ball's trajectory to assess when it significantly deviates from its initial values.Define polar radius r as, using (9): To a first approximation, as we do not know the convergence radius of the solution, we can consider the second term in this series to be below the tolerance value.This leads to: The above estimate is mathematical, as it depends on tolerance e and is thus not related to any system-specific factors, such as the spring length.Therefore, the conclusion, similar to [1] that the ball continues in circular motion until the string attains its unstretched length is not achievable within the scope of this mathematical analysis.However, as can be seen from the graphs in figure 2, in some cases, the deviation of the radius from its initial value when the spring retains its undeformed length can nevertheless be small.

Conclusions
A model has been that accurately predicts the start of circular motion immediately after the string is detached.Prediction relies on the spring possessing mass and finite stiffness.If the string were instead massless, the ball's motion would start tangential to the circle.In the case of an infinitely stiff (i.e.rigid) string, upon release, the string would not follow its original path.Instead, the trajectory's curvature instantly increase, resulting in a sudden in radial acceleration of the ball.
After the initial circular motion, it is important to note that the ball does not just follow the tangent of the circle.Instead, it performs oscillations transverse to a line in to the direction in which the centre of mass is moving.This can be demonstrated by numerically solving the appropriate initial value problem.n 16 min . 1 The solution for ( ) = ℓ ℓ t described as 'approx..' is obtained using (25) whereas the polar radius ( ) = r r t described as 'series' is obtained using (13) with two terms of series expansion.

Appendix
In this appendix, we will examine a solution to the system of equation (4) , which describe the motion of the two-mass system under central force (1).When dealing with central force problem [6] it is convenient to express the unknown coordinates as Here, q is the angle between the y axis and the spring axis (see figure 1), and ( ) x y , C C are the coordinates of the centre of mass of the ball-spring system.Substituting (15) into (4), we obtain the equations of motion The initial conditions are derived from (5) to (6) and (15): Now, integration of the first two equations gives, The system's centre of mass executes uniform motion along the straight line = y y C parallel to the x-axis with a velocity of  w = x y .Which is a inhomogeneous, nonlinear differential equation without analytical solution.
However, it has a stationary solution ¥ ℓ , which is found by solving: To obtain an approximate solution of (21), we set:  The solution of these equations, subject to initial condition ( ) q = 0 0 and ( )  q w = 0 , 0 are: ( ) q w = t 29 Unlike an extensible spring, where the trajectory's radius of curvature remains constant, an inextensible string undergoes an instantaneous change in curvature at the moment of detachment, resulting in a sudden increase in radial acceleration.

C C 0
Integration of the third equation (16) subject to initial condition ( )

2 2, 0 0
The center of mass of the ball oscillates around its linear path of motion = y y C axis with amplitude If the spring is massless, the ball undergoes uniform linear motion tangential to the circle after the string is released.If = ¥ k and = c 0, that is, the ball is attached to the rotating, inextensible rope and initially performs rotational motion with the radius of = r ℓ the equation (4) remains valid.However, force F becomes an unknown reaction force, and the equation (2) with = ℓ ℓ 0 serves as a constraint equation.The form of the solution (15) remains valid.According to (19), which also holds in this case, the centre of mass moves uniformly in a straight line.The remaining equations are ), and (29) into (15) the equations of motion for the ball become to the period of the initial rotation.Using equation (31), the initial radius of curvature of the ball's trajectory is From this analysis, one obtains an approximate time for the spring to reach its unstretched length.For example, if a and z are small then this time is