Circular Motion Analysis

Consider the situation where

$m = 1 kg$ $R = 1 m$ $g = 10 m/s^2$

Consider the system to be centered at the origin of the X-Y plane and the ball to be moving anti-clockwise, at a general instant of time at the first quadrant, making an angle $\theta$ with the horizontal. The equations of motion are:

$a_t=\ddot{\theta} = -10\cos{\theta}$ $a_r=\dot{\theta}^2 = T + 10\sin{\theta}$

Consider that the system is set to motion at $t=0$ and when the string is at rest in its stable equilibrium position. The initial conditions are chosen as:

$(\theta_i,\dot{\theta}_i) = (-\pi/2,10)$

Here, the initial angular speed is chosen so that:

$\dot{\theta}_i > \sqrt{5g}$

This condition ensures that the string does not slack. Proving this result can be treated as a trivial exercise. Now, with the equations of motion, one can see that the magnitude of the tangential acceleration is maximum when $\theta = 0$ . This gives:

$a_{t,max} = 10 m/s^2$

Now, the tension in the string becomes minimum when the ball reaches its highest point during motion and so does the speed of the ball. At the highest point, the radial acceleration can be computed as such:

$a_{r,min} = T_{min} + 10 \sin(\pi/2)= T_{min}+10 = \dot{\theta}_{min}^2$

Applying the conservation of energy principle by taking the origin as the zero PE level, one gets:

$-mgR + \frac{1}{2} mR^2\dot{\theta}_i^2 = mgR + \frac{1}{2} mR^2\dot{\theta}_{min}^2$

Solving for $\dot{\theta}_{min}^2$ gives:

$\dot{\theta}_{min}^2 = 60 m/s^2$

In other words, the minimum radial acceleration is much greater than the maximum tangential acceleration.

Conclusion: At no point in time are the radial and tangential components of acceleration equal.

This has been verified by numerical simulation: Results are as follows: